7.3 Gravitational Potential Energy, Escape Speed and Satellites

When Chandrayaan-3 left Sriharikota in July 2023, it did not fly straight to the Moon. For more than two weeks it looped round the Earth on ever larger ellipses, its engine firing briefly each time it swung closest to the Earth, until it was finally travelling fast enough to coast out to the Moon's orbit. Mission planners did not track every force at every instant to design this. They used energy: how much a spacecraft has, how much it needs to reach a given height, and how much more it needs to break free of the Earth altogether.

The formula $U=mgh$ from Chapter 5 cannot do this job — it assumes $g$ is constant, and Section 7.2 showed that $g$ falls to a quarter of its surface value just one Earth radius up. In this section you will find the potential energy that works at any distance, $U=-\dfrac{GMm}{r}$. With it you can calculate the escape speed from any planet, the speed and period of a satellite at any height, and the total energy of a satellite in orbit — and see why a satellite that loses energy to air drag actually speeds up.

Gravitational potential energy

Gravity is a conservative force: the Earth's pull always points towards its centre and depends only on the distance $r$, so the work it does when a body moves from one place to another depends only on the starting and finishing distances (any sideways part of the path, at constant $r$, is perpendicular to the force). We can therefore define a potential energy, exactly as in Chapter 5 — but now we must allow for the force changing with $r$.

Move a body of mass $m$ directly away from the Earth (mass $M$) from $r_1$ to $r_2$. The force on it is $\dfrac{GMm}{r^2}$ towards the centre, so the work done by gravity is minus the area under the $F$–$r$ graph between $r_1$ and $r_2$:

(The integral of $\dfrac{1}{r^2}$ is $-\dfrac1r$; you will prove this in mathematics.) The potential energy rises by $-W_\text{grav}$, so $U(r_2)-U(r_1)=-\dfrac{GMm}{r_2}+\dfrac{GMm}{r_1}$. The natural reference point is where the force vanishes — infinitely far away.

The minus sign is not a mistake. The force is attractive, so the bodies give out energy as they come together from infinity: their potential energy must end up below its value at infinity, which is zero. $U$ is negative everywhere, and it increases (towards zero) as $r$ increases — just as $mgh$ increases with height.

Escape speed

Throw a ball upwards and it comes back. Throw it faster and it goes higher before returning — and because $g$ weakens with height, each extra metre per second buys more height than the last. Is there a speed at which it never comes back at all?

Only gravity does work, so mechanical energy is conserved. A body that just escapes arrives "at infinity" with $U=0$ and (at the least launch speed) $K=0$, so its total energy is zero:

For the Earth, $v_e=\sqrt{2\times9.8\times6.37\times10^6}$ $=1.12\times10^4\ \text{m/s}$ $=11.2\ \text{km/s}$ — about $40\,000\ \text{km/h}$. A body at a height $h$ needs less: replace $R$ by $R+h$.

Earth satellites: orbital speed and period

Fire a cannonball horizontally from a very high mountain. Faster and faster shots land farther and farther away, as the Earth's surface curves away beneath them. At one particular speed the ball falls round the Earth exactly as fast as the surface curves away: it never lands. It has become a satellite. For a circular orbit of radius $r=R_E+h$, gravity supplies exactly the centripetal force (the same step as Result 7.6):

Energy of an orbiting satellite

From $\dfrac{mv^2}{r}=\dfrac{GMm}{r^2}$, a satellite in a circular orbit has $mv^2=\dfrac{GMm}{r}$, so its kinetic energy is exactly half the size of its potential energy:

This explains a famous puzzle. A satellite in a low orbit brushes the thin upper atmosphere, and air drag removes energy, so $E$ becomes more negative. By Result 7.13 the orbit shrinks — and since $K=-E$, the satellite speeds up. Gravity does more positive work on the inward spiral than drag does negative work. To raise an orbit, a spacecraft must add energy with its engines, after which it ends up moving more slowly.

Exercises

Take $G=6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$, $M_E=5.97\times10^{24}\ \text{kg}$ (so $GM_E=3.98\times10^{14}\ \text{N m}^2/\text{kg}$), $R_E=6.37\times10^6\ \text{m}$ and $g=9.8\ \text{m/s}^2$; ignore air resistance and the Earth's rotation unless told otherwise.