Coaching Centre
Energy in gravitational fields
Gravitational potential energy U = −GMm/r and why it is negative, the total energy of an orbiting satellite, energy changes when a satellite changes orbit, and escape velocity.
Launching a satellite to geostationary orbit costs several times as much as launching the same satellite to low Earth orbit. Yet a geostationary satellite moves at less than half the speed. If it ends up slower, why does it take so much more energy to get it there?
The answer lies in gravitational potential energy, and in a strange-looking fact: far from Earth, potential energy is higher, but it is still negative. This page sorts out the signs and then tracks energy through orbits and escape.
Gravitational potential energy
In Year 11 you used $\Delta U = mg\Delta h$. That works near the surface, where $g$ is constant. Over large distances $g$ changes, and the full expression for the gravitational potential energy of two masses whose centres are $r$ apart is
$U = -\dfrac{GMm}{r}$
Two things about the minus sign:
- We choose $U = 0$ when the masses are infinitely far apart, where there is no force between them.
- Gravity is attractive, so bringing the masses together releases energy. Starting from zero at infinity, the potential energy can only go down, so it is negative everywhere at a finite distance.
Think of it as a well: an object near Earth sits deep in it. Climbing out takes work. Moving outward, $U$ gets less negative (it increases), even though its size gets smaller. Moving inward, $U$ decreases and the lost potential energy becomes kinetic energy.
As a check, $mg\Delta h$ is the near-surface version: for a small rise $h$ above radius $R$, $\Delta U = GMm\left(\frac1R - \frac1{R+h}\right) \approx \frac{GMm}{R^2}h = mgh$.
Energy of a satellite in orbit
For a circular orbit, $\frac{mv^2}{r} = \frac{GMm}{r^2}$, so $mv^2 = \frac{GMm}{r}$ and the kinetic energy is
$K = \tfrac12 mv^2 = \dfrac{GMm}{2r}$
The total mechanical energy is
$E = K + U = \dfrac{GMm}{2r} - \dfrac{GMm}{r} = -\dfrac{GMm}{2r}$
So for any circular orbit, $K = -\tfrac12 U$ and $E = \tfrac12 U = -K$. The total energy is negative: the satellite is bound to the planet. A larger orbit has a total energy closer to zero, which means higher. That's the answer to the opening puzzle: moving out to a higher orbit, kinetic energy drops, but potential energy rises twice as much, so the total goes up. The rockets have to supply that difference.
A 1000 kg satellite is moved from a circular orbit at radius $6.78 \times 10^6$ m (408 km up) to geostationary orbit at $4.23 \times 10^7$ m. Use data-sheet values to find the change in its kinetic, potential and total energy.
The reverse happens with orbital decay (5H). Drag removes energy, $E$ becomes more negative, the orbit shrinks, and the satellite speeds up, with $K$ rising by the same amount that $E$ falls.
In an elliptical orbit the total energy is also constant; it just shifts between $K$ and $U$ around the orbit (Kepler's second law). For an ellipse, $E = -\frac{GMm}{2a}$, where $a$ is the semi-major axis.
Escape velocity
To escape completely, an object must reach infinity (where $U = 0$) with at least zero kinetic energy. So its total energy must be at least zero. Launching from the surface with speed $v_{\text{esc}}$, and ignoring air resistance:
$\tfrac12 mv_{\text{esc}}^2 - \dfrac{GMm}{r} = 0 \quad\Rightarrow\quad v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$
Like orbital speed, this doesn't depend on the mass being launched, or on the direction (as long as it doesn't hit the planet). It is $\sqrt2$ times the speed of a circular orbit at the same radius.
Find the escape velocity from the surface of Earth and from the Moon ($M = 7.35 \times 10^{22}$ kg, $R = 1.737 \times 10^6$ m).
The Moon's low escape velocity is why it has no atmosphere: over billions of years, gas molecules moving at their thermal speeds have leaked away. It also made it possible for the Apollo astronauts to launch home with a small rocket. At the other extreme, if a mass is squeezed so small that $\sqrt{2GM/r}$ exceeds the speed of light, nothing can escape: a black hole.
In practice, rockets don't fire once at 11.2 km/s. They burn steadily, climbing out of the atmosphere first. But the energy needed to escape is the same: $\frac{GMm}{R}$, about 63 MJ for every kilogram.
- Choose Low Earth orbit. Check that K = 2.94 × 10¹⁰ J and U = −5.88 × 10¹⁰ J, so $K = -\tfrac12 U$.
- Press Remember this orbit, then choose Geostationary. The change in total energy is 2.47 × 10¹⁰ J. Is that positive or negative, and what supplies it?
- Move the altitude slider up and watch the speed and the escape speed. Is the escape speed always √2 times the orbital speed?
- Remember the GPS orbit, then return to low Earth orbit. Why is the energy change now negative even though the satellite speeds up?
Test yourself
A satellite moves from a low circular orbit to a higher circular orbit. Which row is correct?
Higher orbits are slower; U and E both become less negative.
A satellite has kinetic energy $3.0 \times 10^{10}$ J in a circular orbit. Its total energy is
In a circular orbit, $E = -K$.
Planet X has the same radius as Earth but four times the mass. Its escape velocity is
$v \propto \sqrt M$, so it doubles.
Why is gravitational potential energy negative?
With zero at infinity, an attractive force means energy is released as the masses approach, so U falls below zero.
Label the energy curves for circular orbits.
Order these 1000 kg satellites from lowest (most negative) to highest total energy.
Gravitational potential energy is defined as zero at , so it is at any finite distance. For a circular orbit, the total energy equals half the energy. When atmospheric drag removes energy, the satellite moves to a lower orbit and its speed . To escape, the total energy must be at least .
(a) Derive an expression for the escape velocity from the surface of a planet of mass $M$ and radius $R$. (b) Calculate the escape velocity from Mars ($M = 6.42 \times 10^{23}$ kg, $R = 3.39 \times 10^6$ m). (c) A probe in a circular orbit around Mars fires its engines to move to a lower circular orbit. Explain the changes in its kinetic, potential and total energy.
(a) To just escape, the object reaches infinite distance ($U = 0$) with zero kinetic energy, so its total energy at launch is zero: $\tfrac12 mv^2 - \dfrac{GMm}{R} = 0$, giving $v_{\text{esc}} = \sqrt{\dfrac{2GM}{R}}$.
(b) $v_{\text{esc}} = \sqrt{\dfrac{2(6.67 \times 10^{-11})(6.42 \times 10^{23})}{3.39 \times 10^6}} = 5.03 \times 10^3$ m s⁻¹ (5.0 km/s).
(c) In a lower orbit $r$ is smaller. Potential energy $U = -GMm/r$ becomes more negative (decreases). Kinetic energy $K = GMm/2r$ increases, so the probe ends up faster. The total energy $E = -GMm/2r$ decreases, by half as much as $U$ does. Since the total energy falls, the engines must do negative work on the probe: they fire against its direction of motion as a braking burn.