Gravitation & Orbits

The Moon is falling and never lands

A banked curve works because the track can tilt and push sideways. The previous lesson was about finding which force, in a real contraption, happens to point at the centre of the circle. Gravity is the case where you do not have to hunt: the force points at the centre because the centre is where the other mass sits.

Here is the question that took Newton twenty years. Drop a stone and it hits the ground in under a second. The Moon is also falling toward Earth (it has to be, or it would fly off in a straight line) and it has been falling for four billion years without arriving. Why does one land and the other not? The answer is not that the Moon is different. It is that the Moon is going sideways fast enough that the ground curves away underneath it exactly as fast as it drops. Get that idea and the rest of this lesson is arithmetic.

The inverse square law

Every pair of masses attracts. Newton's guess, checked against the Moon's motion, was that the pull weakens as the square of the distance:

$F = \frac{Gm_1m_2}{r^2}$

$m_1$ and $m_2$ are the two masses in kilograms, $r$ is the distance between their centres in metres, and $G = 6.67\times10^{-11}\ \mathrm{N\,m^2/kg^2}$ is a constant of nature. The force on each mass has the same size and points along the line joining them, toward the other mass. Double the distance and the force drops to a quarter. Triple it and the force drops to a ninth.

Why centres and not surfaces? Because of a result Newton needed calculus to prove: a uniform spherical shell pulls on anything outside it exactly as though all of its mass were squeezed into a point at its centre. Stack shells and you get a solid sphere, so a planet or a star can be replaced by a point at its centre for any calculation done outside it. The same proof gives a second fact that looks strange the first time: inside a hollow shell, the pull is exactly zero everywhere, not just at the middle. You will use that one at the end.

Start with the smallest case. Two 100 kg people stand 2.0 m apart, centre to centre. Then

That is about the weight of a grain of dust. Gravity is by far the weakest force in physics, and it only runs the universe because it never cancels: there is no negative mass to shield it, so pile up enough and it wins.

Field: dividing the mass out

Carrying the test mass around is a nuisance. Divide the force by the mass being pulled and you get the pull per kilogram, which is a property of the source alone. That is the gravitational field $g$, measured in $\mathrm{N/kg}$, which is the same thing as $\mathrm{m/s^2}$:

$g = \frac{F}{m} = \frac{GM}{r^2}$

At the surface of a planet, $r$ is the planet's radius $R$, and this is where the familiar 9.81 comes from. For Earth, $M = 5.97\times10^{24}$ kg and $R = 6.37\times10^6$ m:

The whole "objects fall at the same rate" fact sits in that division: the mass being pulled cancelled out. A feather and an anvil get the same $g$ because both the pull and the resistance to being pulled scale with the same $m$.

Because $g$ falls off as $1/r^2$, you rarely need $G$ and $M$ separately. If you already know the surface value $g_0$, then at distance $r$ from the centre

$g(r) = g_0\left(\frac{R}{r}\right)^2$

At two Earth radii, $g = 9.81/4 = 2.45\ \mathrm{m/s^2}$. Not zero, not close to zero. Astronauts at 400 km are at $r = 6.77\times10^6$ m, where $g$ is still about 89 per cent of its surface value.

For a small change in height, do not subtract two nearly equal big numbers: you will lose all your significant figures. Expand instead. With $h \ll R$,

The $(1+x)^{-2}\approx 1-2x$ step is the binomial approximation, good whenever $x$ is small. For $h = 10$ km: $\Delta g = 2(9.81)(1.0\times10^4)/(6.37\times10^6) = 0.0308\ \mathrm{m/s^2}$. Climbing a 10 km mountain costs you three parts in a thousand of your weight.

Fields from several sources add as vectors, one per source, each pointing at its own source. Two cases come up constantly. On the line between two masses there is a neutral point where the two fields cancel. Set the magnitudes equal: with the masses a distance $d$ apart and the point a distance $x$ from $M_1$,

For Earth and Moon, $\sqrt{M_2/M_1} = \sqrt{7.35\times10^{22}/5.97\times10^{24}} = \sqrt{0.01231} = 0.1110$, so $x = 3.84\times10^8/1.1110 = 3.456\times10^8$ m from Earth's centre: about 90 per cent of the way to the Moon.

Off the line, you must break the fields into components. Two equal masses, so the fields have equal size, and the point sits on the perpendicular bisector, so the components along the line cancel and the ones toward the line add.

Circular orbits

Now put the Moon back in. A satellite of mass $m$ going in a circle of radius $r$ needs a centripetal force $mv^2/r$ pointing at the centre, and gravity is the only force available. So set them equal:

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

Cancel one $m$ and one $r$:

$v = \sqrt{\frac{GM}{r}}$

The satellite's own mass vanished. A loose bolt and a 400-tonne station in the same orbit move identically, which is why they stay side by side. And notice the sign of the dependence: higher orbits are slower. That is the opposite of what most people guess.

The period comes from the fact that one lap is $2\pi r$ at speed $v$:

Square it because the square root is awkward; what you get is clean. $T^2 \propto r^3$ for every satellite of the same central body.

Worked example: the geostationary radius. A satellite must have $T = 86164$ s so that it hangs over one spot on the equator. Solve the boxed result for $r$:

$r = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3}$

Numerator: $GM = 3.982\times10^{14}$ and $T^2 = (8.6164\times10^4)^2 = 7.424\times10^9$, so $GMT^2 = 2.956\times10^{24}$. Divide by $4\pi^2 = 39.48$ to get $7.488\times10^{22}$. Now the cube root. Write it as $74.88\times10^{21}$ so the exponent divides by three cleanly: $\sqrt[3]{74.88} = 4.215$ and $\sqrt[3]{10^{21}} = 10^7$. So

$r = 4.215\times10^7\ \mathrm{m}$

That is 6.6 Earth radii, about 36 000 km above the surface. Rewriting a power of ten so its exponent is divisible by three is the trick that makes cube roots doable without a calculator, and it is worth practising.

Energy, escape, and why drag speeds you up

Force gets you circles. Energy gets you everything else: anything where the distance changes. Since $F$ falls off as $1/r^2$, the work done pulling two masses apart gives a potential energy that falls off as $1/r$, and the standard choice is to call $U$ zero when the bodies are infinitely far apart:

$U = -\frac{GMm}{r}$

The minus sign is not decoration. It says a bound pair sits in a hole, and you must supply energy to climb out. Bringing a mass in from infinity releases energy; separating them costs it.

Escape speed. To just barely get away, a body must reach $r = \infty$ with nothing left over, so its total energy must be zero:

Escape speed is $\sqrt2$ times the circular speed at the same radius: always, for any body. At Earth's surface, $v_{\text{esc}} = \sqrt{2(3.982\times10^{14})/6.37\times10^6} = \sqrt{1.250\times10^8} = 1.118\times10^4$ m/s. Energy is a scalar, so the direction of launch does not matter, only the speed.

Worked example: falling from two Earth radii. An object is released from rest at $r_i = 2R = 1.274\times10^7$ m and falls to the surface, $r_f = 6.37\times10^6$ m. Energy is conserved, and the starting kinetic energy is zero:

$0 - \frac{GMm}{r_i} = \tfrac12 mv^2 - \frac{GMm}{r_f}$

Every term carries $m$, so divide it out: the answer cannot depend on what you dropped:

So $v = \sqrt{GM/R}$, which is exactly the circular orbital speed at the surface. Numerically $v = \sqrt{3.982\times10^{14}/6.37\times10^6} = \sqrt{6.251\times10^7} = 7.91\times10^3$ m/s. Notice how much cheaper that was than integrating the varying force over the fall.

The total energy of an orbit. In a circular orbit, $v^2 = GM/r$, so $K = \tfrac12 m v^2 = GMm/2r$. Add the potential:

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

Three things worth keeping. $E$ is negative, and negative total energy is what "bound" means. $E = -K$, so the kinetic energy alone tells you the binding. And $E$ depends only on $r$.

That last point has a consequence people find perverse. Air drag removes energy, making $E$ more negative, which means $r$ gets smaller. But $K = +GMm/2r$ rises as $r$ shrinks. Friction makes a satellite go faster. The energy lost to heat comes out of the potential term, which falls twice as fast as the kinetic term rises.

To get the increase in speed as an orbit decays from 300 km to 200 km altitude, just take the difference of $\sqrt{GM/r}$ at the two radii: $r_1 = 6.67\times10^6$ gives 7726 m/s, $r_2 = 6.57\times10^6$ gives 7785 m/s, a gain of about 59 m/s.

Two situations that look different and are not

Binary stars. Two stars of equal mass $M$ separated by $d$ orbit their common centre of mass, which for equal masses is the midpoint. So each star moves in a circle of radius $d/2$, not $d$, but the force between them still uses the full separation $d$. That is the only trap here. For one star,

With $T = 2\pi(d/2)/v = \pi d/v$ and $M = 2.0\times10^{30}$ kg, $d = 1.0\times10^{12}$ m: $v^2 = (6.67\times10^{-11})(2.0\times10^{30})/(2\times10^{12}) = 6.67\times10^7$, so $v = 8.17\times10^3$ m/s and $T = \pi(1.0\times10^{12})/8.17\times10^3 = 3.85\times10^8$ s, roughly 12 years.

A tunnel through the Earth. Drop something down a hole bored clean through the centre. Here the shell theorem earns its keep: at radius $r$ from the centre, every shell outside you pulls with zero net force, and the sphere inside you pulls as a point mass. A uniform Earth has inside-mass $M(r) = M(r/R)^3$, so

$g(r) = \frac{GM(r)}{r^2} = \frac{GM r}{R^3} = \frac{g_0}{R}\,r$

The pull grows straight out of the centre, always pointing back to it. A force proportional to displacement and directed back toward the origin is simple harmonic motion with $\omega^2 = g_0/R$. So

About 84 minutes for a round trip: the same as the period of a satellite skimming the surface, which is not a coincidence.

Recognising the technique

Three questions sort almost every gravitation problem.

Is the distance changing? If the body moves in a circle at fixed $r$, use force: $GMm/r^2 = mv^2/r$, and everything about speed, period and radius follows from it. If $r$ changes (falling, launching, escaping, hopping between orbits) use energy: $\tfrac12 mv^2 - GMm/r$ is conserved.

How many sources? One source means plug into $GM/r^2$. Two or more means add fields as vectors: on the line, set magnitudes equal for a null point; off the line, resolve into components first.

Are you inside or outside? Outside a sphere, treat it as a point at the centre. Inside, only the mass closer to the centre than you counts, which turns the inverse square into a straight line in $r$.

Two small habits save a lot of time. Group $GM$ into a single number for each body ($3.98\times10^{14}$ for Earth) because it appears in every formula and is often known better than $M$ alone. And if the problem gives you surface $g$ rather than $M$, use $GM = g_0R^2$ and skip $G$ entirely.

What you have so far only handles circles. Real orbits are ellipses, comets swing in and back out, and the same energy bookkeeping tells you how. That is Kepler's laws, where $T^2 \propto r^3$ turns out to be the special case of something more general.