Gravitation

A satellite is boosted from a low orbit to a higher one. Does its speed increase or decrease?

Most say increase — you added energy, after all.

Its speed decreases. Both statements are true at once:

$E=-\frac{GMm}{2r}(\text{rises as }r\text{ grows, being less negative}),v_o=\sqrt{\frac{GM}{r}}(\text{falls})$

Higher orbits are slower orbits. The extra energy you supplied went into potential energy — and more of it than the kinetic energy you lost.

That inversion is only visible because $U$ is negative. Get the sign convention right and escape velocity, orbital energy and binding energy all become readings of one equation.

1. Newton's law of gravitation

$F=\frac{G{m}_1{m}_2}{r^2},G=6.674\times 10^{-11}{\text{N m}}^2{\text{kg}}^{-2}$

$G$ is a universal constant — same everywhere, every era. $g$ is a local property of one planet.

Why gravity is weak yet rules the universe — both halves come from one fact:

• Two protons repel electrically $\sim 10^{36}$ times more strongly than they attract gravitationally. Gravity is irrelevant in chemistry.
• But gravity is always attractive, so it never cancels and cannot be shielded. Bulk matter is electrically neutral; mass only accumulates.

Third-law pair: the Earth pulls an apple exactly as hard as the apple pulls the Earth. Only the accelerations differ, by about $10^{25}$.

Superposition holds — a third body never modifies the force between the first two.

Illustration 1

The Earth is 81 times as massive as the Moon and the two are $3.84\times 10^5$ km apart. Where between them does the net gravitational field vanish?

Superposition, so simply set the two fields equal in magnitude at a point $x$ from the Earth:

$\frac{G{M}_E}{x^2}=\frac{G{M}_m}{(d-x{)}^2}\Rightarrow \frac{d-x}{x}=\sqrt{\frac{M_m}{M_E}}=\frac{1}{9}$

$9d-9x=x\Rightarrow x=\frac{9d}{10}=3.46\times 10^5\text{km}$

Read it back: the null point sits 90% of the way to the Moon, because the field ratio depends on the square root of the mass ratio, not the mass ratio itself. A spacecraft that reaches this point coasting is over the hill; beyond it the Moon does the pulling.

Note what is not zero here: the potential. Potentials are negative scalars that add, so they cannot cancel — the potential at this point is a large negative number, which is exactly why arriving here at zero speed still leaves you bound.

2. The shell theorem

The force law is stated for points. Planets are not points, so a theorem is needed before you can write $r$ as the centre-to-centre distance.

Part 1 — a uniform shell attracts an external mass exactly as if all its mass sat at the centre. Exact, not approximate, even at the surface.

Part 2 — the field everywhere inside a uniform shell is exactly zero. Not just at the centre; at every interior point.

Why the interior result holds: take any interior point and draw a narrow double cone through it. The nearer patch is closer, so its pull is stronger by $1/r^2$ — but it is also smaller in area by exactly $r^2$. The two cancel precisely.

That cancellation needs the exponent to be exactly 2. An inverse-cube law would leave a residual field inside, which makes this null result one of the sharpest tests of the inverse-square law ever performed.

Illustration 2

A tunnel is bored straight through the centre of the Earth and a stone is dropped in. Show that it oscillates, and find the period. $R=6400$ km, $g=9.8$.

Inside a uniform sphere only the mass within your radius pulls, so from the shell theorem

$g_{inside}=\frac{GMr}{R^3}=\frac{g}{R}r$

Measuring $x$ from the centre, the acceleration is $a=-\frac{g}{R}x$ — proportional to the displacement and directed back toward the centre, which is precisely the condition for simple harmonic motion.

$\omega =\sqrt{\frac{g}{R}}\Rightarrow T=2\pi \sqrt{\frac{R}{g}}=2\pi \sqrt{\frac{6.4\times 10^6}{9.8}}=5077\text{s}=84.6\text{min}$

The remarkable part: a satellite skimming the surface has $T=2\pi \sqrt{R^3/GM}=2\pi \sqrt{R/g}$ — the same 84.6 minutes. The stone falling through the Earth keeps pace with a satellite racing round the outside, and they meet at the far side together.

A solid sphere is an onion of shells, so both results extend. At depth, only the mass within your radius counts — and for a uniform sphere that goes as $r^3$, so dividing by $r^2$ leaves

$g_{inside}=\frac{GMr}{R^3}$

3. How g varies

Surface: equate $mg=GMm/R^2$, giving $g=GM/R^2$. The falling body's mass cancels — Galileo's result. Rearranged, $M=g{R}^2/G$ weighs the Earth at $5.97\times 10^{24}$ kg, a number unknown until Cavendish measured $G$ in 1798.

Note the factor of 2. Going up costs twice as fast as going down, so $g$ at height $h$ equals $g$ at depth $2h$ (for small $h$).

Rotation. At latitude $\lambda$ part of the true pull supplies centripetal acceleration:

$g_{eff}=g-{\omega }^{2}R{\cos }^2\lambda$

Zero effect at the poles, largest at the equator ($\approx 0.034{\text{m/s}}^2$). If a day were about 1.4 hours, equatorial objects would float.

Shape. The Earth is oblate — equatorial radius is 21 km larger — so $g$ is further reduced there. Both effects agree: $g\approx 9.83$ at the poles, $9.78$ at the equator.

Illustration 3

By how much does the Earth's rotation reduce $g$ at the equator? How short would the day have to be for objects there to float? $R=6400$ km, $g=9.8$.

$\omega =\frac{2\pi }{86400}=7.27\times 10^{-5}\text{rad/s}$

At the equator $\cos \lambda =1$, so the reduction is the full ${\omega }^{2}R$:

${\omega }^{2}R=(7.27\times 10^{-5}{)}^2(6.4\times 10^6)=0.034{\text{m/s}}^2$

That is only 0.35% of $g$ — small, but easily measurable, and it accounts for most of the pole-to-equator difference.

To float, the whole of $g$ must be used up turning you:

${\omega }^{2}R=g\Rightarrow \omega =\sqrt{\frac{9.8}{6.4\times 10^6}}=1.24\times 10^{-3}\text{rad/s}$

$T=\frac{2\pi }{\omega }=5079\text{s}=1.41\text{hours}$

Read it back: the required $\omega$ is about 17 times the actual one, and the effect goes as ${\omega }^2$ — hence a 289-fold gap between a 0.35% correction and total weightlessness.

4. Field and potential

$\vec{E}=\frac{\vec{F}}{m},V=-\frac{GM}{r},E=-\frac{dV}{dr}$

Potential is a scalar — that is its whole practical value. Fields from several masses must be added as vectors with components; potentials just add algebraically. For a symmetric arrangement that turns a page of work into one line.

Inside a shell the field is zero but the potential is not — it is constant at its surface value. Zero field means no change in potential, not zero potential.

Illustration 4

Three particles, each of mass $m$, sit at the corners of an equilateral triangle of side $a$. Find the potential energy of the system and the work needed to separate them to infinity.

Potential energy belongs to pairs, not to particles. Three particles give three pairs:

$U=-\frac{G{m}^2}{a}\times 3=-\frac{3G{m}^2}{a}$

Work to disassemble $=0-U=+\frac{3G{m}^2}{a}$

The standard error is counting six pairs — treating AB and BA as different. With $n$ particles there are $\frac{n(n-1)}{2}$ pairs, so three here and six for four particles.

Contrast with the potential, which is a per-point scalar: at the centroid, each mass is $a/\sqrt{3}$ away, so

$V_{centroid}=-3\frac{Gm}{a/\sqrt{3}}=-\frac{3\sqrt{3}Gm}{a}$

Note this is not $U/m$ for any one particle — potential and potential energy of a system are different questions, and reading which one is asked is half the marks.

5. Energy, and the sign that runs the chapter

$U=-\frac{G{m}_1{m}_2}{r}$

Negative because $U=0$ is chosen at infinity, and gravity does positive work as masses approach.

$U=mgh$ is a near-surface approximation only. The exact difference is $\frac{GMmh}{R(R+h)}$, which reduces to $mgh$ when $h\ll R$. For satellite problems it is simply wrong.

$E=K+U\{\begin{array}{ll}<0& \text{bound — trapped in a closed orbit}\\ \ge 0& \text{escapes, never returns}\end{array}$

Almost every question in this chapter is settled by asking where $E$ sits relative to zero.

6. Escape velocity

The borderline case: arrive at infinity with exactly zero speed, so $E=0$.

$\frac{1}{2}m{v}_e^2=\frac{GMm}{R}$, and the mass cancels:

$v_e=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}\approx 11.2\text{km/s (Earth)}$

• Independent of the escaping mass — a pebble and a spacecraft need the same speed.
• Independent of direction — energy is a scalar. Fired sideways at 11.2 km/s, a body escapes just as surely.
• Depends on where you launch from, since $R$ appears. Escape from a high orbit is cheaper.

Why the Moon has no atmosphere: its $v_e\approx 2.4$ km/s is close to typical molecular speeds, so gas leaks away over geological time. Earth retains N₂ and O₂ but has lost most of its hydrogen and helium.

Push it to the limit: set $v_e=c$ and you get $R_s=2GM/c^2$, the Schwarzschild radius. The Newtonian derivation is not strictly valid there, but it happens to give the right answer.

Illustration 5

A body is projected vertically from the Earth's surface at half the escape velocity. How high does it rise?

Use $GM=g{R}^2$ throughout. Launch speed $v=\frac{1}{2}\sqrt{2gR}$, so

$K_i=\frac{1}{2}m\left(\frac{2gR}{4}\right)=\frac{mgR}{4},U_i=-mgR$

$E=\frac{mgR}{4}-mgR=-\frac{3mgR}{4}$

At the highest point the speed is zero, so all of $E$ is potential:

$-\frac{mg{R}^2}{R+h}=-\frac{3mgR}{4}\Rightarrow R+h=\frac{4R}{3}\Rightarrow h=\frac{R}{3}\approx 2130\text{km}$

Now see why $U=mgh$ would have failed. It gives $h=\frac{v^2}{2g}=\frac{R}{4}=1600$ km — 25% too small, because over 2000 km the field has already weakened appreciably and the near-surface approximation has no business being used.

Note also: half the escape speed does not get you halfway to escaping. Energy goes as $v^2$, so half the speed is a quarter of the kinetic energy.

7. Satellites and orbits

Gravity supplies the centripetal force: $\frac{GMm}{r^2}=\frac{m{v}^2}{r}$

$v_o=\sqrt{\frac{GM}{r}},v_e=\sqrt{2}{v}_o$

Escaping needs exactly $\sqrt{2}$ times the local circular speed. Worth memorising.

From $T=2\pi r/v_o$:

$T^2=\frac{4{\pi }^2{r}^3}{GM}$

That is Kepler's third law — derived, not assumed.

Orbital energy

$K=\frac{GMm}{2r}$ and $U=-\frac{GMm}{r}$, so

$E=-\frac{GMm}{2r}=-K=\frac{U}{2}$

Negative, confirming the orbit is bound. Its magnitude is the binding energy — what must be added to just barely free the satellite.

The counter-intuitive one. Raise the orbit and $E$ increases (less negative) while $v$ decreases. Higher orbits are slower orbits.

Illustration 6

Find the orbital speed and period of a satellite $400\text{km}$ above the surface, and the value of $g$ there. $R=6400$ km, $g_{surface}=9.8$.

$r=6800\text{km}=6.8\times 10^6\text{m}$, and $GM=g{R}^2=9.8(6.4\times 10^6{)}^2=4.01\times 10^{14}$

$v_o=\sqrt{\frac{GM}{r}}=\sqrt{\frac{4.01\times 10^{14}}{6.8\times 10^6}}=7.68\text{km/s}$

$T=\frac{2\pi r}{v_o}=\frac{4.27\times 10^7}{7683}=5561\text{s}=92.7\text{min}$

$g^{\prime }=\frac{GM}{r^2}=\frac{4.01\times 10^{14}}{4.62\times 10^{13}}=8.67{\text{m/s}}^2$, i.e. 88% of the surface value

Check against reality: this is the International Space Station's orbit, and it does circle the Earth about every 93 minutes. The third number is the one worth dwelling on — gravity up there is barely reduced. Astronauts float because they are falling, not because gravity has let go.

Geostationary

Period = one sidereal day (23 h 56 min) $\Rightarrow$ $r\approx 42,400$ km, i.e. $h\approx 36,000$ km.

Must be equatorial and west-to-east. Any inclined orbit traces a figure-eight on the ground, which is why geostationary comms satellites cannot serve the poles.

Weightlessness

An astronaut is not beyond gravity. At the ISS altitude of 400 km, $g$ is still about 89% of its surface value.

They float because they are in free fall along with the station. No floor pushes back, so there is no sensation of weight — a lift with a cut cable, prolonged indefinitely.

8. Kepler's laws

The second law means fastest at perihelion, slowest at aphelion. The Earth is closest to the Sun in early January — seasons come from axial tilt, not distance.

The third law's constant depends only on the central body, which makes it a weighing scale for the cosmos. Observe any satellite's $T$ and $r$ and you have the mass of what it orbits. That is how the masses of the Sun, the planets and distant stars are actually measured.

Illustration 7

A comet is $0.5\text{AU}$ from the Sun at perihelion and $5\text{AU}$ at aphelion. Find the ratio of its speeds at the two points, and its orbital period.

Speeds, from Kepler's second law. Gravity is central, so it exerts no torque about the Sun and angular momentum is conserved. At both perihelion and aphelion the velocity is perpendicular to the radius, so $L=mvr$ directly:

$v_p{r}_p=v_a{r}_a\Rightarrow \frac{v_p}{v_a}=\frac{r_a}{r_p}=\frac{5}{0.5}=10$

Period, from Kepler's third law. The semi-major axis is the average of the two extremes:

$a=\frac{0.5+5}{2}=2.75\text{AU}$

Measuring $T$ in years and $a$ in AU makes the constant equal to 1 for anything orbiting the Sun:

$T=a^{3/2}=2.75^{1.5}=4.56\text{years}$

Read it back: the comet spends the overwhelming majority of those 4.56 years crawling through the outer part of its orbit, and whips past the Sun in a matter of weeks. That is why comets are visible so briefly.

Illustration 8

At what height above the surface is $g$ the same as at depth 2 km? Take $R=6400$ km.

$g\left(1-\frac{2h}{R}\right)=g\left(1-\frac{d}{R}\right)\Rightarrow 2h=d\Rightarrow h=1\text{km}$

The height is half the depth, because the altitude formula carries the factor of 2 and the depth formula does not.

Illustration 9

A planet has twice Earth's radius and the same mean density. Compare its escape velocity with Earth's.

Same density $\Rightarrow M\propto R^3$, so $v_e=\sqrt{\frac{2GM}{R}}\propto \sqrt{\frac{R^3}{R}}=R$.

$v_e$ is twice Earth's, about 22.4 km/s.

Note what happened: it is not $\sqrt{2}$ times. Holding density fixed rather than mass changes the scaling entirely — that substitution is the whole question.

Illustration 10

Find the extra energy needed to move a satellite of mass $m$ from a circular orbit of radius $r$ to one of radius $2r$.

$E=-\frac{GMm}{2r}$, so

$\Delta E=-\frac{GMm}{4r}-\left(-\frac{GMm}{2r}\right)=+\frac{GMm}{4r}$

Positive — energy must be supplied. Meanwhile the speed fell from $\sqrt{GM/r}$ to $\sqrt{GM/2r}$, so kinetic energy halved while potential energy rose by twice as much.

Summary

• One force law, $F=G{m}_1{m}_2/r^2$. Everything else follows.
• The shell theorem is what licenses treating a planet as a point. Inside a shell, the field is exactly zero — and that requires the exponent to be exactly 2.
• $g=GM/R^2$. Up: $g(1-2h/R)$. Down: $g(1-d/R)$. Going up costs twice as fast.
• Rotation and oblateness both make $g$ smaller at the equator.
• Potential is a scalar — add algebraically. Inside a shell, field zero but potential constant and non-zero.
• Fields can cancel between two masses; potentials cannot, since they are all negative.
• Potential energy belongs to pairs: $n$ masses give $n(n-1)/2$ terms, not $n^2$.
• Dropped down a tunnel through the Earth, a stone executes SHM with $T=2\pi \sqrt{R/g}=84.6$ min — the same as a surface-grazing orbit.
• $U=-GMm/r$, zero at infinity. $E<0$ means bound; $E\ge 0$ means escape.
• $v_e=\sqrt{2gR}=\sqrt{2}{v}_o$. Independent of the escaping mass and of the launch direction.
• Orbit: $E=-GMm/2r=-K=U/2$. Raising the orbit raises $E$ and lowers $v$.
• Geostationary: 36,000 km up, equatorial, west-to-east.
• Weightlessness is free fall, not absence of gravity — $g$ is still 89% at the ISS.
• Kepler 2 is angular momentum conservation. Kepler 3's constant depends only on the central body, which is how cosmic masses are measured.