8 concepts27 worked examples33 exercises5 exam-level8 figures
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07Gravitation and Newton’s Synthesis
An apple lets go of a branch and drops 4.9 m in its first second. Over that same second the Moon also falls toward the Earth, by about 1.4 mm, and that tiny fall is exactly why it never arrives. Two motions that look nothing alike, and one number, 3600, turns out to connect them.
By the end of this section you can put a number on the pull between any two masses, work out the strength of gravity at any height above a planet, find the speed and the period of a circular orbit, and weigh a planet using something that goes round it, checking every one of those answers a second independent way.
In 60 seconds
One force law, $F = Gm_1m_2/r^{2}$, fed into the second law you already have, accounts for a dropped stone, the value of $g$, every satellite and every planet, with no new machinery of any kind.
Universal gravitation
$$F = G\frac{m_1 m_2}{r^{2}}$$
any two masses at all; $r$ runs centre to centre and the pull is along that line
Strength of gravity at a distance r from a planet's centre
$$g = \frac{GM}{r^{2}}$$
you need the acceleration due to gravity somewhere other than the surface, or you want to weigh a planet
Circular orbit
$$v = \sqrt{\frac{GM}{r}}$$
a satellite, a moon or a planet on a circular path of radius $r$ around a mass $M$
Kepler's third law with its constant supplied
$$T^{2} = \frac{4\pi^{2}}{GM}\,r^{3}$$
a period is given and a radius is wanted, or two orbits round the same central body are compared
Three most common mistakes
Measuring $r$ from the surface instead of from the centre. A satellite 400 km up is 6780 km from the centre, not 400 km, and the two answers differ by a factor of 280.
Squaring the distance in the force law but forgetting to when the radius changes. Going twice as far out does not halve the pull, it quarters it, and going to three times the distance leaves a ninth.
Reading “weightless” as “no gravity there”. At the height of a low orbit gravity is still about 88 per cent as strong as it is at the ground; what has gone to zero is the contact force from the floor, not the pull of the planet.
The assessment for this course is two midterms at 20 per cent each, a 25 per cent final, quizzes totalling 10 per cent, homework 5 per cent and lab work 20 per cent. The syllabus does not say which topics land on which paper, so nothing here should be read as a prediction about where gravitation will be asked; treat it as examinable on any of them.
How much time do you have?
10 minutes
You leave with the one force law, the formula that turns it into a value of $g$ anywhere, and the single sentence that decides most quiz marks on this topic: $r$ is measured from the centre.
The 60 second card · Formula card · Newton's law of universal gravitation · Gravity near a planet: where g comes from · Mistake ledger
45 minutes
You add the parts that actually earn marks in a problem: how fast the pull falls off with distance, the recipe that turns any circular orbit into one equation, and the difference between having no weight and having no contact force.
The 60 second card · What you should already have · Newton's law of universal gravitation · How fast the pull falls off · Gravity near a planet: where g comes from · Satellites: an orbit is one equation · Weightlessness is about the floor, not the planet · Method boxes · Fading ladder · Practice B (computation) · Check yourself
full read
Everything above plus adding several pulls as vectors and the three laws of planetary motion, which is where the harder exam questions live, and the interleaved set that forces you to decide which tool a question wants before you reach for one.
The opening pages · What you should already have · Notation · All eight concept blocks · Method boxes · Contrast pairs · Fading ladder · Exam level example · Practice A to D · Mistake ledger · Formula card · Glossary · Check yourself
By the end of this section
State the law of universal gravitation, apply it to two masses at a given separation, and show that the same law accounts for a falling apple and for the Moon.
Predict how the gravitational force or the acceleration changes when a distance or a mass is scaled, without recomputing anything from scratch.
Compute the acceleration due to gravity at any distance from a planet's centre, and go backwards from a measured value of that acceleration to the mass of the planet.
Add the gravitational pulls of several bodies as vectors and report the resultant with a magnitude and a direction.
Solve for the speed, the period or the radius of a circular orbit by setting the gravitational force equal to the mass times the centripetal acceleration.
Explain why a body in orbit registers zero on a scale although the pull on it is nearly undiminished, and compute the reading of a scale in an accelerating lift or spacecraft.
Derive the relation between the period and the radius of a circular orbit, use it to compare two orbits round the same body, and use it to find the mass of that central body.
Syllabus coverage
Gravitation
The force law itself: how it depends on the two masses and on the distance, what it gives at the surface of a planet, how several pulls add, and what it does to a satellite
The week line names the topic and gives no chapter numbers, so no chapter number is quoted anywhere in this section. The scope is the standard content of that topic in the set textbook, split across seven of the eight concept blocks.
covered
Newton’s Synthesis
The claim that the sky and the ground obey one law: Kepler's three descriptive laws turned into consequences of universal gravitation plus the second law
The word synthesis is the whole point of the last block, and the apple-and-Moon comparison in the first block is the first instalment of it.
covered
The gravitational field
Reading $g$ as a property of the space around a mass rather than as a force on a particular body
Given two paragraphs because the symbol $g$ is used in this way throughout the section and the wording would be confusing otherwise. No question asks for it and nothing later in this section depends on it, so treat it as background rather than examinable.
off_syllabus
Tides, and why the Moon keeps one face toward us
What happens when the pull on the near side of a body differs from the pull on its far side
Named here and not developed. It is a genuine and popular application of the , but it needs the difference between two nearly equal forces to be handled carefully, and no question in this section asks for one.
off_syllabus
Energy in a gravitational field, and escape speed
How much energy it takes to move a body away from a planet, and the launch speed that never returns
Deferred to the later sections on work and energy. Everything in this section is done with forces, accelerations, speeds and periods in newtons, metres per second squared and seconds; no question here asks for a quantity in joules, and none needs one.
deferred
Non-circular orbits, worked out rather than described
Computing speeds and positions along an ellipse instead of taking the orbit to be a circle
Kepler's first and second laws are stated as he found them and used qualitatively, and the third is derived only for the circular case. Every orbit calculation in this section assumes a circle, which is stated on the page each time it is used.
deferred
Recall first
Newton's second law, written one axis at a time
$\sum F_x = ma_x$ and $\sum F_y = ma_y$, applied to one named body after every force acting on it has been drawn.
Nothing in this section replaces this law. Everything here is one new expression for a force, dropped into the same two equations, so if the bookkeeping worked last week it works now.
Newton's third law
If body A pulls on body B, then B pulls on A with a force of the same size in the opposite direction. The two forces act on different bodies and never cancel each other.
The gravitational pull is a third law pair. The Earth pulls a 60 kg student with about 590 N and the student pulls the Earth with exactly 590 N; the accelerations differ wildly because the masses do, not because the forces do.
Centripetal acceleration
A body on a circular path of radius $r$ at constant speed $v$ has an acceleration of size $a_R = v^{2}/r = 4\pi^{2}r/T^{2}$ pointing at the centre, where $T$ is the time for one full turn.
Every orbit in this section is a circle traced at constant speed, so this is the right-hand side of the equation each time. The left-hand side is the only thing that is new.
Weight, and the normal force that a scale reads
The weight of a body of mass $m$ where the local gravitational acceleration is $g$ is $W = mg$, pointing down. A scale does not read the weight; it reads the contact force $N$ that it pushes up with, and $N$ equals $mg$ only when the body is not accelerating vertically.
The whole of the block on weightlessness is this distinction applied to a body in free fall, and the numbers on a scale in a lift are computed exactly as they were in the section on the laws of motion.
Adding forces as vectors
Forces add by components: $\sum F_x$ from the $x$ parts, $\sum F_y$ from the $y$ parts, then the resultant has size $\sqrt{(\sum F_x)^{2} + (\sum F_y)^{2}}$ and a direction given by the arctangent of the ratio.
Gravitational pulls from several bodies are ordinary forces and add the ordinary way. The only thing worth extra care is that each pull points along its own line joining centres, so the directions are rarely along the axes.
Free fall means gravity is the only force acting
A body is in free fall when nothing but gravity acts on it. Near the surface that gives an acceleration of $9.80\ \mathrm{m/s^{2}}$ downward whatever the mass, and the body need not be moving downward to be in free fall.
A satellite is in free fall by this definition even though it never gets closer to the ground, and that sentence is the whole explanation of .
Try it yourself first (3 questions)
1§07.0 — what a constant speed on a circle implies●●○○○
Before any of the new material, one question about the tool you will be using on every orbit. A 2.0 kg ball is whirled on a string in a horizontal circle of radius 0.80 m, going round at a steady 4.0 m/s.
Given
$m = 2.0\ \mathrm{kg}$
$r = 0.80\ \mathrm{m}$
$v = 4.0\ \mathrm{m/s}$, unchanging
Find
(a) Which statement about the net force on the ball is correct?
Hint 1/4
The question is really asking whether a body on a circle is accelerating, and if so in which direction. Settle that before touching any numbers.
Hint 2/4
For circular motion at constant speed the acceleration is $a_R = v^{2}/r$ directed at the centre, so the net force is $ma_R$ in that same direction.
Hint 3/4
With the numbers as given, $m = 2.0$ kg, $v = 4.0$ m/s and $r = 0.80$ m, the acceleration is $(4.0)^{2}/0.80$.
Hint 4/4
The net force is 40 N and it points at the centre of the circle.
Show solutionDecide whether there is an acceleration at all
the second law does not change form for circular motion; the direction of the net force is the direction of the acceleration
Answer $$\boxed{\;F = 40\ \mathrm{N} \text{ toward the centre}\;}$$
Check
Independent route through the period. One turn covers $2\pi(0.80) = 5.03$ m at 4.0 m/s, so $T = 1.26$ s, and $a_R = 4\pi^{2}r/T^{2} = 4\pi^{2}(0.80)/(1.26)^{2} = 20\ \mathrm{m/s^{2}}$, reached without using the speed formula.
Everything in this section replaces the string with gravity and keeps this equation exactly as it stands.
2§07.0 — the third law applied to a pull you cannot see●●○○○
A student of mass 60 kg stands still on the ground. The Earth pulls the student down with a force of about 590 N. The student also pulls the Earth.
Given
$m_{\text{student}} = 60\ \mathrm{kg}$
The Earth pulls the student with about 590 N
The Earth's mass is $5.97\times10^{24}\ \mathrm{kg}$
Find
(a) How large is the pull the student exerts on the Earth?
Hint 1/4
This is not a calculation. Decide which law tells you about a pair of forces between two bodies, and then read what it actually asserts.
Hint 2/4
Newton's third law: the force A exerts on B and the force B exerts on A are equal in size and opposite in direction, and they act on different bodies.
Hint 3/4
Here A is the Earth and B is the student, and the force of A on B is given as about 590 N.
Hint 4/4
The student pulls the Earth with about 590 N, upward on the Earth.
Show solutionRead the third law off the page
$$F_{\text{student on Earth}} = F_{\text{Earth on student}} = 590\ \mathrm{N}$$
the law is about the pair, and it puts no condition on the sizes of the two bodies; a threshold mass appears nowhere in it
same force, mass larger by twenty three orders of magnitude, so the asymmetry is entirely in the denominator
Answer $$\boxed{\;F = 590\ \mathrm{N} \text{ on the Earth, toward the student}\;}$$
Check
Order of magnitude sanity check on the Earth's acceleration: at $10^{-22}\ \mathrm{m/s^{2}}$, and given that the age of the Earth is under $10^{18}$ s, even acting steadily for the whole of that time this acceleration would move the planet by well under a millimetre. That is what unmeasurable means, and it is not the same as zero.
Keep this in reserve for the block that states the force law: the force law you are about to meet is symmetric in the two masses, and this is the reason it has to be.
3§07.0 — what a scale reads when the floor accelerates●●○○○
A 60.0 kg person stands on a bathroom scale inside a lift. The lift accelerates downward at $2.00\ \mathrm{m/s^{2}}$. The scale reads the force with which it pushes up on the person's feet.
Given
$m = 60.0\ \mathrm{kg}$
$a = 2.00\ \mathrm{m/s^{2}}$ downward
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) What does the scale read, in newtons?
Hint 1/4
The scale reads a contact force, not a weight. So the target is the normal force, and the way to it is the vertical equation for the person.
Hint 2/4
Taking down as positive for a downward acceleration, $mg - N = ma$, so $N = m(g - a)$.
Hint 3/4
With $m = 60.0$ kg, $g = 9.80\ \mathrm{m/s^{2}}$ and $a = 2.00\ \mathrm{m/s^{2}}$ downward, the bracket is $9.80 - 2.00$.
Hint 4/4
The scale reads 468 N.
Show solutionWrite the vertical equation for the person alone
$$mg - N = ma$$
two forces act on the person, the Earth's pull down and the scale's push up, and down is taken positive because that is the direction of the acceleration, which keeps every term positive
solving for the quantity actually asked for rather than for the acceleration
Answer $$\boxed{\;N = 468\ \mathrm{N}\;}$$
Check
Two limiting cases. With $a = 0$ the formula gives $N = mg = 588$ N, the ordinary standing reading. With $a = g$, a lift in free fall, it gives $N = 0$, a scale reading nothing at all. Both extremes come out right, so the middle is likely to as well.
That second limiting case is the whole content of the block on weightlessness later in this section; the only change will be that the lift is replaced by a spacecraft and nothing else at all.
Notation
symbol
reads as
means
watch out
$G$
capital G, the
the universal constant $6.674\times10^{-11}\ \mathrm{N\cdot m^{2}/kg^{2}}$ in the force law
it is not $g$ and the two are never interchangeable: $G$ is the same on every planet in the universe, $g$ changes if you climb a ladder
$g$
little g
the acceleration a freely falling body has at the place in question, $9.80\ \mathrm{m/s^{2}}$ at the Earth's surface
it is a local value, not a constant of nature; when this section uses it away from the surface it says where
$M$
capital M
the mass of the central body, the planet or star being orbited
in the orbit formulas only $M$ appears and the orbiting mass has cancelled, which is why a bolt and a space station at the same radius orbit at the same speed
$r$
r
the distance between the two centres, in metres
not the altitude and not the surface separation; for a body at height $h$ above a planet of radius $R$ it is $R+h$
$R_E$
R sub E
the radius of the Earth, $6.38\times10^{6}\ \mathrm{m}$
used only as the distance from the centre to the surface; an orbit radius is always larger than this and any answer smaller than it describes a hole in the ground
$h$
h
the height of a body above the surface of a planet
it never appears in the force law by itself; it is only ever the second half of $r = R + h$
$T$
capital T
the period, the time for one complete orbit, in seconds
days and years have to be converted before use; $T^{2}$ has units of seconds squared and a stray day will show up as a factor of about $7.5\times10^{9}$
$a_R$
a sub R
the radial, or centripetal, component of the acceleration, pointing at the centre
it is the whole acceleration only when the speed is constant, which is the case for every circular orbit in this section
$F_G$
F sub G
the gravitational force between two bodies, in newtons
it is a single force with a partner acting on the other body; it is not a net force unless gravity is the only thing acting
Conventions used here
The two constants that appear in every gravitation formula here
$G = 6.674\times10^{-11}\ \mathrm{N\cdot m^{2}/kg^{2}}$ and, at the Earth's surface, $g = 9.80\ \mathrm{m/s^{2}}$. $G$ is the same everywhere in the universe and is never adjusted; $g$ is a local value that this section will change on purpose whenever the distance from the centre changes, and when it does it is written $g$ with a note saying where.
Which distance the symbol r stands for
Always the distance between the two centres, never the gap between the surfaces and never the height above the ground. When a problem gives an altitude $h$ above a planet of radius $R$, the first line of the solution is $r = R + h$, written out, before any substitution. This one line is where most of the marks are lost.
Which way is positive in a radial equation
For anything moving on a circle the useful axis is the radial one, and the positive radial direction points toward the centre, the same choice as in the section on circular motion. With that choice $\sum F_R = mv^{2}/r$ carries no minus signs, and any force pointing away from the centre enters with one.
The astronomical data used throughout this section
Earth: mass $5.97\times10^{24}\ \mathrm{kg}$, radius $6.38\times10^{6}\ \mathrm{m}$. Moon: mass $7.35\times10^{22}\ \mathrm{kg}$, orbit radius $3.84\times10^{8}\ \mathrm{m}$, period 27.3 days. Sun: mass $1.99\times10^{30}\ \mathrm{kg}$. Earth's orbit: radius $1.50\times10^{11}\ \mathrm{m}$, period $3.156\times10^{7}\ \mathrm{s}$. Any question that needs one of these says so; none of them is expected from memory.
What is idealised away in every orbit problem here
Orbits are taken to be circles, the central body is treated as a sphere whose mass is spread evenly, the orbiting body is treated as a point, and the pull of every third body is ignored. Air resistance is ignored above the atmosphere and no orbit in this section decays. Each of these is stated in the problem that uses it rather than assumed silently.
Digits kept in a gravitation answer
Three significant figures throughout, because $G$ and the astronomical data are quoted to three. Unrounded values are carried through the middle of a calculation and the rounding is done once, at the end; this matters more here than usual because distances get cubed and squared and a rounding error grows with them.
What the word weightless is allowed to mean
In this section weightless never means that gravity has stopped. It means that the contact force on the body, the thing a scale actually measures, is zero. The phrase used for that is apparent weightlessness, and where the two could be confused the sentence says which is meant.
7.1The Moon test: how the pull falls off with distance
The Moon's own orbit measures how fast gravity weakens with distance: it goes as one over the square.
In the last section every circular motion was held on its circle by something you could point at: a road, a string, a seat.
Solvable with what we have
Find a body's acceleration on a circular path from its radius and period, using $a_R = 4\pi^{2}r/T^{2}$.
Find the net force that must be acting on it, using $\sum F_R = ma_R$.
Compute the weight of anything near the ground as $mg$ with $g = 9.80\ \mathrm{m/s^{2}}$.
Name what supplies the force on a car going round a bend: friction, a rope, or a banked road.
Not solvable yet
Say what supplies the force on the Moon, since nothing is in contact with it.
Say how strong gravity is 400 km up, or out at the Moon.
Predict the speed a satellite must have to stay on a given orbit.
Explain why an astronaut floats although the Earth is plainly still there underneath.
Try the tool you already have. The Earth pulls things with $mg$, so let us assume it pulls the Moon with $mg$ too, giving the Moon an acceleration of $9.80\ \mathrm{m/s^{2}}$ toward us. The Moon starts $3.84\times10^{8}$ m away, so the time to arrive would be $t = \sqrt{2r/g} = \sqrt{2(3.84\times10^{8})/9.80} = 8.85\times10^{3}\ \mathrm{s}$, under two and a half hours.
Why it fails
The Moon has been up there rather longer than two and a half hours. The arithmetic is fine; what broke is the assumption underneath it, that gravity has the same strength wherever you go. The Moon's own orbit is enough to measure how much weaker it gets, and that measurement is the rest of this block.
The hook of this section, answered. Both bodies are pulled toward the same centre, and in one second the $\textcolor{#1f6feb}{\text{apple falls 4.9 m}}$ while the $\textcolor{#1f6feb}{\text{Moon falls 1.4 mm}}$; the only thing that differs between them is the $\textcolor{#d1690a}{\text{distance}}$ from that centre. Not to scale: drawn with the Earth at this size, the Moon would sit about four page widths to the right.
Looks like this, but is not
The Moon falls 1.4 mm every second, so it must be creeping toward us; a millimetre a second adds up.
It falls 1.4 mm away from the straight line it would have taken if nothing pulled on it, not 1.4 mm closer to us. In that same second its sideways motion carries it just far enough for the two to cancel, which is why the distance never changes; that cancellation is what an orbit is.
quantity
an apple at the surface
the Moon on its orbit
distance from the Earth's centre
6.38 × 10⁶ m
3.84 × 10⁸ m
that distance in Earth radii
1
60.2
measured acceleration toward the Earth
9.80 m/s²
0.00272 m/s²
how many times smaller than the apple's
1
3600
Read the last two rows together. The distance grew by a factor of 60 and the acceleration shrank by a factor of 3600, and 3600 is 60 squared to within the precision of the data. Nothing in this table was adjusted to make that come out; the two distances come from surveying and from the Moon's orbit, and the two accelerations from a stopwatch and from the Moon's period. The exponent in the force law is a measurement. Two rows on their own cannot show that the law is a power law at all; they can only report which exponent fits. What makes 2 convincing is that it landed on 60 squared to three figures here, and that the same exponent keeps working for every planet in the solar system, which is the last block of this section.
One law for the apple and for the Moon
The Moon goes round the Earth on a circle of radius $3.84\times10^{8}$ m, taking 27.3 days for one turn. Find its acceleration toward the Earth, and compare it with the acceleration of a falling apple at the Earth's surface, where the distance from the centre is $6.38\times10^{6}$ m.
Given
$r_{\text{Moon}} = 3.84\times10^{8}\ \mathrm{m}$
$T = 27.3\ \mathrm{days}$
$R_E = 6.38\times10^{6}\ \mathrm{m}$ and $g = 9.80\ \mathrm{m/s^{2}}$ at the surface
Find
the Moon's acceleration, and the ratio of the two accelerations compared with the ratio of the two distances
SolutionGet the period into seconds before anything else
the acceleration formula divides by $T^{2}$, so a period left in days would be wrong by a factor of about $7.5\times10^{9}$, which is not the sort of error that shows up as a slightly odd answer
Use the circular motion result from the previous section
$$a_{\text{Moon}} = \frac{4\pi^{2}r}{T^{2}}$$
this form is chosen over $v^{2}/r$ because a period is what the data gives; going through the speed first would add a step and a rounding
Independent route to the same acceleration, avoiding the period formula entirely. The Moon's speed is $v = 2\pi r/T = 2\pi(3.84\times10^{8})/(2.36\times10^{6}) = 1.02\times10^{3}\ \mathrm{m/s}$, and then $a = v^{2}/r = (1.02\times10^{3})^{2}/(3.84\times10^{8}) = 2.72\times10^{-3}\ \mathrm{m/s^{2}}$. The agreement is genuine because the second route squares a speed instead of squaring a period.
One unit conversion, one substitution and two divisions. The entire argument for an inverse square law rests on this much arithmetic.
This is the promise at the start of the section, cashed. The apple falls 4.9 m in the first second and the Moon falls $\tfrac{1}{2}(2.72\times10^{-3})(1)^{2} = 1.4$ mm in the same second, and both numbers come out of the same law once the distance is put in properly. The general lesson to carry forward: when a prediction fails by a clean factor, the factor is data. Here it was 3600, and 3600 asked to be read as 60 squared.
Checkpoint
§07.1 — predicting the Moon's fall from the apple's●○○○○
Thirty seconds and one division. The Moon's orbit puts it 60.2 Earth radii from the Earth's centre, and at the surface, one Earth radius out, a falling body accelerates at $9.80\ \mathrm{m/s^{2}}$.
Given
$r_{\text{Moon}} = 60.2\,R_E$
$g = 9.80\ \mathrm{m/s^{2}}$ at $r = R_E$
the acceleration the Earth produces falls off as $1/r^{2}$
Find
(a) Predict the Moon's acceleration toward the Earth, without using $G$ or the mass of the Earth.
Hint 1/4
Only the distance differs between the apple and the Moon, so this is a comparison rather than a fresh calculation, and no constant has to be substituted.
Hint 2/4
If the acceleration falls off as $1/r^{2}$, then $a_{\text{Moon}}/g = (R_E/r_{\text{Moon}})^{2}$.
Hint 3/4
With $r_{\text{Moon}} = 60.2\,R_E$ the bracket is $1/60.2$, its square is $1/(3.62\times10^{3})$, and $g = 9.80\ \mathrm{m/s^{2}}$.
Hint 4/4
Dividing gives $2.70\times10^{-3}\ \mathrm{m/s^{2}}$, pointing at the Earth's centre.
Show solutionWrite the two situations as one ratio
the distance is given as a multiple of $R_E$, so $R_E$ itself is never needed; putting both distances in metres would be two more keystrokes for the same number
three significant figures, which is what the two data values carry
Answer $$\boxed{\;a_{\text{Moon}} = 2.70\times10^{-3}\ \mathrm{m/s^{2}} \text{ toward the Earth}\;}$$
Check
The orbit gives the same number by a route that shares no step with this one: $a = 4\pi^{2}r/T^{2}$ with $r = 3.84\times10^{8}$ m and $T = 2.36\times10^{6}$ s comes out at $2.72\times10^{-3}\ \mathrm{m/s^{2}}$. This prediction used no orbital data at all, only $g$ and a distance ratio, so agreement to under one per cent is a test and not a restatement.
That is the Moon test in a single line: predicted from the ground, checked against the sky, and the two agree to under one per cent. It is also the reason the exponent in the force law is a 2 and not a 1.
⚠ Measuring the apple's distance from the Earth's surface instead of from its centre
the apple starts on the ground, so its distance from the Earth feels like zero, and the comparison with the Moon is then a division by nothing
One formula gives the pull between any two masses, and it is the only genuinely new thing in this section.
The last block measured how the pull falls off with distance; it said nothing about how strong that pull is, or what the two bodies have to do with it.
RuleLaw 7.1: Newton's law of universal gravitation
Conditions
The two bodies are treated as points, or as uniform spheres, in which case all the mass of each may be placed at its centre
$r$ is the distance between the two centres, in metres, never the gap between the surfaces
The force acts along the line joining the two centres and is always an attraction; there is no repulsive case
The pull on the first body and the pull on the second are a third law pair: equal in size, opposite in direction
$$\boxed{\;F = G\,\frac{m_1 m_2}{r^{2}}, \qquad G = 6.674\times10^{-11}\ \mathrm{N\cdot m^{2}/kg^{2}}\;}$$
Any two lumps of matter pull on each other. The pull grows in proportion to each of the two masses, so doubling either one doubles it, and it shrinks with the square of the distance between their centres, so moving them three times as far apart leaves a ninth. The constant $G$ is spectacularly small, which is why gravity only matters when at least one of the two masses is planet sized.
The anatomy of the formula. The $\textcolor{#1f6feb}{\text{two pulls}}$ have the same size however unequal the masses are, they lie along the line joining the centres, and the $\textcolor{#d1690a}{\text{distance}}$ in the denominator is measured centre to centre and not between the surfaces.
Looks like this, but is not
The Earth is enormous and I am not, so the Earth pulls me far harder than I pull it. Everyone believes this the first time, and the first half of the sentence is true.
The formula is symmetric: swap $m_1$ and $m_2$ and nothing changes. The Earth pulls a 60 kg student with about 590 N and the student pulls the Earth with the same 590 N, as the third law requires. What is wildly asymmetric is the response, because 590 N divided by 60 kg is an obvious acceleration and divided by $5.97\times10^{24}$ kg is unmeasurable.
The pull between two people standing a metre apart
A 70 kg person and a 60 kg person stand with their centres 1.0 m apart. Find the gravitational force between them, and say why nobody has ever noticed it.
Given
$m_1 = 70\ \mathrm{kg}$
$m_2 = 60\ \mathrm{kg}$
$r = 1.0\ \mathrm{m}$
Find
the force between them, and something to compare it with
Dimension check on the constant, which is the useful check here. $G$ carries $\mathrm{N\cdot m^{2}/kg^{2}}$, the masses supply $\mathrm{kg^{2}}$ and the denominator supplies $\mathrm{m^{2}}$, so everything cancels except the newton. Had the answer come out in newtons per metre, the exponent on $r$ would have been wrong.
One substitution. The work is in the interpretation, not the arithmetic.
About one seventieth of the weight of a mosquito, between two adults standing close together. This is why $G$ has to be measured with a torsion balance rather than with a scale, and it is also the answer to the question of why gravity, the weakest force in this course by an enormous margin, is the one that runs the solar system: it never cancels, because there is no negative mass to cancel it with.
Getting a bathroom scale reading out of the force law
A 60.0 kg student stands on the ground. Using $M_E = 5.97\times10^{24}$ kg and $R_E = 6.38\times10^{6}$ m, compute the Earth's gravitational pull on the student from the force law, and compare it with $mg$.
the same physical quantity computed the way it was computed in earlier sections
Answer $$\boxed{\;F = 587\ \mathrm{N} \text{ from the law}, \qquad mg = 588\ \mathrm{N}\;}$$
Check
The two answers agree to within 0.2 per cent, which is the level at which the quoted value of $g$ and the quoted values of $M_E$ and $R_E$ are consistent with each other. If they had disagreed by a factor of a thousand, the likely culprit would have been $r$: using $R_E$ in kilometres rather than metres changes the answer by exactly $10^{6}$.
Two routes to one number, and that is the whole point of the example.
$mg$ was never a separate law. It is the universal law with the distance frozen at one Earth radius and the two constants $G$ and $M_E$ bundled together into a single number that happens to be 9.80. The next two blocks unfreeze that distance.
Checkpoint
§07.1 — the law applied to two ordinary spheres●○○○○
Thirty seconds, straight substitution. Two uniform lead spheres of mass 8.00 kg each are placed on a bench with their centres 0.500 m apart.
the separation is squared, so halving it would quadruple the answer rather than double it
Answer $$\boxed{\;F = 1.71\times10^{-8}\ \mathrm{N} \text{ on each sphere}\;}$$
Check
Scaling check against the worked example with two people. There the masses were 70 and 60 kg at 1.0 m and the answer was $2.8\times10^{-7}$ N. Here the mass product is smaller by a factor of 65.6 and the distance is halved, which multiplies by 4, so the expected answer is $2.8\times10^{-7}\times4/65.6 = 1.7\times10^{-8}$ N. It matches.
A pull of this size is what a Cavendish torsion balance is built to detect, and it is why the experiment was famous.
⚠ Using the gap between the surfaces as r
the problem usually describes the arrangement the way you would see it, as two balls with a gap, and the word distance in ordinary speech means the gap
wrong$$F = \frac{Gm_1m_2}{(0.50)^{2}} \quad \text{for two spheres of radius }0.30\ \mathrm{m}\text{ with a }0.50\ \mathrm{m}\text{ gap}$$
To compare two gravitational situations you never need the constant. Multiply by the factor each mass has grown by, then multiply by the square of the factor the distance has shrunk by. A mass twice as large doubles the pull; a distance twice as large leaves a quarter of it, because the distance ratio enters squared and upside down.
The shape behind every scaling question in this block. The $\textcolor{#1f6feb}{\text{force}}$ on a fixed mass against its distance from the centre, with the two $\textcolor{#d1690a}{\text{readings}}$ that get quoted most often marked on the curve: a quarter at two radii, a ninth at three.
Looks like this, but is not
Twice as far away, so half as strong. This is the answer most people give before they think, and it is the answer an inverse first power law would give.
The exponent is two, and it was measured, not chosen for tidiness. The apple and the Moon differ in distance by 60 and in acceleration by 3600, and $3600$ is $60^{2}$, not $60$. A useful habit: read $1/r^{2}$ out loud as one over r squared every time, so that the squaring is part of the sentence rather than an extra step you might skip.
distance from the centre
in Earth radii
force on a 1.00 kg mass
6.38 × 10⁶ m, the surface
1.00
9.79 N
1.28 × 10⁷ m
2.00
2.45 N
1.91 × 10⁷ m
3.00
1.09 N
4.22 × 10⁷ m
6.62
0.223 N
3.84 × 10⁸ m, the Moon
60.2
0.00272 N
The second and third rows are the ones to memorise the shape from: doubling the distance did not halve the 9.79, it quartered it, and tripling it left a ninth. Notice also how slowly the force dies at large distances, which is the reason gravity organises the solar system at all: at sixty Earth radii it is down by a factor of 3600 but it has not gone away, and nothing screens it.
The height at which the Earth's pull has dropped to a quarter
At what height above the Earth's surface is the gravitational force on a spacecraft one quarter of its value at the surface? Take $R_E = 6.38\times10^{6}$ m.
Substitute back through the full law rather than through the ratio. At $r = 2R_E$ the acceleration is $GM_E/(2R_E)^{2} = (6.674\times10^{-11})(5.97\times10^{24})/(1.276\times10^{7})^{2} = 2.45\ \mathrm{m/s^{2}}$, which is a quarter of 9.79 as required, and the route used neither of the ratio steps.
Two lines and no constants. Ratio methods are worth reaching for whenever a question compares two situations rather than asking for one absolute value.
The trap in the question is the difference between $r$ and $h$, and it is not incidental: every altitude question in this section hides that same subtraction. Write $r = R + h$ as an explicit line before substituting anything, and the trap cannot spring.
Surface gravity on a planet with twice the radius and eight times the mass
A planet has eight times the Earth's mass and twice the Earth's radius. How does the acceleration due to gravity at its surface compare with $9.80\ \mathrm{m/s^{2}}$?
Given
$M_p = 8M_E$
$R_p = 2R_E$
$g_E = 9.80\ \mathrm{m/s^{2}}$
Find
the surface value of $g$ on that planet
SolutionWrite the surface acceleration as a formula so the scaling is visible
$$g = \frac{GM}{R^{2}}$$
cancelling the mass of whatever is falling, which is why $g$ does not depend on it; the derivation of this line is the next block, and here only its shape is used
Check by the density route, which is independent of the ratio algebra. Eight times the mass in a body twice as wide means eight times the mass in eight times the volume, so this planet has exactly the Earth's density. For bodies of equal density $g$ is proportional to the radius, and this one has twice the radius, so twice the $g$. The two arguments share no algebra and agree.
Three lines, no constants substituted anywhere.
Notice which way round the two effects pull. Mass helps and distance hurts, and because the distance enters squared it usually wins: a planet with a hundred times the Earth's mass but eleven times its radius would have a weaker surface gravity than ours.
How far out the Earth's pull is one per cent of its surface value
At what distance from the Earth's centre, and at what height above its surface, is the gravitational force on a given body one per cent of its value at the surface?
Given
$F_2/F_1 = 0.0100$
$R_E = 6.38\times10^{6}\ \mathrm{m}$
Find
the distance from the centre and the height above the surface
square rooting a hundredth gives a tenth, and the choice to write the ratio this way up keeps $r$ in the denominator where the square root is easy to read
$$r = 10R_E = 6.38\times10^{7}\ \mathrm{m}$$
the force is down by a hundred when the distance is up by ten, which is the inverse square law read backwards
Report the height as well, since that is what a launch question would ask
$$h = r - R_E = 9R_E = 5.74\times10^{7}\ \mathrm{m}$$
nine radii, not ten; the surface is already one radius out from the centre
Answer $$\boxed{\;r = 6.38\times10^{7}\ \mathrm{m}, \qquad h = 5.74\times10^{7}\ \mathrm{m}\;}$$
Check
Plausibility against a known distance. The Moon sits at about 60 Earth radii, where the pull is down by 3600. Ten radii should therefore give a much larger fraction than the Moon feels, and a hundredth is indeed 36 times the Moon's $1/3600$. The two figures are consistent.
One square root and one subtraction.
This is worth holding on to as a mental yardstick: the Earth's pull is still one per cent as strong as at the ground more than nine Earth radii out, roughly a sixth of the way to the Moon. Gravity has no edge, and nothing about the formula suggests a distance beyond which it stops.
Checkpoint
§07.2 — scaling the distance and the mass at once●●○○○
Thirty seconds, no calculator. A satellite feels a certain gravitational force at a distance $r$ from the centre of a planet. It is then moved to a distance $3r$ from the centre of a different planet, one whose mass is three times as large.
Given
the satellite's own mass does not change
the distance goes from $r$ to $3r$
the central mass goes from $M$ to $3M$
Find
(a) The new force is what fraction of the old one?
Hint 1/4
Two things changed at once, so handle them separately and multiply the two factors at the end rather than trying to see the answer whole.
Hint 2/4
The ratio rule is $F_2/F_1 = (M'/M)(r_1/r_2)^{2}$, with the mass factor upright and the distance factor inverted and squared.
Hint 3/4
Here the mass factor is 3 and the distance goes from $r$ to $3r$, so the distance factor is $(1/3)^{2}$.
Hint 4/4
The product is $3\times 1/9 = 1/3$, so the new force is a third of the old one.
the mass is in the numerator of the law so its factor goes in upright; the distance is squared in the denominator so its factor is inverted and squared
$$\frac{F_2}{F_1} = \frac{1}{3}$$
one number, and no constants were needed at any point
Sanity check on the direction of the change. Tripling the distance costs a factor of nine while tripling the mass only buys a factor of three, so the force must end up smaller. A third is smaller, so the sign of the effect is right, which rules out any answer larger than one.
Whenever two quantities change together, get one factor per quantity and multiply. Trying to see the combined answer in one move is where the ninth and the one both come from.
⚠ Forgetting to square the distance ratio
the mass factors go in without any exponent, so the hand keeps going in the same rhythm when it reaches the distance
wrong$$\frac{F_2}{F_1} = \frac{r_1}{r_2} = \frac{1}{2} \quad \text{for a doubled distance}$$
right$$\frac{F_2}{F_1} = \left(\frac{r_1}{r_2}\right)^{2} = \frac{1}{4} \quad \text{for a doubled distance}$$
⚠ Doubling the height when the question doubled the distance from the centre
altitudes are what launch questions quote, so the height feels like the natural variable, but the law only ever sees the distance from the centre
wrong$$h \to 2h \;\Rightarrow\; F \to \frac{F}{4}$$
right$$r = R + h, \qquad r \to 2r \;\Rightarrow\; F \to \frac{F}{4}$$
7.4Gravity near a planet: where g comes from
The number 9.80 is not fundamental; it is what the universal law gives for this planet at this radius.
Two blocks of scaling have compared one situation with another. Now we cash the law for an absolute number, and the number turns out to be one you have been using since the first week.
TheoremResult 7.3: the acceleration due to gravity at a distance r
Conditions
$M$ is the mass of the planet and $r$ the distance from its centre
The falling body's own mass has cancelled, so it does not appear
The planet is treated as a uniform sphere and its rotation is ignored
At the surface $r = R$; at a height $h$ above the surface $r = R + h$
The acceleration a freely falling body has at a given place is the planet's mass times the gravitational constant, divided by the square of the distance from the planet's centre. Nothing about the falling body survives into the formula, which is the reason a feather and a hammer fall together, and the reason $g$ is a property of the place rather than of the thing being dropped.
Where this comes from
Take a body of mass $m$ at a distance $r$ from the centre of a planet of mass $M$, with gravity as the only force on it.
The force on it, from the law of universal gravitation, is $F = GMm/r^{2}$.
The second law says that force equals $ma$, so $ma = GMm/r^{2}$.
The mass $m$ appears on both sides and cancels, leaving $a = GM/r^{2}$.
That acceleration is what the symbol $g$ has meant all along, so $g = GM/r^{2}$.
The cancellation in step 3 is the whole content of the result. It happens only because the same $m$ that measures how hard a body is to accelerate also measures how hard gravity pulls it, which is a fact about the world rather than a fact about algebra.
Why the strength of gravity barely changes at the height of a low orbit. The $\textcolor{#128a5a}{\text{orbit}}$ and the surface are drawn at one scale, so the 400 km altitude is the thin gap it really is against a $\textcolor{#d1690a}{\text{radius}}$ of 6380 km.
Looks like this, but is not
$g$ is a constant of nature, like the speed of light. It certainly behaves like one in the first six weeks of this course, where it never changes.
$G$ is the constant of nature; $g$ is a local reading. Two planets with different masses or different radii have different values of $g$, and even on this one the value falls with height: it is 9.79 at sea level and 8.67 at the height of a low satellite orbit. The reason it looks constant in kinematics problems is that a building 100 m tall changes $r$ by 0.0016 per cent, and no experiment in a first year laboratory could see the difference.
place
distance from the centre
g there
the ground
6.38 × 10⁶ m
9.79 m/s²
the top of a 400 m tower
6.38 × 10⁶ m
9.79 m/s²
a low satellite orbit, 400 km up
6.78 × 10⁶ m
8.67 m/s²
2000 km up
8.38 × 10⁶ m
5.68 m/s²
a , 35 900 km up
4.22 × 10⁷ m
0.223 m/s²
the distance of the Moon
3.84 × 10⁸ m
0.00272 m/s²
Two things to take from this table. First, the second row is the reason every kinematics problem you have done so far was allowed to treat $g$ as fixed: a 400 m tower does not move the third significant figure. Second, the third row is the one people find hardest to believe. Four hundred kilometres up, where astronauts are photographed floating, $g$ is still 8.67, or 88 per cent of its value at the ground. Whatever floating is, it is not the absence of gravity.
Producing 9.80 from the Earth's mass and radius
Using $G = 6.674\times10^{-11}\ \mathrm{N\cdot m^{2}/kg^{2}}$, $M_E = 5.97\times10^{24}$ kg and $R_E = 6.38\times10^{6}$ m, compute the acceleration due to gravity at the Earth's surface.
the numerator and the denominator are each evaluated before the division, so that a misplaced exponent shows up as an absurd intermediate rather than as a plausible final answer
the exponents differ by one, so the answer should be somewhere near ten, and it is
Answer $$\boxed{\;g = 9.79\ \mathrm{m/s^{2}}\;}$$
Check
Independent check against something measured a completely different way. A stone dropped from rest falls $\tfrac{1}{2}gt^{2}$, and a stopwatch gives about 4.9 m in the first second, which needs $g = 9.8\ \mathrm{m/s^{2}}$. That measurement involves no astronomy at all, and it agrees with the astronomy to three figures.
One substitution with four powers of ten in it, which is where the errors are.
The value 9.79 rather than 9.80 is not a mistake; it is what three figure data gives, and the difference is a tenth of a per cent. The section keeps using 9.80 because the syllabus does, and states so rather than quietly changing the number. Sanity check on the order of magnitude: $g$ near ten metres per second squared is a car going from rest to 35 km/h in one second, which is roughly what falling feels like.
The strength of gravity at the height of a low satellite orbit
Find the acceleration due to gravity 400 km above the Earth's surface, and express it as a percentage of the value at the ground.
the same grouped product $GM_E$ from the previous example, reused rather than recomputed
$$\frac{8.67}{9.79} = 0.886$$
the question asked for the comparison, and a bare 8.67 does not answer it
Answer $$\boxed{\;g = 8.67\ \mathrm{m/s^{2}}, \text{ which is } 88.6\% \text{ of the surface value}\;}$$
Check
Check with the ratio method, which uses none of the constants. $g_2/g_1 = (R_E/r)^{2} = (6.38/6.78)^{2} = (0.941)^{2} = 0.885$, so $g_2 = 0.885(9.79) = 8.66\ \mathrm{m/s^{2}}$. The two routes agree in the third figure.
One addition, one square, one division. The addition is the one that decides whether the rest is worth anything.
Hold on to this number. Nearly nine tenths of the Earth's pull is still acting on everyone aboard a low orbiting spacecraft, and yet they float. Whatever the explanation for the floating is, it cannot be that gravity has gone, and the block on weightlessness later in this section is where it gets settled.
Weighing the Earth with a stopwatch and a surveyor's measurement
The acceleration due to gravity at the Earth's surface is measured as $9.80\ \mathrm{m/s^{2}}$ and the Earth's radius is known from surveying to be $6.38\times10^{6}$ m. Using $G = 6.674\times10^{-11}\ \mathrm{N\cdot m^{2}/kg^{2}}$, find the mass of the Earth.
solving in symbols first means the substitution happens once, into a formula whose shape can be checked: a bigger planet of the same size must have a bigger $g$, and $M$ is indeed proportional to $g$ here
Independent check through density, using nothing from the calculation above. The Earth's volume is $\tfrac{4}{3}\pi R_E^{3} = 1.09\times10^{21}\ \mathrm{m^{3}}$, so this mass implies an average density of $5.5\times10^{3}\ \mathrm{kg/m^{3}}$, about five and a half times that of water. Surface rock is about 2.7 times water and iron is 7.9, so a planet with a rocky crust and an iron core landing between the two is exactly right. Had the answer come out at the density of water, the calculation would have been wrong.
One rearrangement and one substitution, to obtain a quantity that cannot be weighed by any other means.
This is why the mattered so much. Measuring the tiny attraction between two lead spheres in a laboratory fixes $G$, and once $G$ is known this two line calculation gives the mass of the planet. Before that experiment nobody had a number for it. The same trick, run with an orbit instead of a dropped stone, will give the mass of the Sun at the end of this section.
Checkpoint
§07.3 — surface gravity on another planet●●○○○
Thirty seconds. A planet has a mass of $6.40\times10^{23}$ kg and a radius of $3.39\times10^{6}$ m.
(a) Find the acceleration due to gravity at its surface.
(b) A 70.0 kg astronaut stands on it. What is the Earth's pull on that astronaut compared with this planet's pull, as a fraction?
Hint 1/4
Part (a) needs a value of $g$ from a mass and a radius, and part (b) is then a comparison of two weights of the same person, which is a comparison of two values of $g$.
Hint 2/4
$g = GM/R^{2}$ at a surface, and the weight of a body there is $mg$ with that local $g$.
Hint 3/4
With $M = 6.40\times10^{23}$ kg and $R = 3.39\times10^{6}$ m, the numerator is $(6.674\times10^{-11})(6.40\times10^{23})$ and the denominator is $(3.39\times10^{6})^{2}$.
Hint 4/4
The surface value is $3.72\ \mathrm{m/s^{2}}$, and the astronaut weighs about 0.38 of what they weigh on Earth.
Show solutionSurface gravity from the two given quantities
Scaling check against the Earth, done with ratios instead of constants. This planet has $0.107$ of the Earth's mass and $0.531$ of its radius, so $g$ should be $0.107/(0.531)^{2} = 0.379$ of the Earth's, which is $3.71\ \mathrm{m/s^{2}}$. That matches the direct calculation and used neither $G$ nor any power of ten.
Those numbers are Mars, and the 0.38 is the figure quoted whenever someone says you would weigh a third as much there. Notice what did not change: the mass, and therefore how hard the astronaut is to push sideways.
⚠ Using the altitude in place of the distance from the centre
the question gives an altitude and the formula wants a distance, and the two are both lengths in metres so nothing looks wrong
right$$r = R_E + h = 6.78\times10^{6}\ \mathrm{m}, \qquad g = \frac{GM_E}{(6.78\times10^{6})^{2}} = 8.67\ \mathrm{m/s^{2}}$$
⚠ Leaving the mass of the falling body in the formula for g
the derivation starts with $F = GMm/r^{2}$ and the $m$ is written twice before it cancels, so it is easy to carry one copy through
wrong$$g = \frac{GMm}{r^{2}}$$
right$$g = \frac{GM}{r^{2}}$$
7.5When more than one body pulls
Gravitational pulls are ordinary forces: one per body, each along its own line, added as vectors.
Every calculation so far has had exactly two bodies in it. Real problems put a spacecraft between a planet and a moon, and the law has to be applied once per pulling body.
RuleRule 7.4: the
Conditions
Apply the force law separately to each pair, once for every body that pulls
Each pull points from the body being acted on toward the body doing the pulling
The presence of a third body does not change the pull between the first two; nothing shields gravity
Add the results as vectors, by components, exactly as any other forces are added
To find the total gravitational force on a body, work out the pull of each other body separately, as though the rest were not there, and then add all those pulls as vectors. No pull is weakened by a mass sitting in the way, and no new term appears from the bodies interacting with each other.
Two $\textcolor{#1f6feb}{\text{pulls}}$ on the same body, one from each of the two neighbours, and the $\textcolor{#128a5a}{\text{resultant}}$ obtained by completing the rectangle. The 10.0 kg mass is nearer in effect than the 6.0 kg one not because of the distance, which is the same, but because of the mass.
Looks like this, but is not
Two masses pull on a third, so the total pull is the sum of the two numbers. This is right exactly when the two pulls point the same way, and questions are rarely that kind.
Each pull points along its own line joining centres, and those lines are almost never parallel. Two pulls of 6.4 nN and 10.7 nN at right angles give 12.5 nN, not 17.1 nN, and if they point in opposite directions they give 4.3 nN. Adding sizes before checking directions is the most reliable way to get a number that is too big.
The pull on a 4.0 kg mass from two neighbours at right angles
A 4.0 kg mass sits at the origin. A 6.0 kg mass sits at $x = 0.50$ m and a 10.0 kg mass sits at $y = 0.50$ m. Find the size and direction of the net gravitational force on the 4.0 kg mass.
Given
$m_0 = 4.0\ \mathrm{kg}$ at the origin
$m_1 = 6.0\ \mathrm{kg}$ at $(0.50, 0)$ m
$m_2 = 10.0\ \mathrm{kg}$ at $(0, 0.50)$ m
Find
the magnitude and direction of the resultant pull on the 4.0 kg mass
SolutionOne pull at a time, with its direction named
the 6.0 kg mass sits along the positive $x$ axis, so its pull on the body at the origin points along positive $x$; there is no component of it in $y$ at all
measured from the positive $x$ axis, which is the convention declared at the start of the section; a direction without a stated reference is not an answer
Answer $$\boxed{\;F = 1.25\times10^{-8}\ \mathrm{N} \text{ at } 59^{\circ} \text{ above the } +x \text{ axis}\;}$$
Check
Bracketing check that uses no trigonometry. The resultant of two perpendicular forces must be larger than either one on its own and smaller than their arithmetic sum, so it has to lie between 10.7 nN and 17.1 nN. It is 12.5 nN. The angle must also be past $45^{\circ}$ because the $y$ pull is the larger, and $59^{\circ}$ is.
Two applications of the force law and one vector addition. The law was applied twice because two bodies pull, not once with the masses added.
Notice what would have happened with the masses added first: $G(4.0)(16.0)/0.25 = 1.71\times10^{-8}$ N, which is not the answer and is not even in the right direction. Superposition adds forces, never masses.
The point between the Earth and the Moon where the two pulls cancel
Find the distance from the Earth's centre, along the line joining the two centres, at which the Earth's pull on a spacecraft and the Moon's pull on it are equal in size and opposite in direction. Take the centre to centre separation as $3.84\times10^{8}$ m.
Given
$M_E = 5.97\times10^{24}\ \mathrm{kg}$
$M_M = 7.35\times10^{22}\ \mathrm{kg}$
$d = 3.84\times10^{8}\ \mathrm{m}$ between the centres
Find
the distance $x$ from the Earth's centre where the two pulls cancel
SolutionSet the two magnitudes equal and let the spacecraft's mass go
the two pulls point in opposite directions along the same line, so cancelling means equal magnitudes; the spacecraft's own mass and $G$ appear on both sides and are gone in the next line
$$\frac{M_E}{x^{2}} = \frac{M_M}{(d-x)^{2}}$$
solving for a position rather than for a force, which is why nothing needs to be substituted for $G$ at any stage
Take a square root rather than expanding the brackets
square rooting both sides turns a quadratic into a linear equation; expanding instead gives a quadratic with a second root outside the two bodies, which then has to be discarded by hand
one line of algebra, and the answer is a distance from the Earth's centre as the question asked, not from the Moon
Answer $$\boxed{\;x = 3.46\times10^{8}\ \mathrm{m} \text{ from the Earth's centre}\;}$$
Check
Substitute the answer back into the two separate pulls on a 1.00 kg mass. Earth: $GM_E/(3.46\times10^{8})^{2} = 3.33\times10^{-3}\ \mathrm{N}$. Moon: $GM_M/(3.84\times10^{8}-3.46\times10^{8})^{2} = 3.36\times10^{-3}\ \mathrm{N}$. They agree to the rounding, and neither number was used in getting the answer.
The whole problem is one square root; the temptation to expand $(d-x)^{2}$ costs four extra lines and introduces a spurious root.
The cancellation point is 90 per cent of the way to the Moon, not halfway, and the reason is the square root: the Earth is 81 times more massive but only $\sqrt{81} = 9$ times further in its reach along this line. Whenever a ratio of masses controls a distance, expect a square root to soften it.
Two pulls in a straight line on a spacecraft halfway to the Moon
A 1500 kg spacecraft is at the midpoint of the line joining the centres of the Earth and the Moon, $1.92\times10^{8}$ m from each. Find the net gravitational force on it.
Given
$m = 1500\ \mathrm{kg}$
distance to each centre $1.92\times10^{8}\ \mathrm{m}$
directed toward the Moon, that is, exactly opposite to the first; the distances are equal so the ratio of the two pulls is the ratio of the two masses, 81 to 1
opposite directions along one line means the components have opposite signs, so the vector sum is a difference of magnitudes
Answer $$\boxed{\;F_{\text{net}} = 16.0\ \mathrm{N} \text{ toward the Earth}\;}$$
Check
Ratio check that avoids the arithmetic entirely. At equal distances the two pulls are in the ratio of the masses, $5.97\times10^{24}/7.35\times10^{22} = 81.2$, so the Moon's pull should be 1.2 per cent of the Earth's. And $0.200/16.2 = 1.23$ per cent.
Two substitutions and a subtraction, with the only judgement being the sign.
Halfway to the Moon the Earth still wins by a factor of eighty. That is why the cancellation point in the previous example sits so far out, and it is a useful check on any answer of this kind: the balance point is always much closer to the lighter body.
Checkpoint
§07.4 — two pulls in opposite directions●●●○○
Thirty seconds. Three spheres lie on a straight line. A 20.0 kg sphere is at $x = 0$, a 5.00 kg sphere at $x = 0.200$ m and a 10.0 kg sphere at $x = 0.600$ m.
Given
$20.0\ \mathrm{kg}$ at $x = 0$
$5.00\ \mathrm{kg}$ at $x = 0.200\ \mathrm{m}$
$10.0\ \mathrm{kg}$ at $x = 0.600\ \mathrm{m}$
Find
(a) Find the net gravitational force on the middle sphere, giving its direction.
Hint 1/4
Two bodies pull the middle sphere and they pull it opposite ways, so the job is two applications of the law and one subtraction, with the direction decided by which pull is larger.
Hint 2/4
Superposition: $F_{\text{net}} = F_1 - F_2$ when the two pulls are along the same line in opposite senses, with each $F$ from $Gm_1m_2/r^{2}$.
Hint 3/4
The middle sphere is 0.200 m from the 20.0 kg one and 0.400 m from the 10.0 kg one, so the two separations are not equal and neither are the two masses.
Hint 4/4
The net force is $1.46\times10^{-7}\ \mathrm{N}$ pointing in the negative $x$ direction, toward the 20.0 kg sphere.
the two are antiparallel, so the vector sum is the difference and the direction is that of the larger
Answer $$\boxed{\;F_{\text{net}} = 1.46\times10^{-7}\ \mathrm{N} \text{ in the } -x \text{ direction}\;}$$
Check
Ratio check without arithmetic. The first sphere is twice as massive and half as far, so its pull should be $2\times 2^{2} = 8$ times the other one. And $1.67\times10^{-7}/2.09\times10^{-8} = 8.0$.
Being twice as close is worth more than being twice as heavy, because the distance is squared and the mass is not. That single sentence answers a large fraction of the qualitative questions on this topic.
⚠ Adding the masses of two pulling bodies instead of adding their two pulls
the formula has a product of masses in it, so replacing two bodies by one of their combined mass feels like a legitimate shortcut
Put the gravitational pull on the left of the radial equation and $mv^{2}/r$ on the right, and the orbit is solved.
Nothing new is needed now. The left-hand side of the radial equation is the force law, the right-hand side is the circular motion result, and putting the two together is the whole of satellite physics at this level.
TheoremResult 7.5: speed and period of a circular orbit
Conditions
The orbit is a circle of radius $r$ measured from the centre of the central body
Gravity is the only force acting, so no engine is firing and there is no air
$M$ is the mass of the central body; the orbiting mass cancels and does not appear
The speed is constant, which follows because the pull is always perpendicular to the motion
$$\boxed{\;v = \sqrt{\frac{GM}{r}}, \qquad T = 2\pi\sqrt{\frac{r^{3}}{GM}}\;}$$
For a circular orbit of a given radius there is exactly one speed that works, and it depends on the mass of the body being orbited and on the radius, but not at all on what is doing the orbiting. Further out means slower, in proportion to one over the square root of the radius, and it also means a longer path, so the period grows even faster than the radius does.
Where these come from
Take a satellite of mass $m$ on a circle of radius $r$ round a body of mass $M$, with gravity the only force acting.
Along the radial direction, positive toward the centre, the only force is the gravitational pull: $\sum F_R = GMm/r^{2}$.
The radial acceleration for circular motion at constant speed is $a_R = v^{2}/r$, from the previous section.
The second law then reads $GMm/r^{2} = mv^{2}/r$.
The satellite's mass $m$ cancels, and one power of $r$ cancels, leaving $v^{2} = GM/r$, so $v = \sqrt{GM/r}$.
For the period, one revolution covers $2\pi r$ at that speed, so $T = 2\pi r/v = 2\pi r\sqrt{r/GM} = 2\pi\sqrt{r^{3}/GM}$.
Step 4 is why a bolt that falls off a space station keeps pace with it exactly. Nothing about the orbiting body survives the cancellation.
Why a steady speed is possible at all. The $\textcolor{#1f6feb}{\text{pull}}$ points at the centre and the $\textcolor{#128a5a}{\text{velocity}}$ lies along the circle, so the force is perpendicular to the motion at every instant and can change the direction of the velocity without ever changing its length.
Looks like this, but is not
A heavier satellite needs to be launched to a higher speed to stay on the same orbit. It sounds like the kind of thing that should be true, and it is how a heavier vehicle behaves on a bend in the road.
The satellite's mass cancels in step four of the derivation, so the required speed at a given radius is the same for a bolt, a person and a fifty tonne station. It cancels because gravity is exactly proportional to mass: a heavier satellite is pulled harder in precisely the proportion that it is harder to turn.
Speed and period of a satellite 400 km up
A satellite is in a circular orbit 400 km above the Earth's surface. Find its and the time it takes to go round once. Take $R_E = 6.38\times10^{6}$ m and $GM_E = 3.98\times10^{14}\ \mathrm{N\cdot m^{2}/kg}$.
the same first line as every altitude problem in this section; using $4.00\times10^{5}$ as $r$ here would give a speed of 31 km/s, four times too large
Independent route to the period through the second boxed formula, which does not use the speed. $T = 2\pi\sqrt{r^{3}/GM_E} = 2\pi\sqrt{(6.78\times10^{6})^{3}/3.98\times10^{14}} = 5.56\times10^{3}\ \mathrm{s}$. A separate plausibility check: a low orbit period of about an hour and a half is what a photograph of the Earth from a space station implies, since the sunrise comes round roughly sixteen times a day.
One addition, one square root, one division. The addition is the only step that can silently ruin the rest.
Two numbers worth remembering as anchors for this whole block: a low orbit means about 7.7 km/s and about 90 minutes. Any answer far from those, for an orbit a few hundred kilometres up, has an error in it, and the error is nearly always the radius.
The radius at which a satellite keeps pace with the ground below it
A communications satellite must stay above the same point on the equator, so its period has to equal one rotation of the Earth, taken here as 24.0 h. Find the radius of its orbit, its height above the surface and its speed.
using the definition of the period rather than the speed formula, so that the two are not both derived from the same rearrangement
Answer $$\boxed{\;r = 4.22\times10^{7}\ \mathrm{m}, \qquad h = 3.59\times10^{7}\ \mathrm{m}, \qquad v = 3.07\ \mathrm{km/s}\;}$$
Check
Check the speed a second way, through the orbit condition instead of through the geometry. $v = \sqrt{GM_E/r} = \sqrt{3.98\times10^{14}/4.22\times10^{7}} = 3.07\times10^{3}\ \mathrm{m/s}$, which agrees. And a scaling check against the low orbit: this radius is 6.2 times larger, so the speed should be $\sqrt{6.2} = 2.5$ times smaller, and $7.67/2.5 = 3.07$.
One conversion, one rearrangement, one cube root and one subtraction. The cube root is the only unfamiliar operation and it is the one worth practising.
This radius is about 5.6 Earth radii above the surface, which is why a dish pointed at such a satellite never has to move, and also why the signal takes a quarter of a second to get there and back. Notice that the mass of the satellite never entered: a one tonne satellite and a five tonne one share this orbit exactly.
The fastest possible orbit: skimming the surface
Ignoring the atmosphere, find the speed and period of a satellite in a circular orbit just above the Earth's surface, at $r = R_E = 6.38\times10^{6}$ m.
the required centripetal acceleration at this radius is exactly the surface value of $g$, which it has to be, since gravity is the only thing supplying it
Answer $$\boxed{\;v = 7.90\ \mathrm{km/s}, \qquad T = 84.5\ \mathrm{min}\;}$$
Check
The last line of the calculation is itself the check: the centripetal acceleration needed comes out at 9.79, the independently known surface value of $g$. It could not have been anything else, and if it had been, the arithmetic would be wrong.
One square root and one division, to produce a number that bounds every other answer in this block.
No circular orbit of the Earth can be faster than 7.90 km/s or shorter than 84.5 minutes, because no orbit can have a radius smaller than the planet. That makes a good final check on any orbit answer: a period of 40 minutes is not a rounding error, it is an impossibility.
Checkpoint
§07.5 — speed and period from an orbit radius●●○○○
Thirty seconds with a calculator. A satellite moves in a circular orbit of radius $2.00\times10^{7}$ m measured from the Earth's centre.
Given
$r = 2.00\times10^{7}\ \mathrm{m}$ from the centre
the unrounded speed is used, since the period is obtained by dividing by it
Answer $$\boxed{\;v = 4.46\times10^{3}\ \mathrm{m/s}, \qquad T = 7.82\ \mathrm{h}\;}$$
Check
Scaling check against the low orbit worked example. This radius is 2.95 times the 400 km orbit radius, so the speed should fall by $\sqrt{2.95} = 1.72$, giving $7.67/1.72 = 4.46$ km/s, and the period should grow by $2.95^{3/2} = 5.07$, giving $92.6 \times 5.07 = 469$ min, which is 7.82 h. Both agree.
The scaling check is worth more than the answer. Once one orbit is known, every other orbit round the same planet follows from it by powers of the radius ratio, with no constants at all.
⚠ Putting the altitude into the orbit formulas instead of the distance from the centre
the question quotes an altitude because that is how orbits are described in the news, and the formula wants a different length that is never quoted
right$$r = R_E + h = 6.78\times10^{6}\ \mathrm{m}, \qquad v = \sqrt{\frac{GM_E}{6.78\times10^{6}}} = 7.67\times10^{3}\ \mathrm{m/s}$$
⚠ Keeping the satellite's mass in the orbit condition
the equation starts with an $m$ on both sides, and cancelling a symbol that appears twice feels like losing information
wrong$$v = \sqrt{\frac{GMm}{r}}$$
right$$\frac{GMm}{r^{2}} = \frac{mv^{2}}{r} \;\Rightarrow\; v = \sqrt{\frac{GM}{r}}$$
7.7Weightlessness is about the floor, not the planet
A scale reads the contact force, and in free fall that force is zero while gravity is still nearly full strength.
The last block produced an orbit in which gravity is the only force acting. That is the definition of free fall, and applying it to a person rather than to a spacecraft settles the most persistent misconception in the whole of mechanics.
RuleRule 7.6: apparent weight is the contact force
Conditions
The body rests on a scale, a floor or a seat, and $N$ is the contact force that surface exerts
The whole system has a vertical acceleration $a$, taken positive downward here
$g$ is the local value of the gravitational acceleration, which need not be 9.80
The reading is zero exactly when $a = g$, that is, when gravity is the only force acting
$$\boxed{\;N = m(g - a), \qquad a = g \;\Rightarrow\; N = 0\;}$$
What a scale reports is not the pull of the planet but the push of the scale. If the floor is accelerating downward, it needs to push less hard to keep the body with it, so the reading drops; if the floor accelerates downward at exactly the rate gravity alone would give, it does not need to push at all and the reading is zero. Nothing in that sentence requires the pull of the planet to have changed, and in orbit it has barely changed at all.
The same astronaut, two force diagrams. On the ground there is a $\textcolor{#1f6feb}{\text{gravitational pull}}$ and a $\textcolor{#8250df}{\text{contact force}}$ and they balance. In orbit 400 km up the pull is still 607 N, twelve per cent smaller than on the ground, and it is the contact force that has gone, taking the scale reading with it.
Looks like this, but is not
Astronauts float because there is no gravity in space. Every photograph appears to confirm it, and the phrase zero gravity is printed on the pictures.
At the height of a low orbit the gravitational acceleration is $8.67\ \mathrm{m/s^{2}}$, which is 88 per cent of its value at the ground. A 70.0 kg astronaut is pulled with 607 N there, against 686 N on the ground. If gravity had really stopped, the spacecraft would travel in a straight line and leave the Earth behind, which is precisely what it does not do. What is zero is the contact force, because the floor is falling at the same rate as the person standing on it, and a floor that is not pressing on you is a floor you cannot feel.
One person, one scale, three different readings
A 60.0 kg person stands on a scale in a lift. Find the reading when the lift accelerates upward at $2.00\ \mathrm{m/s^{2}}$, when it accelerates downward at $2.00\ \mathrm{m/s^{2}}$, and when the cable breaks so that the lift is in free fall. Take $g = 9.80\ \mathrm{m/s^{2}}$.
Given
$m = 60.0\ \mathrm{kg}$
$a = 2.00\ \mathrm{m/s^{2}}$ up, then down, then $a = g$
Consistency check across the three cases. The Earth's pull is 588 N throughout, and the three readings sit at 120, 80 and 0 per cent of it, with the deviations from 588 being $\pm ma = \pm 120$ N in the first two. That symmetry is what the single formula predicts, and reading it off the three answers uses none of the algebra that produced them.
One equation, three substitutions. Writing three separate free body diagrams would give the same answers and take three times as long.
The third case is the whole of weightlessness, arrived at without leaving the building. A person in a falling lift is weightless in exactly the sense an astronaut is, for exactly the same reason, and for a much shorter time.
Why the scale on an orbiting spacecraft reads zero
A 70.0 kg astronaut stands on a scale bolted to the floor of a spacecraft in a circular orbit 400 km above the Earth, where $g = 8.67\ \mathrm{m/s^{2}}$. Find the Earth's pull on the astronaut, the acceleration of the spacecraft, and the scale reading.
Given
$m = 70.0\ \mathrm{kg}$
$r = 6.78\times10^{6}\ \mathrm{m}$ and $g = 8.67\ \mathrm{m/s^{2}}$ there
the orbit is circular and the engines are off
Find
the gravitational pull, the acceleration, and the reading on the scale
SolutionThe pull, which is not zero and not the surface value either
$$F_G = mg = (70.0)(8.67) = 607\ \mathrm{N}$$
the local value of $g$ is used, not 9.80; that is the only concession the altitude makes
The acceleration, from the fact that gravity is the only force
Cross-check the acceleration a completely different way, from the orbit rather than from gravity. The orbital speed at this radius was found earlier to be $7.67\times10^{3}\ \mathrm{m/s}$, so $a_R = v^{2}/r = (7.666\times10^{3})^{2}/(6.78\times10^{6}) = 8.67\ \mathrm{m/s^{2}}$. The kinematic route and the gravitational route give the same number, which is the condition for the orbit to be possible at all.
Three one-line calculations, of which the middle one carries the whole idea.
607 N of pull and a scale reading zero, in the same problem, on the same person, at the same instant. Whenever a question uses the word weightless, ask which of those two numbers it is talking about; the answer is always the second one.
Spinning a station fast enough to fake ordinary weight
A ring shaped space station of radius 100 m is set spinning so that an astronaut standing on the inside of the rim has an apparent weight equal to their weight on Earth. Find the required period of rotation and the speed of a point on the rim.
Given
$r = 100\ \mathrm{m}$
required apparent weight equals $mg$ with $g = 9.80\ \mathrm{m/s^{2}}$
the station is far from any planet, so the gravitational pull on the astronaut is negligible
Find
the rotation period and the rim speed
SolutionSay what the scale would read here
$$\sum F_R = N = ma_R$$
far from any planet the only force on the astronaut is the push of the floor, and that push is what a scale would read, so this time the contact force is not left over but is the whole story
$$N = mg \;\Rightarrow\; a_R = g = 9.80\ \mathrm{m/s^{2}}$$
requiring the reading to match the Earth reading fixes the acceleration, which then fixes the spin
Turn that acceleration into a period
$$a_R = \frac{4\pi^{2}r}{T^{2}} \;\Rightarrow\; T = 2\pi\sqrt{\frac{r}{g}}$$
the period form is chosen because a rotation rate is what an engineer would set, rather than a speed
the rim speed follows from the geometry once the period is known
Answer $$\boxed{\;T = 20.1\ \mathrm{s}, \qquad v = 31.3\ \mathrm{m/s}\;}$$
Check
Substitute back through the other form of the centripetal acceleration, which does not use the period. $a_R = v^{2}/r = (31.30)^{2}/100 = 9.80\ \mathrm{m/s^{2}}$, as required. Order of magnitude sanity: three turns a minute on a ring 200 m across is a rim speed of about 110 km/h, which is fast for a structure but not absurd.
Two lines, both of them borrowed unchanged from the previous section on circular motion.
This is the mirror image of the orbit case and worth holding beside it. In orbit the floor accelerates away from the astronaut and the reading is zero; in the spinning station the floor accelerates into the astronaut and the reading is whatever the designer chose. In neither case did the pull of a planet decide the number on the scale.
Checkpoint
§07.6 — what has actually gone to zero●●○○○
Thirty seconds, no calculation. A 70.0 kg astronaut is aboard a spacecraft in a circular orbit 400 km above the Earth, where the gravitational acceleration is $8.67\ \mathrm{m/s^{2}}$. She is seen floating in the middle of the cabin.
Given
$m = 70.0\ \mathrm{kg}$
$g = 8.67\ \mathrm{m/s^{2}}$ at that height
the engines are off
Find
(a) Which statement describes her situation correctly?
Hint 1/4
Two separate quantities are involved, the pull of the planet and the push of the floor, and the question is what each of them is. Decide them one at a time.
Hint 2/4
The pull is $mg$ with the local value of $g$, and the push is $N = m(g - a)$, where $a$ is the acceleration of the spacecraft.
Hint 3/4
Here $m = 70.0$ kg and $g = 8.67\ \mathrm{m/s^{2}}$, and with the engines off gravity is the only force on the ship, so its acceleration is also $8.67\ \mathrm{m/s^{2}}$.
Hint 4/4
The pull is 607 N and the push is zero, so she floats while being pulled hard.
Show solutionThe pull
$$F_G = mg = (70.0)(8.67) = 607\ \mathrm{N}$$
the local $g$ is 88 per cent of the ground value, so the pull is 88 per cent of the ground weight, not zero
The push
$$a = g = 8.67\ \mathrm{m/s^{2}} \;\Rightarrow\; N = m(g-a) = 0$$
the ship falls at the same rate as its occupant, so the two never press on each other
Answer $$\boxed{\;F_G = 607\ \mathrm{N}, \qquad N = 0\;}$$
Check
Test the alternative by its consequences. If the pull really were zero, the net force on the spacecraft would be zero, and a body with no net force moves in a straight line at constant velocity. The spacecraft goes round a circle, so the pull cannot be zero.
Carry the question with you rather than the answer: whenever something is called weightless, ask which force is being said to vanish. It is always the contact force.
⚠ Reading weightless as no gravity
the word contains weight, and the pictures show people floating, which is what genuinely zero gravity would also look like
wrong$$\text{in orbit at }400\ \mathrm{km}: \quad g = 0, \quad F_G = 0$$
right$$\text{in orbit at }400\ \mathrm{km}: \quad g = 8.67\ \mathrm{m/s^{2}}, \quad F_G = mg \ne 0, \quad N = 0$$
⚠ Using the surface value of g at orbital height
9.80 has been the value in every problem for six weeks, so it is written down before the altitude is noticed
Three rules read off astronomical data become consequences of one force law and one law of motion.
The period formula in the previous block does not know what is orbiting what. Applied to planets it turns out to be a relation astronomers had already read off their data, decades before any force law existed to explain it.
TheoremResult 7.7: Kepler's third law, with its constant supplied
Conditions
$T$ is the period and $r$ the orbit radius, both for a circular orbit
$M$ is the mass of the central body, the same for all the orbits being compared
The orbiting masses are small enough not to disturb each other
For an elliptical orbit the same relation holds with $r$ replaced by the , which is stated here and not derived
The square of the time a body takes to go round is proportional to the cube of the size of its orbit, and the constant of proportionality depends only on the mass of the body at the centre. Two consequences follow immediately: any two objects orbiting the same central body satisfy the same ratio, so one orbit can be got from another with no constants at all; and measuring any single orbit gives the mass of the thing at the centre.
Where this comes from
This is the period formula from the previous block, squared.
For a circular orbit the gravitational pull supplies the centripetal force: $GMm/r^{2} = mv^{2}/r$.
The orbiting mass cancels, giving $v^{2} = GM/r$.
The speed is the circumference over the period, $v = 2\pi r/T$, so $4\pi^{2}r^{2}/T^{2} = GM/r$.
Rearranging, $T^{2} = (4\pi^{2}/GM)\,r^{3}$.
The whole of the synthesis is in step 4. Kepler had the shape of this relation from twenty years of Tycho Brahe's observations and no explanation for it; here it drops out of two lines of algebra applied to a force law that was invented to describe falling apples.
Kepler's first law, drawn with a far more elongated orbit than any planet actually has so that the two foci can be separated on the page. The $\textcolor{#128a5a}{\text{Sun}}$ sits at one focus and nothing at all sits at the other, and the $\textcolor{#d1690a}{\text{semi major axis}}$ is the length that takes the place of the radius in the third law.
Looks like this, but is not
Kepler's third law says a planet twice as far out takes twice as long. The law does connect period with distance, and this is the connection people assume it makes.
The relation is between the square of one and the cube of the other, so twice the radius means $2^{3/2} = 2.83$ times the period, not twice. Check it against data: Mars is 1.52 times as far out as the Earth and takes 1.88 years, and $1.52^{3/2} = 1.87$, while a linear rule predicts 1.52. A second warning: the law only compares orbits around the same central body.
planet
orbit radius r (m)
period T (s)
T²/r³ (s²/m³)
Mercury
5.79 × 10¹⁰
7.61 × 10⁶
2.98 × 10⁻¹⁹
Venus
1.08 × 10¹¹
1.94 × 10⁷
2.99 × 10⁻¹⁹
Earth
1.50 × 10¹¹
3.16 × 10⁷
2.95 × 10⁻¹⁹
Mars
2.28 × 10¹¹
5.93 × 10⁷
2.97 × 10⁻¹⁹
Jupiter
7.78 × 10¹¹
3.76 × 10⁸
3.00 × 10⁻¹⁹
Saturn
1.43 × 10¹²
9.31 × 10⁸
2.96 × 10⁻¹⁹
The orbit radii in the first column span a factor of 25 and the periods a factor of 122, and the last column is constant to about one per cent. That is the content of the third law as Kepler found it, and it is an experimental fact with no theory attached. Now compute the constant that the derivation predicts, $4\pi^{2}/GM_{\text{Sun}}$, using the Sun's mass: it comes to $2.97\times10^{-19}\ \mathrm{s^{2}/m^{3}}$. The number in the last column was measured with telescopes and the number just computed comes from a law written down to describe falling bodies on Earth. They agree.
The length of the Martian year from the size of its orbit
Mars orbits the Sun at an average distance of $2.28\times10^{11}$ m, and the Earth at $1.50\times10^{11}$ m with a period of 1.00 year. Find the period of Mars in years, without using $G$ or the mass of the Sun.
the constant $4\pi^{2}/GM_{\text{Sun}}$ is the same for both orbits and divides out, which is what makes this a two line problem instead of a five line one
which is 1.88 years. The two routes share no numbers except the orbit radius.
Two lines and no constants. The absolute route in the verification takes six lines and needs the mass of the Sun.
Whenever two bodies orbit the same centre, use the ratio form. It is shorter, it cannot go wrong through a misplaced power of ten, and it gives the answer in whatever unit the given period was in.
Weighing the Sun with the Earth's orbit
The Earth orbits the Sun at a radius of $1.50\times10^{11}$ m with a period of $3.156\times10^{7}$ s. Find the mass of the Sun.
SolutionRearrange the third law for the central mass
$$T^{2} = \frac{4\pi^{2}r^{3}}{GM} \;\Rightarrow\; M = \frac{4\pi^{2}r^{3}}{GT^{2}}$$
the central mass is the only unknown, and solving in symbols first means the shape can be checked: a heavier Sun would pull harder and give a shorter year, and $M$ is indeed inversely proportional to $T^{2}$ here
Check against a quantity that was not used. The Earth's mass was found earlier to be $5.98\times10^{24}$ kg, so this makes the Sun about 335 000 times more massive than the Earth. A second, sharper check: put this mass back into $4\pi^{2}/GM$ and it gives $2.97\times10^{-19}\ \mathrm{s^{2}/m^{3}}$, matching the constant measured from six separate planetary orbits in the table above.
One rearrangement and one substitution, to weigh an object nobody will ever put on a scale.
The general trick is worth naming, because half the exam questions on this topic are it in disguise: anything with something in orbit around it can be weighed, and only the orbit of the satellite is needed, never anything about the satellite itself.
The Earth's mass from the Moon's orbit, and why it is one per cent out
The Moon orbits the Earth at $3.84\times10^{8}$ m with a period of 27.3 days. Use this to find the mass of the Earth, and compare with the value $5.97\times10^{24}$ kg obtained earlier from surface gravity.
the same rearrangement as for the Sun, with the Moon playing the part the Earth played there
Compare honestly rather than rounding the difference away
$$\frac{6.02 - 5.97}{5.97} = 0.008$$
a difference of under one per cent, which is about what three figure inputs and a circular orbit assumption can be expected to deliver
Answer $$\boxed{\;M_E = 6.02\times10^{24}\ \mathrm{kg}, \text{ within } 1\% \text{ of } 5.97\times10^{24}\ \mathrm{kg}\;}$$
Check
The check is the comparison itself, and it is a strong one because the two routes have nothing in common. One used a stopwatch, a dropped stone and the radius of the Earth; the other used the Moon's distance and the length of a month. Agreement to one per cent between measurements that share no data is the sort of thing that makes a law believable.
One conversion and one substitution, plus the comparison, which is the part worth the marks.
The residual one per cent is not a mistake to be hidden. The Moon's orbit is not exactly circular, the Earth and Moon both move about their common centre, and the data is quoted to three figures. Say which of those you think is responsible and the answer becomes a scientific statement rather than a number.
Checkpoint
§07.7 — comparing two orbits round one planet●●○○○
Thirty seconds, no calculator. Satellite A orbits a planet at radius $r$ and satellite B orbits the same planet at radius $4r$.
Given
both orbits are circular and round the same planet
$r_B = 4r_A$
the two satellites have different masses
Find
(a) The period of B is how many times the period of A?
Hint 1/4
Both satellites orbit the same planet, so the constant is shared and this is a pure ratio question with no constants in it.
Hint 2/4
$T^{2} \propto r^{3}$, so $T_B/T_A = (r_B/r_A)^{3/2}$.
Hint 3/4
Here the radius ratio is 4, so the period ratio is $4^{3/2}$, which is $4\sqrt{4}$.
the shared constant cancels, which is also why the satellites' masses never appear
Answer $$\boxed{\;T_B = 8\,T_A\;}$$
Check
Numerical spot check with real numbers. An orbit at $8.00\times10^{6}$ m has a period of 119 min, and one at $3.20\times10^{7}$ m has a period of 950 min. The ratio is 7.98, which is 8 to the precision of the rounding.
Powers of three halves come up constantly in this block. It is worth being able to do $4^{3/2}$, $9^{3/2}$ and $2^{3/2}$ without a calculator: 8, 27 and 2.83.
⚠ Treating the period as proportional to the radius
the law is remembered as a connection between period and distance, and the exponents are the part that fades first
right$$\frac{T^{2}}{r^{3}} = \frac{4\pi^{2}}{GM}, \quad \text{equal only for orbits round the same } M$$
Computing a gravitational force or a value of g
Any question that asks for the pull between two bodies, or for the acceleration due to gravity somewhere. The arithmetic is one substitution; everything that goes wrong goes wrong before it.
Name the two bodies whose interaction is wanted
The law is about a pair. If three bodies are in the problem, this step is done once per pair and the results are added at the end, as vectors.
Write down r as a separate line
Not the altitude, not the surface gap. If the problem gives a height $h$ above a planet of radius $R$, write $r = R + h$ and evaluate it before going on. More marks are lost here than anywhere else in this topic.
Decide whether you want a force or an acceleration
A force needs both masses: $F = Gm_1m_2/r^{2}$. An acceleration needs only the mass of the body doing the pulling: $g = GM/r^{2}$. Putting the falling body's mass into the second formula is the standard error.
Substitute with the powers of ten grouped
Evaluate the numerator and the denominator separately before dividing. For the Earth it is worth using the product $GM_E = 3.98\times10^{14}$ directly, since it appears in every question about this planet.
Check the size against something known
At the surface $g$ must come out near 9.8; in low orbit near 8.7; at the Moon's distance near 0.0027. An answer of $2.5\times10^{3}$ means the altitude was used as $r$.
Where it goes wrong
Using $h$ instead of $R + h$, which multiplies the answer by roughly 280 for a low orbit.
Writing $g$ where $G$ belongs, which changes the answer by eleven orders of magnitude and is invisible on the page.
Leaving the falling body's mass in the formula for $g$.
Forgetting to square the distance, which for a low orbit gives an answer that is out by a factor of about 6.8 million.
Turning any circular orbit into one equation
Anything going round anything under gravity alone: a satellite, a moon, a planet, a star in a galaxy. Whatever is asked for, the route starts the same way.
Draw the body and the single force on it
One arrow, from the orbiting body toward the centre of the central body. If you find yourself drawing a second arrow pointing outward, stop: nothing pushes a satellite outward.
Write the radial equation with the pull on the left
$GMm/r^{2} = mv^{2}/r$, or $GMm/r^{2} = m\,4\pi^{2}r/T^{2}$ if a period is what the problem gives. Choosing the right hand side to match the data given saves a whole step.
Cancel the orbiting mass immediately
It appears on both sides and always goes. Cancelling it early makes it obvious that the answer cannot depend on it, which kills a whole family of wrong answers about heavy satellites.
Solve in symbols, then substitute once
$v = \sqrt{GM/r}$, $T = 2\pi\sqrt{r^{3}/GM}$, or $M = 4\pi^{2}r^{3}/GT^{2}$ if the central mass is the unknown. All three are the same equation rearranged.
Test the answer against the two anchors
A low Earth orbit is 7.7 km/s and 90 minutes; a geosynchronous one is 3.1 km/s and 24 hours. Any Earth orbit answer must lie between the surface limit of 84.5 minutes and whatever the problem's radius implies.
Where it goes wrong
Adding an outward force to the diagram, which makes the net radial force zero and the orbit impossible.
Using the altitude as the orbit radius.
Keeping the satellite's mass in the final formula.
Leaving a period in hours or days when the formula squares it.
Deciding which of Kepler's three laws a question wants
Whenever a question mentions planets, ellipses, orbital periods or the size of an orbit, and it is not obvious which relation is being asked for.
First law, the shape
Every planet moves on an ellipse with the Sun at one focus, not at the centre. Use it when a question asks about the shape of an orbit, about the closest and farthest points, or about what sits at the second focus, which is nothing at all. It gives no numbers on its own.
Second law, the pacing
A line drawn from the Sun to a planet sweeps out equal areas in equal times, so a planet moves faster when it is nearer the Sun. Use it to say which of two positions has the greater speed. It is stated here as Kepler found it in the data and is not derived in this section, and no calculation here needs it.
Third law, the numbers
$T^{2} = (4\pi^{2}/GM)r^{3}$. This is the one that answers computational questions. Use the ratio form when two bodies orbit the same centre, and the absolute form when the mass of the central body is wanted or given.
Check that both orbits share a central body
The ratio form only holds within one system. Two moons of Jupiter, yes; a moon of Jupiter and a planet of the Sun, no.
Where it goes wrong
Using the ratio form across two different central bodies.
Reading the third law as a proportionality between $T$ and $r$.
Putting the Sun at the centre of the ellipse rather than at a focus.
Mixing units, most often a period in days against a radius in metres.
The force on a 2000 kg satellite at 8.00 × 10⁶ m
A 2000 kg satellite orbits the Earth at a radius of $8.00\times10^{6}$ m. Find the gravitational force on it.
The local $g$ at that radius is $3.98\times10^{14}/(8.00\times10^{6})^{2} = 6.22\ \mathrm{m/s^{2}}$, and $mg = 2000 \times 6.22 = 1.24\times10^{4}$ N, reached without the force formula.
Double the satellite's mass and this answer doubles.
The speed of a 2000 kg satellite at 8.00 × 10⁶ m
A 2000 kg satellite orbits the Earth at a radius of $8.00\times10^{6}$ m. Find its orbital speed.
Given
$m = 2000\ \mathrm{kg}$
$r = 8.00\times10^{6}\ \mathrm{m}$
$GM_E = 3.98\times10^{14}$
Find
the orbital speed
SolutionUse the form in which the satellite's mass has cancelled
Cross-check through the force from the other example: $a = F/m = 1.244\times10^{4}/2000 = 6.22\ \mathrm{m/s^{2}}$, and $v = \sqrt{a r} = \sqrt{(6.22)(8.00\times10^{6})} = 7.06\times10^{3}\ \mathrm{m/s}$.
Double the satellite's mass and this answer does not change at all.
The two problems hand you exactly the same three numbers, and the mass matters in one of them and is irrelevant in the other.
How to tell them apart
If the question asks for a force, the orbiting mass stays. If it asks for a speed, a period, a radius or an acceleration, the orbiting mass cancels and any value given for it is there to see whether you know that.
The Earth's pull on an astronaut 400 km up
Find the gravitational force the Earth exerts on a 70.0 kg astronaut in a spacecraft 400 km above the surface, where $g = 8.67\ \mathrm{m/s^{2}}$.
Given
$m = 70.0\ \mathrm{kg}$
$g = 8.67\ \mathrm{m/s^{2}}$ at that height
Find
the gravitational force on the astronaut
SolutionOne line, with the local value of g
$$F_G = mg = (70.0)(8.67) = 607\ \mathrm{N}$$
the altitude has already been accounted for in the value of $g$ that was supplied
Answer $$\boxed{\;F_G = 607\ \mathrm{N}\;}$$
Check
Ratio check: 8.67/9.80 is 0.885, and 0.885 of the ground level 686 N is 607 N.
This is a large force. It is 88 per cent of what the same astronaut feels standing on the ground.
What the astronaut's bathroom scale reads 400 km up
The same 70.0 kg astronaut, in the same spacecraft on the same circular orbit, stands on a bathroom scale bolted to the floor. Find the reading.
Given
$m = 70.0\ \mathrm{kg}$
$g = 8.67\ \mathrm{m/s^{2}}$ at that height
the spacecraft is in free fall on a circular orbit
Find
the reading on the scale
SolutionAsk what the scale measures, then write that body's equation
$$a = g = 8.67\ \mathrm{m/s^{2}}$$
gravity is the only force on the spacecraft, so it and everything in it have the same acceleration
$$N = m(g-a) = 70.0(8.67-8.67) = 0$$
the floor never has to push, because it is already going where the astronaut is going
Answer $$\boxed{\;N = 0\;}$$
Check
Limiting case check against the lift: a lift in free fall gives $N = m(g-g) = 0$ by the same equation, and everyone accepts that a falling lift makes you float.
Zero, in a situation where the pull is 607 N.
Same astronaut, same instant, same orbit: 607 N of gravitational pull and a scale reading of zero.
How to tell them apart
Ask which force the question is about. The word weight, in ordinary speech, points at the scale reading, which is the contact force $N$. The pull of the planet is $mg$ with the local $g$, and it is only equal to the scale reading when the vertical acceleration is zero.
Scaffolding comes off
The common skeleton
Name the orbiting body and the central body, and say which mass is which
Turn whatever length the problem gives into a distance from the centre: $r = R + h$ if an altitude is quoted
Draw the one force, pointing from the orbiting body at the centre, and write the radial equation with it on the left
Put the right hand side in the form that matches the data: $mv^{2}/r$ if a speed is involved, $m\,4\pi^{2}r/T^{2}$ if a period is
Cancel the orbiting mass, rearrange in symbols for the unknown, and substitute only at the end
Convert back to whatever the question asked for, and test the answer against a known orbit
1 · fully worked
Speed and period of a satellite 1000 km above the Earth
A satellite is in a circular orbit 1000 km above the Earth's surface. Find its orbital speed and its period. Take $R_E = 6.38\times10^{6}$ m and $GM_E = 3.98\times10^{14}\ \mathrm{N\cdot m^{2}/kg}$.
the unrounded speed is carried in, since the period is obtained by dividing by it
$$T = 105\ \mathrm{min}$$
converted to minutes, the unit orbital periods are quoted in
Answer $$\boxed{\;v = 7.35\ \mathrm{km/s}, \qquad T = 105\ \mathrm{min}\;}$$
Check
Independent route to the period through $T = 2\pi\sqrt{r^{3}/GM_E}$, which never mentions the speed: $2\pi\sqrt{(7.38\times10^{6})^{3}/3.98\times10^{14}} = 6.31\times10^{3}\ \mathrm{s}$. And a scaling check against the 400 km orbit: the radius is 1.088 times larger, so the period should be $1.088^{3/2} = 1.135$ times longer, and $92.6 \times 1.135 = 105$ min.
One addition, one square root, one division and a conversion.
Note how little the speed changed for an altitude two and a half times larger. Orbital speed goes as one over the square root of the radius, so it is remarkably insensitive to height near the planet.
2 · you write the reasoning
Easier this time, because the radius is given directly from the centre and only one quantity is wanted. A satellite orbits the Earth at $r = 8.00\times10^{6}$ m from the centre, with $GM_E = 3.98\times10^{14}\ \mathrm{N\cdot m^{2}/kg}$. Find its orbital speed. The three lines below are correct. Your job is to say why each one is allowed, in your own words, before opening the model reasons.
reasoning
The left side is the only force acting on the satellite, since the engines are off and there is no air at this height, and it points at the centre. The right side is what the second law requires of any body moving on a circle at constant speed, from the previous section. Setting them equal is not a new law; it is the second law with a particular force substituted in.
reasoning
Two cancellations happen here at once, and both are worth naming. The satellite's mass $m$ divides out of both sides, which is why nothing about the satellite appears in the answer. One factor of $r$ also cancels, leaving $r$ to the first power rather than the second, which is why the speed falls off as one over the square root of the radius rather than as one over the radius.
reasoning
Taking the positive square root, because a speed is the magnitude of a vector and cannot be negative. The value should be checked against the anchor for this planet: the fastest possible circular orbit is 7.90 km/s at the surface, and this orbit is higher than the surface, so anything above 7.90 would be impossible.
3 · find the buried error
Harder than rung 2, because the unknown is now inside a cube root and the answer has to be converted at the end. The problem: a satellite is in a circular orbit around the Earth with a period of exactly 2.00 h. Find its height above the surface, given $GM_E = 3.98\times10^{14}\ \mathrm{N\cdot m^{2}/kg}$ and $R_E = 6.38\times10^{6}$ m. A student's solution is written out below and reaches $1.18\times10^{7}$ m. Exactly two of its four steps are faulty. Find them.
the two buried errors (2)
⚠ step 2
The $2\pi$ outside the square root becomes $4\pi^{2}$ when it is squared and moved, not $4\pi$. The correct rearrangement is $r^{3} = GM_E T^{2}/4\pi^{2}$, with $4\pi^{2} = 39.5$ rather than 12.57, and it gives $r^{3} = 5.23\times10^{20}\ \mathrm{m^{3}}$.
Squaring a product means squaring both factors, and the $\pi$ is the one that gets left behind because the 2 is the visible part. The error survives step 3 because 12.57 is a plausible looking number and the arithmetic that follows is perfectly correct.
right
Keep the constant written as $4\pi^{2}$ all the way through the algebra and only turn it into 39.5 at the moment of substitution. A quick numerical guard: $4\pi^{2}$ is about 40, and any constant near 12 in one of these rearrangements has lost a $\pi$.
⚠ step 4
The cube root gives the distance from the Earth's centre, not the height above the surface. The height is $h = r - R_E$. With the corrected $r^{3}$ this is $r = 8.06\times10^{6}$ m and $h = 8.06\times10^{6} - 6.38\times10^{6} = 1.68\times10^{6}$ m, about 1680 km.
The question asked for a height and the last line produced a length in metres, so the two are matched up without checking which length it is. Nothing on the page looks wrong at that moment.
right
Write the target as $h = r - R_E$ at the top of the page, before starting, so that the final line has somewhere to go. A guard on the size: any orbit radius must be larger than $6.38\times10^{6}$ m, and any altitude for a two hour orbit must be a good deal smaller than the geosynchronous 3.59 × 10⁷ m.
4 · the bare problem
§07.5 — from a period to an altitude, unaided●●●○○
No scaffolding this time. A satellite is in a circular orbit around the Earth and takes exactly 3.00 h to go round once.
(a) Find the radius of its orbit, measured from the Earth's centre.
(b) Find its height above the surface.
(c) Find its orbital speed.
Hint 1/4
A period is given and a length is wanted, so the unknown is inside a cube root. Decide which of the three rearrangements of the orbit condition puts $r$ on its own, and note that part (b) needs one more line after that.
Hint 2/4
$T = 2\pi\sqrt{r^{3}/GM}$ rearranges to $r^{3} = GM\,T^{2}/4\pi^{2}$, and then $h = r - R_E$ and $v = 2\pi r/T$.
Hint 3/4
With $T = 3.00\ \mathrm{h} = 1.08\times10^{4}$ s and $GM_E = 3.98\times10^{14}$, the numerator is $(3.98\times10^{14})(1.08\times10^{4})^{2}$ and the constant on the bottom is $4\pi^{2} = 39.5$.
Hint 4/4
The orbit radius is $1.06\times10^{7}$ m, the height is $4.18\times10^{6}$ m and the speed is $6.14\times10^{3}$ m/s.
using the definition of the period rather than $\sqrt{GM/r}$, so that the two available routes stay independent
Answer $$\boxed{\;r = 1.06\times10^{7}\ \mathrm{m}, \qquad h = 4.18\times10^{6}\ \mathrm{m}, \qquad v = 6.14\times10^{3}\ \mathrm{m/s}\;}$$
Check
Check the speed the other way: $\sqrt{GM_E/r} = \sqrt{3.98\times10^{14}/1.056\times10^{7}} = 6.14\times10^{3}\ \mathrm{m/s}$, agreeing with the geometric route. And a scaling check against the two hour orbit worked in rung 3: the period ratio is 1.5, so the radius ratio should be $1.5^{2/3} = 1.31$, and $8.06\times10^{6} \times 1.31 = 1.06\times10^{7}$ m.
Three separate traps sat in this one question: the conversion, the $4\pi^{2}$, and the difference between $r$ and $h$. None of them is hard, and all three are silent.
Full exam-style question
Weighing a planet from one of its moons, then predicting its surface gravityexam format
A large planet has a moon in a circular orbit of radius $4.22\times10^{8}$ m, with a period of 1.769 days. The planet's own radius is $7.15\times10^{7}$ m. Take $G = 6.674\times10^{-11}\ \mathrm{N\cdot m^{2}/kg^{2}}$.
(a) Find the mass of the planet. (b) Find the acceleration due to gravity at the planet's surface. (c) A second moon of the same planet orbits at a radius of $1.07\times10^{9}$ m. Find its period, in days, without using the mass again.
Given
first moon: $r_1 = 4.22\times10^{8}\ \mathrm{m}$, $T_1 = 1.769\ \mathrm{days}$
both moons orbit the same planet, so the constant $4\pi^{2}/GM$ cancels and part (a) does not need to be reused; if part (a) were wrong, this part could still be right
$$(2.536)^{3/2} = 2.536\sqrt{2.536} = 4.037$$
splitting the power keeps it doable without a calculator and makes the size checkable
Three independent checks, one per part. (a) The answer is about 320 times the Earth's mass; the planet's radius is 11.2 Earth radii, giving a mean density of about $1.2\times10^{3}\ \mathrm{kg/m^{3}}$, a little above water, which is what a gas giant should give and a rocky planet should not. (b) Compute $g$ from the surface orbit condition instead: a satellite skimming this planet would need $v = \sqrt{GM/R} = 4.21\times10^{4}\ \mathrm{m/s}$, and $v^{2}/R = 24.8\ \mathrm{m/s^{2}}$. (c) Substitute the second moon's numbers into the absolute form with the mass from part (a): $T_2 = 2\pi\sqrt{r_2^{3}/GM} = 6.17\times10^{5}\ \mathrm{s} = 7.14$ days, matching the ratio route.
Nine lines in total, and the third part deliberately avoids the answer to the first so that one mistake cannot propagate into three lost marks.
The structure of this question is the structure of most exam questions on this topic: an orbit gives you a central mass, a central mass gives you a surface gravity, and a second orbit round the same body is always a ratio. Answer part (c) by ratio rather than by reusing part (a), and a slip in part (a) costs one mark instead of three.
Practice
A · concept 4 questions
1§07.1 — which of the two pulls is larger●●○○○
A statement of the kind that turns up as a one mark item and that most people get wrong on a first reading. The Earth is about 81 times more massive than the Moon, and the two are held together by their mutual gravitational attraction.
Given
$M_E = 5.97\times10^{24}\ \mathrm{kg}$
$M_M = 7.35\times10^{22}\ \mathrm{kg}$
the two are $3.84\times10^{8}$ m apart
Find
(a) True or false: the Earth pulls the Moon with a greater force than the Moon pulls the Earth. Give your reason in one sentence.
Hint 1/4
No numbers are needed. The question is whether the expression for the force treats the two masses differently.
Hint 2/4
$F = Gm_1m_2/r^{2}$ is symmetric under exchange of the two masses, and Newton's third law says the same thing independently.
Hint 3/4
With $M_E = 5.97\times10^{24}$ kg, $M_M = 7.35\times10^{22}$ kg and $r = 3.84\times10^{8}$ m, the same product $M_EM_M$ appears whichever body you say is doing the pulling.
Hint 4/4
False: the two forces are equal, at $1.99\times10^{20}$ N each.
Show solutionRead the formula for what it says
$$F_{E \text{ on } M} = G\frac{M_EM_M}{r^{2}} = F_{M \text{ on } E}$$
the same product of masses appears in both, so the two expressions are not merely close but identical; the third law says the same thing without any formula at all
81 times smaller, exactly the mass ratio, which is where the intuition about the Earth winning actually comes from
Answer $$\boxed{\;F_{E \text{ on } M} = F_{M \text{ on } E} = 1.99\times10^{20}\ \mathrm{N}\;}$$
Check
Cross-check the Moon's acceleration against the value computed in the first block from its orbit alone, $2.72\times10^{-3}\ \mathrm{m/s^{2}}$. That number came from the orbit radius and the period, with no force law involved, and it agrees with the force route to three figures.
The Earth does move in response to the Moon; both bodies circle their common centre. The reason nobody notices is the factor of 81 in the accelerations, not any asymmetry in the forces.
2§07.5 — does a heavier satellite need a different speed●●○○○
A design question of the kind that gets asked as a quick multiple choice. Two satellites are to be placed in the same circular orbit, at a radius of $8.00\times10^{6}$ m from the Earth's centre. One has a mass of 500 kg and the other 5000 kg.
Given
both orbits are circular with $r = 8.00\times10^{6}\ \mathrm{m}$
(a) How do the two required orbital speeds compare?
Hint 1/4
Before comparing anything, ask whether the satellite's mass can appear in the answer at all. Write the equation the orbit satisfies and look for it.
Hint 2/4
$GMm/r^{2} = mv^{2}/r$, and the $m$ on each side is the satellite's own mass.
Hint 3/4
Cancelling gives $v = \sqrt{GM_E/r}$ with $GM_E = 3.98\times10^{14}$ and $r = 8.00\times10^{6}$ m, and neither 500 kg nor 5000 kg is anywhere in it.
Hint 4/4
Both need the same speed, $7.06\times10^{3}\ \mathrm{m/s}$.
Show solutionWrite the condition and cancel
$$\frac{GM_Em}{r^{2}} = \frac{mv^{2}}{r} \;\Rightarrow\; v = \sqrt{\frac{GM_E}{r}}$$
the satellite's mass appears once on each side and divides out; the reason it does is that gravity is exactly proportional to mass, so a heavier satellite is pulled harder in exactly the proportion that it is harder to turn
Check by computing the forces separately, which do differ. On the 500 kg satellite the pull is $GM_Em/r^{2} = 3.11\times10^{3}$ N and on the 5000 kg one it is $3.11\times10^{4}$ N, ten times larger. Dividing each by its own mass gives the same acceleration, $6.22\ \mathrm{m/s^{2}}$, and equal accelerations at equal radii mean equal speeds.
The same cancellation is why a spanner released by an astronaut drifts alongside the station instead of falling away, and it is worth recognising as the same fact in a different costume.
3§07.4 — combining two pulls on one body●●○○○
A small spacecraft is pulled by two nearby asteroids at the same time. One pull has a size of 4.0 N and the other 3.0 N, and the two asteroids lie in directions at right angles to each other as seen from the spacecraft.
Given
$F_1 = 4.0\ \mathrm{N}$
$F_2 = 3.0\ \mathrm{N}$
the two pulls are at right angles
Find
(a) True or false: the total gravitational force on the spacecraft is 7.0 N. Give your reason in one sentence.
Hint 1/4
The two pulls are forces, so the question is really about how forces combine, not about gravity in particular.
Hint 2/4
Forces add as vectors. For two at right angles the resultant has size $\sqrt{F_1^{2}+F_2^{2}}$, and it equals $F_1+F_2$ only when they are parallel.
Hint 3/4
Here $F_1 = 4.0$ N and $F_2 = 3.0$ N at right angles, so the resultant is $\sqrt{16+9}$.
the components are already perpendicular so no resolving is needed, and the Pythagorean form is the vector sum written out
$$\theta = \arctan\frac{3.0}{4.0} = 37^{\circ}$$
measured from the 4.0 N pull, because a resultant without a direction is only half an answer
Answer $$\boxed{\;F = 5.0\ \mathrm{N} \text{ at } 37^{\circ} \text{ from the } 4.0\ \mathrm{N} \text{ pull}\;}$$
Check
Bounds check with no arithmetic. Two forces of 4.0 N and 3.0 N must combine to something between 1.0 N, when they oppose, and 7.0 N, when they align. The answer 5.0 N sits inside that range, and 7.0 N sits exactly at the edge case that the geometry rules out.
Gravitational forces obey no special addition rule. Everything you learned about combining forces in the section on the laws of motion applies here unchanged.
4§07.5 — why the Moon does not arrive●●●○○
A question that sounds philosophical and has a precise answer. The Earth pulls the Moon with a force of about $2\times10^{20}$ N, and that force points straight at the Earth's centre and never lets up.
Given
the pull on the Moon is about $2\times10^{20}$ N
it points at the Earth at every instant
the Moon's speed on its orbit is about 1.0 km/s
Find
(a) Why has the Moon not fallen into the Earth?
Hint 1/4
Ask what the pull actually does to the Moon's velocity. A force changes velocity, and velocity has a direction as well as a size.
Hint 2/4
A net force perpendicular to the velocity changes the direction of motion without changing the speed; that is precisely the condition for circular motion.
Hint 3/4
Here the pull is $2\times10^{20}$ N pointing at the Earth while the Moon moves at about 1.0 km/s along its orbit, at right angles to that pull.
Hint 4/4
The Moon is falling toward the Earth continuously, and its sideways motion carries it past, so the fall becomes a circle.
Show solutionQuantify the fall
$$\Delta y = \tfrac{1}{2}at^{2} = \tfrac{1}{2}(2.7\times10^{-3})(1.00)^{2} = 1.4\times10^{-3}\ \mathrm{m}$$
the same kinematics as a dropped stone, applied for one second, because there is nothing special about falling in space
Quantify the sideways motion over the same second
$$\Delta x = vt = (1.0\times10^{3})(1.00) = 1.0\times10^{3}\ \mathrm{m}$$
the tangential motion is unaffected by a force at right angles to it
the amount the circular path drops below the tangent line over that horizontal distance, computed from geometry alone with no dynamics in it
Answer $$\boxed{\;\text{fall per second} \approx 1.4\ \mathrm{mm} \approx \text{the amount the circle curves away}\;}$$
Check
The two numbers, 1.4 mm from the dynamics and 1.3 mm from the geometry, were computed by completely different routes and agree to the precision of the rounded inputs. That agreement is the condition for a circular orbit, and it is what the equation $GM/r^{2} = v^{2}/r$ says in one line.
Newton's own way of putting it was a cannon on a mountain: fire slowly and the ball lands, fire fast enough and the ground curves away as fast as the ball falls. Nothing in the physics distinguishes the two cases.
B · computation 8 questions
1§07.1 — the pull that holds the Moon●○○○○
The Earth and the Moon are held together by their mutual attraction, and the size of that force is worth knowing because it is the largest single force acting on either body.
Given
$M_E = 5.97\times10^{24}\ \mathrm{kg}$
$M_M = 7.35\times10^{22}\ \mathrm{kg}$
$r = 3.84\times10^{8}\ \mathrm{m}$ centre to centre
Find
(a) Find the gravitational force between the Earth and the Moon.
Hint 1/4
Two bodies, one separation, one force. Nothing has to be decided except keeping the powers of ten straight.
Hint 2/4
$F = Gm_1m_2/r^{2}$, with $r$ measured centre to centre, which is what the problem has given.
Hint 3/4
With $M_E = 5.97\times10^{24}$ kg, $M_M = 7.35\times10^{22}$ kg and $r = 3.84\times10^{8}$ m, the numerator is $(6.674\times10^{-11})(5.97\times10^{24})(7.35\times10^{22})$ and the denominator $(3.84\times10^{8})^{2}$.
Hint 4/4
The force is $1.99\times10^{20}\ \mathrm{N}$, acting on each body toward the other.
Show solutionSubstitute, evaluating top and bottom separately
Independent route through the Moon's known acceleration. The Moon's centripetal acceleration is $2.72\times10^{-3}\ \mathrm{m/s^{2}}$, measured from its orbit alone, so the force on it must be $M_Ma = (7.35\times10^{22})(2.72\times10^{-3}) = 2.00\times10^{20}$ N. That route uses the period and the orbit radius rather than the two masses.
Order of magnitude check on what a number like this means: it is roughly the weight of $2\times10^{19}$ kg of material at the Earth's surface, or a cube of rock about 20 km on a side. Large, but not mysterious.
2§07.3 — finding the distance at which g takes a given value●●○○○
A probe drifts away from the Earth and its onboard accelerometer, in free fall, is used to infer the local strength of gravity. At one moment the reading corresponds to $1.00\ \mathrm{m/s^{2}}$.
the question asked for a height as well, and forgetting this subtraction is worth a mark in every version of this problem
Answer $$\boxed{\;r = 2.00\times10^{7}\ \mathrm{m}, \qquad h = 1.36\times10^{7}\ \mathrm{m}\;}$$
Check
Ratio check against the surface. This $r$ is 3.13 Earth radii, so $g$ there should be $9.79/(3.13)^{2} = 1.00\ \mathrm{m/s^{2}}$, which is what was given. The check used only the surface value and the radius ratio.
Notice the shape of the answer: to weaken gravity by a factor of about ten you have to get about three times further from the centre, which is well beyond most satellites and nowhere near the Moon.
3§07.3 — surface gravity of the Moon●●○○○
Film of the Apollo landings shows astronauts bouncing rather than walking, and the reason is a number that can be computed from two pieces of data about the Moon.
(a) Find the acceleration due to gravity at the Moon's surface.
(b) An 80.0 kg astronaut wears a suit and pack of total mass 45.0 kg. Find the combined weight on the Moon and on the Earth.
Hint 1/4
Part (a) is a surface gravity from a mass and a radius. Part (b) is then two weights of the same total mass in two different places.
Hint 2/4
$g = GM/R^{2}$ at a surface, and the weight of a mass $m$ there is $W = mg$ with the local $g$.
Hint 3/4
With $M_M = 7.35\times10^{22}$ kg and $R_M = 1.74\times10^{6}$ m the numerator is $(6.674\times10^{-11})(7.35\times10^{22})$, and the total mass in part (b) is $80.0+45.0 = 125.0$ kg.
Hint 4/4
The Moon's surface gravity is $1.62\ \mathrm{m/s^{2}}$, and 125.0 kg weighs 203 N there against 1225 N on Earth.
Ratio check without the constants. The Moon has 0.0123 of the Earth's mass and 0.273 of its radius, so $g_M/g_E = 0.0123/(0.273)^{2} = 0.165$, and $0.165\times9.79 = 1.62\ \mathrm{m/s^{2}}$.
The 203 N is roughly what 21 kg weighs on Earth, which is why a fully suited astronaut can hop. The inertia has not changed though: stopping suddenly is exactly as hard on the Moon as it is here, because that depends on the 125 kg and not on the 1.62.
4§07.3 — weighing a planet from its surface gravity●●○○○
A probe lands on a planet, releases a ball and times its fall, from which the local gravitational acceleration is found to be $8.87\ \mathrm{m/s^{2}}$. The planet's radius is already known from imaging as $6.05\times10^{6}$ m.
(b) Express it as a fraction of the Earth's mass, $5.97\times10^{24}$ kg.
Hint 1/4
The unknown is the planet's mass and everything else in the surface gravity relation is given, so this is one rearrangement.
Hint 2/4
$g = GM/R^{2}$ rearranges to $M = gR^{2}/G$.
Hint 3/4
With $g = 8.87\ \mathrm{m/s^{2}}$ and $R = 6.05\times10^{6}$ m, the numerator is $(8.87)(6.05\times10^{6})^{2}$ and the denominator $6.674\times10^{-11}$.
Hint 4/4
The mass is $4.86\times10^{24}$ kg, which is 0.81 of the Earth's.
Consistency between the two parts. If the planet has 0.81 of the Earth's mass and 0.948 of its radius, then $g$ should be $9.79 \times 0.81/(0.948)^{2} = 8.82\ \mathrm{m/s^{2}}$, which is the value given to within the rounding. The two parts therefore agree with each other rather than merely being computed.
Every question of this shape is the same rearrangement. If a planet has something on its surface you can time, or something in orbit you can watch, its mass follows.
5§07.4 — two pulls at right angles●●●○○
Three small spheres are fixed in a laboratory. A 3.00 kg sphere sits at the origin, a 7.00 kg sphere at $x = 0.300$ m and a 4.00 kg sphere at $y = 0.400$ m.
Given
$3.00\ \mathrm{kg}$ at the origin
$7.00\ \mathrm{kg}$ at $(0.300, 0)$ m
$4.00\ \mathrm{kg}$ at $(0, 0.400)$ m
Find
(a) Find the magnitude of the net gravitational force on the 3.00 kg sphere.
(b) Find its direction, as an angle from the positive $x$ axis.
Hint 1/4
Two bodies pull the sphere at the origin and their pulls are along different axes, so this is two applications of the law followed by one vector addition.
Hint 2/4
Each pull is $Gm_1m_2/r^{2}$ along the line to the pulling body, and the resultant is $\sqrt{F_x^{2}+F_y^{2}}$ at $\arctan(F_y/F_x)$.
Hint 3/4
The 7.00 kg sphere is 0.300 m away along $x$ and the 4.00 kg sphere is 0.400 m away along $y$, so the two separations differ as well as the two masses.
Hint 4/4
The resultant is $1.64\times10^{-8}\ \mathrm{N}$ at $17.8^{\circ}$ above the positive $x$ axis.
directed along the positive $y$ axis; this one is smaller both because the mass is smaller and because the distance is larger, and the distance is the stronger effect
the angle is small because the $x$ pull dominates, which is the right qualitative outcome
Answer $$\boxed{\;F = 1.64\times10^{-8}\ \mathrm{N} \text{ at } 17.8^{\circ} \text{ from the } +x \text{ axis}\;}$$
Check
Bounds and ratio checks. The resultant must exceed the larger component, $1.56\times10^{-8}$ N, and be under the arithmetic sum, $2.06\times10^{-8}$ N; it is. And the ratio of the components should be $(7.00/4.00)\times(0.400/0.300)^{2} = 3.11$, while $1.557/0.5006 = 3.11$.
The angle is nowhere near $45^{\circ}$ even though the masses are of the same order, because the distances are squared and the masses are not. When judging which pull will dominate, look at the distances first.
6§07.5 — an orbit given as an altitude●●●○○
An Earth observation satellite is placed in a circular orbit 600 km above the surface.
both times in seconds; a fractional answer is expected and correct, since the orbit does not divide the day evenly
Answer $$\boxed{\;v = 7.55\times10^{3}\ \mathrm{m/s}, \qquad T = 96.8\ \mathrm{min}, \qquad N = 14.9\ \text{orbits per day}\;}$$
Check
Independent route to the period from the third law: $T = 2\pi\sqrt{r^{3}/GM_E} = 2\pi\sqrt{(6.98\times10^{6})^{3}/3.984\times10^{14}} = 5.81\times10^{3}$ s. And a scaling check against the 400 km orbit: $(6.98/6.78)^{3/2} = 1.045$, and $92.6\times1.045 = 96.8$ min.
Fourteen or fifteen orbits a day is characteristic of everything in low Earth orbit, and it is a fast way to sanity check an answer: a period of a few hours means the satellite is much higher than this, and a period of a few days means it is out near the geosynchronous ring or beyond.
7§07.6 — apparent weight during a launch●●○○○
A 65.0 kg passenger sits on a scale inside a vehicle that accelerates straight upward at $3.20\ \mathrm{m/s^{2}}$ near the Earth's surface.
Given
$m = 65.0\ \mathrm{kg}$
$a = 3.20\ \mathrm{m/s^{2}}$ upward
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the reading on the scale.
(b) Express it as a multiple of the passenger's ordinary weight.
Hint 1/4
The scale reads a contact force, so write the vertical equation for the passenger alone and solve it for that force.
Hint 2/4
With up positive, $N - mg = ma$, so $N = m(g+a)$ for an upward acceleration.
Hint 3/4
With $m = 65.0$ kg, $g = 9.80\ \mathrm{m/s^{2}}$ and $a = 3.20\ \mathrm{m/s^{2}}$ upward, the bracket is $9.80+3.20 = 13.00$.
Hint 4/4
The scale reads 845 N, which is 1.3 times the ordinary weight of 637 N.
Show solutionThe passenger's vertical equation
$$N - mg = ma \;\Rightarrow\; N = m(g+a)$$
up is taken positive because that is the direction of the acceleration, which keeps every term positive and makes a sign error visible
Two limiting cases of the same formula. With $a = 0$ it gives $N = 637$ N, the ordinary reading; with $a = -g$, free fall, it gives zero. The answer sits above the first and far from the second, in the direction an upward acceleration requires.
Pilots and astronauts quote this ratio rather than the force, calling it a number of g. Here it is 1.33 g, and the ratio is the useful quantity precisely because the mass has cancelled out of it.
8§07.7 — comparing two orbits with the third law●●●○○
Two satellites circle the Earth. Satellite A has an orbit radius of $8.00\times10^{6}$ m and a period of 119 min. Satellite B is placed in an orbit of radius $3.20\times10^{7}$ m.
(a) Find the period of satellite B, without using $G$ or the Earth's mass.
(b) Find the ratio of their orbital speeds, $v_B/v_A$.
Hint 1/4
Both satellites go round the same planet, so the constant in the third law is shared and cancels. That makes this a ratio problem with no constants in it at all.
Hint 2/4
$T_B/T_A = (r_B/r_A)^{3/2}$ for the periods, and $v = \sqrt{GM/r}$ gives $v_B/v_A = (r_A/r_B)^{1/2}$ for the speeds.
Hint 3/4
The radius ratio is $3.20\times10^{7}/8.00\times10^{6} = 4.00$, and satellite A's period is 119 min.
Hint 4/4
B's period is $8\times119 = 952$ min, about 15.9 h, and its speed is half that of A.
Internal consistency check between the two answers. The circumference of B's orbit is 4 times A's and B moves at half the speed, so its period should be $4/0.5 = 8$ times A's. That is what part (a) gave, by a completely different route.
Two exponents worth keeping straight, because questions swap between them: period goes as $r^{3/2}$ and speed as $r^{-1/2}$. Their ratio, $r^{3/2}\cdot r^{-1/2}\cdot$, recovers the circumference dependence $r$, which is a quick way to check you have both the right way up.
C · exam level 5 questions
1§07.5 — the orbit that stays above one place●●●●○
A television company wants a satellite that appears fixed in the sky from the ground, so its period must match one rotation of the Earth, 24.0 h. An engineer proposes four possible orbit radii, measured from the Earth's centre. Take $GM_E = 3.98\times10^{14}\ \mathrm{N\cdot m^{2}/kg}$.
A period is given and a radius is wanted, so the unknown sits inside a cube root. It is also worth deciding in advance whether the answer should be a distance from the centre or a height.
Hint 2/4
$T = 2\pi\sqrt{r^{3}/GM}$ rearranges to $r^{3} = GM\,T^{2}/4\pi^{2}$, with $4\pi^{2} = 39.5$.
Hint 3/4
With $T = 8.64\times10^{4}$ s and $GM_E = 3.98\times10^{14}$, the numerator is $(3.98\times10^{14})(7.46\times10^{9})$ and the constant on the bottom is 39.5.
Hint 4/4
The radius is $4.22\times10^{7}$ m, which is 6.6 Earth radii from the centre and 35 900 km above the ground.
Show solutionRearrange for the cube of the radius
$$r^{3} = \frac{GM_E\,T^{2}}{4\pi^{2}}$$
solving for $r^{3}$ rather than $r$ leaves the cube root as one final operation and keeps the constant visible as $4\pi^{2}$
rewriting with an exponent divisible by three before taking the root
$$h = r - R_E = 3.59\times10^{7}\ \mathrm{m}$$
the height, which is the number usually quoted, and the reason two of the four options differ by exactly one Earth radius
Answer $$\boxed{\;r = 4.22\times10^{7}\ \mathrm{m}, \qquad h = 3.59\times10^{7}\ \mathrm{m}\;}$$
Check
Check by going forwards. At this radius the speed is $\sqrt{GM_E/r} = 3.07\times10^{3}\ \mathrm{m/s}$ and the circumference is $2\pi r = 2.65\times10^{8}$ m, so $T = 2.65\times10^{8}/3.07\times10^{3} = 8.63\times10^{4}$ s, which is 24.0 h.
Two of the four options were the same calculation, one of them stopping a line early. That is a common design in exam questions on this topic, and the defence is to write down at the start which length is wanted.
2§07.6 — a scale in a spacecraft with the engine on●●●●○
A spacecraft is far from any planet, so gravitational forces on it are negligible. Its engine fires and gives it a steady acceleration of $4.90\ \mathrm{m/s^{2}}$. A 70.0 kg astronaut stands on a scale mounted on the wall the engine is pushing against.
Given
$m = 70.0\ \mathrm{kg}$
$a = 4.90\ \mathrm{m/s^{2}}$
gravitational forces are negligible here
Find
(a) What does the scale read?
Hint 1/4
Forget gravity entirely for this question and ask a simpler one: what force is needed to give a 70.0 kg body an acceleration of $4.90\ \mathrm{m/s^{2}}$, and what supplies it?
Hint 2/4
The second law for the astronaut alone: the only force on her is the push of the scale, so $N = ma$.
Hint 3/4
With $m = 70.0$ kg and $a = 4.90\ \mathrm{m/s^{2}}$, the product is $70.0\times4.90$.
Hint 4/4
The scale reads 343 N, which happens to be half her Earth reading because the acceleration is half of $g$.
Show solutionList the forces on the astronaut alone
$$\sum F = N = ma$$
she is not touching anything except the scale and there is no gravity here, so the scale's push is the entire left hand side
$$N = (70.0)(4.90) = 343\ \mathrm{N}$$
the answer, and note that no value of $g$ entered the calculation at any point
Consistency with the orbit case, which is the same equation with the numbers swapped. There, gravity acted and the acceleration equalled $g$, so $N = m(g-a) = 0$. Here gravity does not act and the acceleration is supplied by the scale itself, so $N = ma \ne 0$. Both come from writing the astronaut's own equation and reading off the contact force.
This and the orbit case together make the point of the whole block: the scale reading is decided by the acceleration and the contact, never directly by how much gravity is around.
3§07.7 — a moon system tested against the third law●●●●○
A planet has two moons in circular orbits. The inner moon has an orbit radius of $1.85\times10^{8}$ m and a period of 0.942 days. The outer moon has a period of 3.55 days.
Both moons orbit the same planet, so no constants are needed. Decide which power of the period ratio gives the radius ratio before touching a calculator.
Hint 2/4
$T^{2} \propto r^{3}$, so $r \propto T^{2/3}$ and $r_2 = r_1(T_2/T_1)^{2/3}$.
Hint 3/4
The period ratio is $3.55/0.942 = 3.77$, and the inner radius is $1.85\times10^{8}$ m.
Hint 4/4
The radius ratio is $3.77^{2/3} = 2.42$, so the outer radius is $4.48\times10^{8}$ m.
Show solutionGet the exponent right before anything else
$$T^{2} \propto r^{3} \;\Rightarrow\; r \propto T^{2/3}$$
solving the proportionality for the quantity actually wanted, which is where two of the wrong options come from
$$\frac{T_2}{T_1} = \frac{3.55}{0.942} = 3.77$$
a ratio of two periods in the same unit, so the days cancel and no conversion is needed
Apply it
$$\frac{r_2}{r_1} = (3.77)^{2/3} = 2.42$$
the cube root of $3.77^{2} = 14.2$, which is 2.42, or equivalently the square of $3.77^{1/3} = 1.556$
Check by running the third law backwards, radius to period, which is the inverse exponent and so an independent route through the arithmetic. If $r_2 = 4.48\times10^{8}$ m then the period ratio has to be $(4.48/1.85)^{3/2} = (2.42)^{3/2} = 3.77$, and $3.77\times0.942 = 3.55$ days, which is exactly the outer period given in the problem.
Both exponents come up in exam questions and they are inverses of each other: period from radius is a three halves power, radius from period is a two thirds power. Writing down which one you need before substituting is the whole defence against the three wrong options here.
4§07.3 — finding the fault in a solution for g at altitude●●●●○
A student is asked to find the acceleration due to gravity 300 km above the Earth's surface, given $GM_E = 3.98\times10^{14}\ \mathrm{N\cdot m^{2}/kg}$ and $R_E = 6.38\times10^{6}$ m. Their work is reproduced below and concludes that gravity at that height is essentially unchanged from its value at the ground. Exactly one step is faulty.
(a) Which step contains the fault, and what is the correct value of $g$ at that height?
Hint 1/4
Check each line for whether it does what it claims, starting with the units. A step can be arithmetically perfect and still be wrong if the numbers entering it are in different units.
Hint 2/4
Two quantities can only be added when they are in the same unit, and $r = R + h$ needs both terms in metres.
Hint 3/4
Here $h = 300$ km, which is $3.00\times10^{5}$ m, and $R_E = 6.38\times10^{6}$ m, so the correct sum is $6.68\times10^{6}$ m rather than $6.38\times10^{6}$ m.
Hint 4/4
Step 1 is the faulty one, and the correct value is $8.93\ \mathrm{m/s^{2}}$.
the conversion the student skipped; adding 300 to $6.38\times10^{6}$ changes nothing at three significant figures, which is exactly why the error is invisible on the page
steps 2 and 3 were sound; they simply had the wrong number handed to them
Answer $$\boxed{\;\text{Step 1 is faulty}; \quad g = 8.93\ \mathrm{m/s^{2}}\;}$$
Check
Independent ratio check on the corrected answer. $g$ should be $9.79\times(6.38/6.68)^{2} = 9.79\times0.912 = 8.93\ \mathrm{m/s^{2}}$, matching. And it sits sensibly between the surface value of 9.79 and the 400 km value of 8.67 computed earlier.
This is the most dangerous class of error in the whole section, because nothing on the page looks wrong: the arithmetic is right, the formula is right, and the answer is plausible. The only defence is to convert every length to metres on the line where it first appears.
5§07.7 — a full exam question on an unnamed planet●●●●●
A survey probe is placed in a circular orbit 630 km above the surface of a planet and is observed to complete one orbit every 97.0 min. Imaging gives the planet's radius as $6.05\times10^{6}$ m. Take $G = 6.674\times10^{-11}\ \mathrm{N\cdot m^{2}/kg^{2}}$.
(c) Find the acceleration due to gravity at the planet's surface.
Hint 1/4
Three parts, in a chain: an orbit radius from an altitude, then a central mass from that orbit, then a surface gravity from that mass. Note that (c) uses the planet's own radius, not the orbit radius.
Hint 2/4
$r = R + h$, then $M = 4\pi^{2}r^{3}/GT^{2}$, then $g = GM/R^{2}$.
Hint 3/4
With $h = 6.30\times10^{5}$ m and $R = 6.05\times10^{6}$ m the orbit radius is $6.68\times10^{6}$ m, and $T = 97.0$ min is $5.82\times10^{3}$ s.
Hint 4/4
The orbit radius is $6.68\times10^{6}$ m, the planet's mass is $5.21\times10^{24}$ kg, and its surface gravity is $9.49\ \mathrm{m/s^{2}}$.
Show solution(a) The orbit radius
$$r = R + h = 6.05\times10^{6} + 6.30\times10^{5} = 6.68\times10^{6}\ \mathrm{m}$$
both lengths already in metres, so the addition is safe; this is the length the orbit formulas want and it is not the length part (c) wants
swapping $R$ for $r$ here is the single most likely error in the whole question, and it would give $7.79$ instead
Answer $$\boxed{\;r = 6.68\times10^{6}\ \mathrm{m}, \qquad M = 5.21\times10^{24}\ \mathrm{kg}, \qquad g = 9.49\ \mathrm{m/s^{2}}\;}$$
Check
Check part (c) against the orbit directly, bypassing the mass. The probe's centripetal acceleration is $a = 4\pi^{2}r/T^{2} = 4\pi^{2}(6.68\times10^{6})/(5.82\times10^{3})^{2} = 7.79\ \mathrm{m/s^{2}}$, which is the local $g$ at the orbit. Scaling that back to the surface gives $7.79\times(6.68/6.05)^{2} = 9.49\ \mathrm{m/s^{2}}$, matching part (c) without using $G$ or the mass at all.
Notice that the verification produced $7.79\ \mathrm{m/s^{2}}$ as an intermediate. That is exactly the wrong answer a student gets by putting the orbit radius into part (c), so seeing it appear legitimately here is a good way to remember which length belongs where.
D · interleaved 4 questions
1§07.3 — a throw on a planet with a different g●●●○○
An astronaut on the surface of a planet throws a small rock at $12.0\ \mathrm{m/s}$ at $40.0^{\circ}$ above the horizontal, over flat ground. The planet has no atmosphere and its surface gravitational acceleration is $3.72\ \mathrm{m/s^{2}}$.
Given
$v_0 = 12.0\ \mathrm{m/s}$
$\theta = 40.0^{\circ}$
$g = 3.72\ \mathrm{m/s^{2}}$
the rock lands at the height it left
Find
(a) How far from the astronaut does the rock land?
(b) How long is it in flight?
Hint 1/4
Decide first which part of this is new and which is not. The value of $g$ is unfamiliar, but nothing else about the situation is.
Hint 2/4
For a projectile landing at its launch height, $R = v_0^{2}\sin 2\theta/g$ and $t = 2v_0\sin\theta/g$, with whatever local $g$ applies.
Hint 3/4
Here $v_0 = 12.0\ \mathrm{m/s}$, $\theta = 40.0^{\circ}$ so $2\theta = 80.0^{\circ}$, and $g = 3.72\ \mathrm{m/s^{2}}$ rather than 9.80.
Hint 4/4
The range is 38.1 m and the flight lasts 4.15 s.
Show solutionRecognise the problem type before choosing a formula
$$a_x = 0, \qquad a_y = -3.72\ \mathrm{m/s^{2}}$$
this is an ordinary projectile problem; the only thing the gravitation topic has contributed is the number 3.72 in place of 9.80
the vertical motion decides the time, and it is the only part of the motion that $g$ touches
Answer $$\boxed{\;R = 38.1\ \mathrm{m}, \qquad t = 4.15\ \mathrm{s}\;}$$
Check
Consistency between the two answers. The horizontal speed is $v_0\cos\theta = 9.19\ \mathrm{m/s}$ and it never changes, so the range should be $9.19\times4.15 = 38.1$ m, which it is. And a scaling check: everything here is inversely proportional to $g$, so the range is $9.80/3.72 = 2.63$ times the Earth range of 14.5 m.
The interleaving is the point. Nothing here needed the new material except the number 3.72, and recognising that quickly is worth more marks than any formula on the page.
2§07.3 — dragging a crate where g is small●●●○○
On the surface of a body where the gravitational acceleration is $1.62\ \mathrm{m/s^{2}}$, a 40.0 kg crate is dragged across level ground by a horizontal rope. The coefficient of kinetic friction between the crate and the ground is 0.300 and the rope tension is 25.0 N.
Given
$m = 40.0\ \mathrm{kg}$
$g = 1.62\ \mathrm{m/s^{2}}$
$\mu_k = 0.300$
rope tension 25.0 N, horizontal
Find
(a) Find the normal force on the crate.
(b) Find its acceleration.
Hint 1/4
This is a friction problem, and the only thing the current topic supplies is the value of $g$ that goes into the vertical equation.
Hint 2/4
Vertical: $N = mg$ for a level surface with a horizontal rope. Then $F_{fr} = \mu_k N$, and horizontally $T - F_{fr} = ma$.
Hint 3/4
With $m = 40.0$ kg and $g = 1.62\ \mathrm{m/s^{2}}$ the normal force is $40.0\times1.62$, and the rope pulls with 25.0 N against friction of $0.300N$.
Hint 4/4
The normal force is 64.8 N and the acceleration is $0.139\ \mathrm{m/s^{2}}$.
Show solutionVertical equation first, because friction depends on it
the surface is level and the rope is horizontal, so nothing else has a vertical component; the local $g$ is what makes this different from an Earth problem
Friction, then the horizontal equation
$$F_{fr} = \mu_k N = (0.300)(64.8) = 19.4\ \mathrm{N}$$
kinetic, because the crate is being dragged and is therefore already sliding
dividing by the mass in kilograms, not by the weight in newtons; here the two differ by a factor of 1.62 rather than 9.80, which makes the slip harder to spot
Answer $$\boxed{\;N = 64.8\ \mathrm{N}, \qquad a = 0.139\ \mathrm{m/s^{2}}\;}$$
Check
Check the same setup on Earth as a limiting comparison. There $N = 392$ N and $F_{fr} = 118$ N, which exceeds the 25.0 N pull, so the crate would stay put and the acceleration would be zero rather than negative. The low gravity case gives a small positive acceleration, which is the sensible ordering.
Two things change together when $g$ drops and they are worth separating. The crate becomes easy to hold up and easy to drag, because both depend on the weight. It does not become easy to stop once moving, because that depends on the mass, which has not changed.
3§07.3 — timing a fall in low gravity●●○○○
An astronaut on the Moon drops a hammer from rest at a height of 2.00 m above the surface, where the gravitational acceleration is $1.62\ \mathrm{m/s^{2}}$. There is no atmosphere.
Given
$h = 2.00\ \mathrm{m}$
$v_0 = 0$
$g = 1.62\ \mathrm{m/s^{2}}$
Find
(a) How long does the hammer take to reach the surface?
(b) How fast is it moving when it lands?
Hint 1/4
Nothing about this is new except one number. Identify which of the constant acceleration relations connects a distance and a time when the initial speed is zero.
Hint 2/4
From rest, $h = \tfrac{1}{2}gt^{2}$ and $v = gt$, with the local value of $g$.
Hint 3/4
With $h = 2.00$ m and $g = 1.62\ \mathrm{m/s^{2}}$, the time satisfies $2.00 = \tfrac{1}{2}(1.62)t^{2}$.
Hint 4/4
The fall takes 1.57 s and the landing speed is $2.55\ \mathrm{m/s}$.
the initial speed is zero so the linear term drops out; rearranging in symbols first makes the square root a single final operation
Speed from the time
$$v = gt = (1.62)(1.571) = 2.55\ \mathrm{m/s}$$
unrounded time carried in
Answer $$\boxed{\;t = 1.57\ \mathrm{s}, \qquad v = 2.55\ \mathrm{m/s}\;}$$
Check
Independent route to the landing speed that avoids the time entirely: $v^{2} = 2gh = 2(1.62)(2.00) = 6.48$, so $v = 2.55\ \mathrm{m/s}$. And a scaling check: times go as $1/\sqrt{g}$, so the Moon time should be $\sqrt{9.80/1.62} = 2.46$ times the Earth time of 0.639 s, which gives 1.57 s.
This is the calculation behind the famous hammer and feather demonstration. The absence of an atmosphere is what makes the two land together; the low $g$ only makes it slow enough to film.
4§07.6 — spinning a ring to replace a planet●●●●○
A ring shaped module of radius 12.0 m rotates about its centre, far from any planet, so that an 80.0 kg astronaut standing on the inside of the rim presses on the floor with the same force as she would on the Earth's surface.
Given
$r = 12.0\ \mathrm{m}$
$m = 80.0\ \mathrm{kg}$
required floor force equal to her Earth weight, with $g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) What force does the floor exert on her?
(b) What must the rotation period be?
(c) What is the speed of a point on the rim?
Hint 1/4
Far from any planet the only force on the astronaut is the floor's push, so that push is the entire net force and it must be doing the job of keeping her on a circle.
Hint 2/4
$N = ma_R$ with $a_R = 4\pi^{2}r/T^{2}$, and the requirement is $N = mg$, which fixes $a_R = g$.
Hint 3/4
With $m = 80.0$ kg the required force is $80.0\times9.80$, and with $r = 12.0$ m the period follows from $a_R = 9.80\ \mathrm{m/s^{2}}$.
Hint 4/4
The floor pushes with 784 N, the period must be 6.95 s, and the rim moves at $10.8\ \mathrm{m/s}$.
Show solution(a) The force, from the requirement
$$N = mg = (80.0)(9.80) = 784\ \mathrm{N}$$
the design requirement is that a scale should read what it reads on Earth, and a scale reads the contact force
Answer $$\boxed{\;N = 784\ \mathrm{N}, \qquad T = 6.95\ \mathrm{s}, \qquad v = 10.8\ \mathrm{m/s}\;}$$
Check
Substitute back through the other form of the centripetal acceleration: $a_R = v^{2}/r = (10.84)^{2}/12.0 = 9.80\ \mathrm{m/s^{2}}$, as required. Plausibility: about nine revolutions per minute on a ring 24 m across, with a rim speed of 39 km/h.
Put this beside the orbiting astronaut who read zero. In both cases the planet's pull was irrelevant to the scale; what decided the reading was the acceleration of the floor relative to the person standing on it.
Mistake ledger (19 entries)
⚠ Measuring the apple's distance from the Earth's surface instead of from its centre
the apple starts on the ground, so its distance from the Earth feels like zero, and the comparison with the Moon is then a division by nothing
the problem usually describes the arrangement the way you would see it, as two balls with a gap, and the word distance in ordinary speech means the gap
wrong$$F = \frac{Gm_1m_2}{(0.50)^{2}} \quad \text{for two spheres of radius }0.30\ \mathrm{m}\text{ with a }0.50\ \mathrm{m}\text{ gap}$$
right$$\frac{T^{2}}{r^{3}} = \frac{4\pi^{2}}{GM}, \quad \text{equal only for orbits round the same } M$$
⚠ Writing 4π where 4π² belongs when rearranging the period formula
squaring $2\pi$ has to square the $\pi$ as well, and the 2 is the visible half of the product; the resulting constant, 12.6 instead of 39.5, looks perfectly reasonable on the page
wrong$$r^{3} = \frac{GM\,T^{2}}{4\pi}$$
right$$r^{3} = \frac{GM\,T^{2}}{4\pi^{2}}$$
⚠ Reporting the distance from the centre when the question asked for a height
the calculation ends with a length in metres and the question wanted a length in metres, so the two get matched up without checking which length it is
wrong$$r = 8.06\times10^{6}\ \mathrm{m} \;\Rightarrow\; h = 8.06\times10^{6}\ \mathrm{m}$$
⚠ Adding an altitude in kilometres to a radius in metres
the unconverted altitude vanishes into the rounding instead of producing an absurd number, so nothing on the page looks wrong and the final answer is merely a little too large
the product of $G$ and the Earth's mass, quoted to three figures
Check yourself
Close the page and write out, from memory: the force law with every symbol named, what the letter $r$ is measured between, the formula for $g$ at a distance from a planet's centre and where the falling body's mass went, the two orbit formulas and which mass survives in them, the equation for what a scale reads, and Kepler's third law with its constant. Then open the formula card and mark only the ones you could not produce. Those are the blocks to reread, and there will usually be two of them rather than eight.
Write down the law of universal gravitation, say what $r$ is measured between, and compute the pull between two given masses without being told which distance to use?
c-universal-gravitation
Say instantly what happens to a gravitational force when the distance triples and the mass doubles, without substituting a single constant?
c-inverse-square
Compute $g$ at a stated height above a planet, and run the same relation backwards to get the planet's mass from a measured $g$?
c-g-from-G
Find the resultant pull on a body from two others that lie in different directions, and give both a size and an angle?
c-superposition
Start from an altitude, produce an orbit radius, and get a speed and a period from it, and say why the satellite's mass never appeared?
c-satellites
State both the gravitational pull on an orbiting astronaut and the reading on their scale, and explain in one sentence why only one of the two is zero?
c-weightlessness
Get the period of one orbit from another round the same body using only a ratio, and get a central mass from a single orbit using the full relation?
c-kepler
Glossary (17 terms)
universal gravitationevrensel kütle çekimi
The statement that every mass attracts every other mass with a force proportional to each mass and inversely proportional to the square of the distance between their centres.
gravitational constantkütle çekim sabiti
The constant $G$ in that law, $6.674\times10^{-11}\ \mathrm{N\cdot m^{2}/kg^{2}}$, the same throughout the universe and measured in a laboratory rather than derived.
inverse square lawters kare yasası
Any law in which a quantity falls off as one over the square of the distance, so that trebling the distance leaves one ninth of the effect.
point massnoktasal kütle
A body whose size is ignored so that all its mass can be taken to sit at a single point. A uniform sphere may be treated this way for anything outside it, with the point at its centre.
principle of superpositionüst üste binme ilkesi
The rule that the total gravitational force on a body is the vector sum of the separate pulls from each other body, none of which is altered by the presence of the others.
Cavendish experiment
The laboratory measurement of the attraction between two pairs of lead spheres using a torsion balance, which fixes the value of $G$ and thereby makes the mass of the Earth computable.
kütle çekim alan şiddeti
The gravitational force per unit mass at a point, equal to $GM/r^{2}$ and numerically the same as the acceleration a freely falling body would have there.
satelliteuydu
Any body held on a closed path around another by gravity alone, whether it was put there deliberately or not.
orbital speedyörünge hızı
The single speed at which a body can hold a circular orbit of a given radius, $\sqrt{GM/r}$, which does not depend on the orbiting body's own mass.
geosynchronous orbityer eşzamanlı yörünge
A circular orbit whose period equals one rotation of the Earth, at a radius of $4.22\times10^{7}$ m, so that a satellite on it stays above the same longitude.
apparent weightlessnessgörünürde ağırlıksızlık
The condition of a body whose supporting contact force is zero, so that a scale under it reads nothing. It occurs whenever the body and its support are in free fall together, and it does not require the gravitational pull to be small.
Kepler's first lawKepler'in birinci yasası
Each planet moves on an ellipse with the Sun at one focus. The other focus is empty and the Sun is not at the centre of the ellipse.
Kepler's second lawKepler'in ikinci yasası
The line from the Sun to a planet sweeps out equal areas in equal times, so a planet travels faster when it is nearer the Sun. It is stated here as an observation and is not derived.
Kepler's third lawKepler'in üçüncü yasası
The square of a planet's period is proportional to the cube of its orbit size, with a constant that depends only on the mass of the central body.
semi major axisbüyük yarı eksen
Half of the longest diameter of an ellipse. It plays the part the radius plays for a circle, and it is the length that appears in Kepler's third law for a non circular orbit.
günberi
The point on an orbit around the Sun at which the orbiting body is closest to it. The farthest point is the aphelion.
Newton's synthesisNewton sentezi
The demonstration that the motion of bodies in the sky and the motion of bodies on the ground follow the same laws, so that Kepler's descriptive rules become consequences of universal gravitation together with the laws of motion.
What comes next
§08 · Work and Energy
Everything in this section was done with forces, accelerations, speeds and periods, and one question was carefully never asked: how much effort does it take to get a body from one place to another against a pull that changes with distance. The next section builds the tool for that, and it turns out to answer several questions that were awkward here in a single line.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition The set textbook for the course. Its chapter on gravitation covers the same ground as this section, and its end of chapter problems are harder than the ones here, deliberately, and are the right next step once this set feels comfortable.
Course syllabus, week 7 line and assessment table The scope of this section comes from the week line, which names gravitation and Newton's synthesis and gives no chapter numbers; no chapter number is quoted anywhere here as a result. The weightings quoted on the summary card come from the assessment table and nothing beyond them is claimed.
Accepted values of the gravitational constant and of the standard astronomical data $G = 6.674\times10^{-11}\ \mathrm{N\cdot m^{2}/kg^{2}}$, Earth mass $5.97\times10^{24}\ \mathrm{kg}$, Earth radius $6.38\times10^{6}\ \mathrm{m}$, Moon mass $7.35\times10^{22}\ \mathrm{kg}$ at $3.84\times10^{8}\ \mathrm{m}$ with a 27.3 day period, Sun mass $1.99\times10^{30}\ \mathrm{kg}$. All are quoted to three significant figures, which is why every answer in this section is given to three.