Apply the concepts of gravitational potential energy and kinetic energy to determine the total energy of a planet or satellite in its orbit, and the energy changes that occur when satellites move between orbits

What this dot point is asking

NESA wants you to combine gravitational potential energy and orbital kinetic energy to find the total mechanical energy of a satellite, derive the relationship $E = -G M m / (2 r)$ for circular orbits, and analyse energy changes when a satellite moves between orbits. This dot point pulls together everything from Module 5 and is a frequent extended-response topic.

The answer

Kinetic energy in a circular orbit

For a satellite of mass $m$ in a circular orbit at radius $r$ around a central body of mass $M$, gravity provides the centripetal force:

So:

Gravitational potential energy

From the radial-field formula:

Total mechanical energy

Three things to notice:

1. $E$ is negative. The satellite is gravitationally bound.
2. $|U| = 2 K$ (the virial relation for inverse-square gravity).
3. $E = -K$. The total energy is the negative of the kinetic energy.

Energy changes between orbits

Moving from a circular orbit at $r_1$ to one at $r_2$ requires a change in total energy:

If $r_2 > r_1$ (higher orbit), $\Delta E > 0$: the rocket must do positive work. This is supplied by the propulsion system (chemical, ion, or otherwise).

The counter-intuitive result

When the satellite moves to a higher orbit:

• Kinetic energy decreases (it moves more slowly).
• Potential energy increases (less negative).
• Total energy increases (less negative).

The increase in $U$ is twice the magnitude of the decrease in $K$. So although the satellite slows down, it has more total energy at the higher orbit, because the larger gain in $U$ outweighs the loss in $K$.

Non-circular orbits

For an elliptical orbit with semi-major axis $a$:

Replacing $r$ with $a$. Speed varies around the orbit (faster at perihelion, slower at aphelion) according to conservation of energy, but the total $E$ is constant.

Escape condition

If $E \geq 0$, the satellite is unbound and will escape to infinity. The boundary $E = 0$ corresponds to escape velocity:

Examples in context

Example 1. Raising a satellite from low Earth orbit to geostationary. A $1500 \text{ kg}$ satellite is moved from low Earth orbit ($r_1 = 7.0 \times 10^6 \text{ m}$) to geostationary ($r_2 = 4.22 \times 10^7 \text{ m}$). Total mechanical energy is $E = -G M_E m / (2 r)$. Initially $E_1 = -6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 1500 / (2 \times 7.0 \times 10^6) = -4.27 \times 10^{10} \text{ J}$. Finally $E_2 = -7.08 \times 10^9 \text{ J}$. The work required is $\Delta E = E_2 - E_1 = +3.56 \times 10^{10} \text{ J}$. Despite the orbit being "higher up", the satellite is slower at GEO ($K$ drops) and $U$ increases by twice that amount, so total $E$ still rises.

Example 2. Hohmann transfer to ANU's hypothetical Mars probe. From Earth orbit ($r_1 = 1.50 \times 10^{11} \text{ m}$) to Mars orbit ($r_2 = 2.28 \times 10^{11} \text{ m}$), the elliptical transfer has semi-major axis $a = (r_1 + r_2)/2 = 1.89 \times 10^{11} \text{ m}$. At Earth, the probe needs the vis-viva speed $v_1 = \sqrt{G M_{\odot} (2/r_1 - 1/a)} = \sqrt{1.327 \times 10^{20} (2/1.5 \times 10^{11} - 1/1.89 \times 10^{11})} = 3.28 \times 10^4 \text{ m/s}$, requiring a $\Delta v = 2.94 \times 10^3 \text{ m/s}$ above Earth's $2.98 \times 10^4 \text{ m/s}$ orbital speed. Energy conservation along the ellipse swaps KE for PE as the probe coasts outward.

Try this

Q1. Write down expressions for the kinetic energy, gravitational potential energy and total mechanical energy of a satellite in a circular orbit of radius $r$. State the relationship between $K$ and $U$. [3 marks]

• Cue. $K = G M m / (2r)$, $U = -G M m / r$, $E = -G M m / (2r)$, and $U = -2K$.

Q2. A $200 \text{ kg}$ satellite at $r = 8.0 \times 10^6 \text{ m}$ is boosted to $r = 1.0 \times 10^7 \text{ m}$. Calculate the change in total mechanical energy. [3 marks]

• Cue. $\Delta E = -G M_E m (1/(2 r_2) - 1/(2 r_1))$, evaluate to $\approx +7.46 \times 10^8 \text{ J}$ (work done on satellite).

Q3. A satellite in elliptical orbit has perigee $r_p = 7000 \text{ km}$ and apogee $r_a = 12000 \text{ km}$. (a) Find the semi-major axis. (b) Use conservation of energy to compare KE at perigee vs apogee. (c) State whether total energy is conserved as the satellite orbits, and justify. [2+2+2 marks]

• Cue. (a) $a = 9500 \text{ km}$. (b) KE larger at perigee (closer, faster); from $E$ constant, $K_p - K_a = U_a - U_p > 0$. (c) Yes - only the conservative gravitational force acts.