Gravitation

For a planet of mass $m$ in a (nearly) circular orbit of radius $r$ around the Sun, mass $M$, gravity alone has to supply the centripetal force:

Cancel $m$ and rearrange for the period:

which is Kepler’s third law, with the proportionality constant pinned down explicitly as $4\pi^2/GM$, depending only on the central mass $M$. That it works out this cleanly, from a force law Newton proposed for entirely separate reasons, is the strongest evidence that the law of gravitation is correct.

A test mass $m$ resting at the Earth’s surface feels a gravitational pull equal to its weight, $mg$, and treating the Earth as a uniform sphere of mass $M$ and radius $R$ (concentrated, by the shell theorem, as if at its centre), Newton’s law gives that same force as $GMm/R^2$. Equating the two:

Notice $g$ drops out of $m$ entirely, which is exactly why all objects, heavy or light, fall at the same rate in a vacuum.

At a height $h$ above the surface, the distance from Earth’s centre becomes $R+h$, so:

For $h \ll R$, the binomial approximation $(1+x)^{-2}\approx 1-2x$ turns this into a much more usable linear form:

so $g$ falls off roughly twice as fast, per metre climbed, as it will per metre descended below the surface, which is the whole point of the next subsection.

Go below the surface instead, to a depth $d$, and a different rule applies. By the shell theorem, a uniform spherical shell of matter exerts zero gravitational force on anything strictly inside it, so only the mass enclosed within radius $R-d$ contributes at all; everything farther out simply cancels out. For a uniform-density Earth, that enclosed mass scales as the cube of the radius, $M' = M\left(\frac{R-d}{R}\right)^3$, which gives:

a straight line in $d$, not a curve, reaching exactly zero at the centre ($d=R$), where every direction is equally “down” and the pulls from all sides cancel completely.

Because gravity is attractive, pulling a mass $m$ away from $M$ takes positive work done against the force. Moving it from infinity, where it is conventional to set the potential energy to zero, in to a distance $r$ therefore requires negative net work (gravity itself is doing positive work, pulling it in), so:

The minus sign is not a convention to memorise, it is a genuine physical fact: any bound system, where the mass cannot escape to infinity on its own, must have negative total energy, with zero as the energy of a mass placed infinitely far away and released from rest.

“Escaping” means reaching infinity with (at worst) zero speed left over, i.e. arriving with exactly zero total mechanical energy. Launched from the surface with speed $v_e$, that condition reads:

Note what $v_e$ does not depend on: the launched mass $m$ cancels out completely, and the direction of the launch does not appear anywhere either (ignoring atmospheric drag and the ground itself getting in the way), only the launch speed matters.

A satellite of mass $m$ coasting in a circular orbit of radius $r$ needs gravity to supply exactly the centripetal force, with nothing else pushing or pulling it:

Larger orbits therefore mean slower satellites, the opposite of what intuition about “falling faster from higher up” might suggest; a satellite further out has farther to fall but a gentler pull doing the falling. The orbital period follows directly, $T = 2\pi r / v_{\text{orbital}}$, which rearranges into the same $T^2\propto r^3$ law derived in §7.2.1.

The kinetic energy follows straight from $v_{\text{orbital}}^2=GM/r$:

and adding the potential energy from §7.4.1, $U=-GMm/r$, gives the total mechanical energy of the orbit:

Compare the three quantities and a pattern worth memorising falls out: $E=-K$, and also $E=U/2$. A bound orbit’s total energy is always negative (confirming it is bound), equals minus its kinetic energy, and equals exactly half of its potential energy, for every circular orbit, regardless of $r$, $m$, or $M$.

A satellite is geostationary when it appears fixed over one spot on the ground, which requires three conditions together: its orbit must lie in the equatorial plane, it must orbit west to east (the same way Earth spins), and its period must equal Earth’s rotation period, about 24 hours. Only one radius satisfies $T^2=(4\pi^2/GM)r^3$ for that particular $T$, so there is exactly one geostationary orbit, not a family of them. These satellites are what television and communication relays depend on, since a ground dish can point at one fixed spot in the sky permanently.

A polar satellite instead flies a low orbit (a few hundred to around a thousand kilometres up) passing near both poles on every revolution. Because this orbit is low, its period is only on the order of 100 minutes, far shorter than a day, so as Earth rotates underneath it, each successive pass sweeps over a new strip of the surface. Over enough orbits it images the entire planet, which is exactly why remote-sensing, weather, and mapping satellites use this kind of orbit rather than a geostationary one.

It is tempting to say astronauts float because “there is no gravity up there,” but at ISS altitude Earth’s gravity is still roughly 90% as strong as at the surface, more than enough to bend the station’s path into a circle. What actually vanishes is the normal force: the astronaut, the station, and everything loose inside it are all in free fall together, accelerating toward Earth at exactly the same rate, so nothing presses against anything else, and a scale under the astronaut’s feet would read zero. It is the identical situation as a lift in free fall in §4.13 of the previous chapter, just falling around the Earth rather than straight down into it.