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. IMATH_11 is negative. The satellite is gravitationally bound.
2. IMATH_12 (the virial relation for inverse-square gravity).
3. IMATH_13 . 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:

Worked example with numbers

A $1000$ kg satellite is to be moved from a circular orbit at $r_1 = 7.0 \times 10^6$ m to a higher orbit at $r_2 = 1.0 \times 10^7$ m. Calculate the work required. Use $M_E = 5.97 \times 10^{24}$ kg, $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$.

$\Delta E = \frac{G M m}{2} \left(\frac{1}{r_1} - \frac{1}{r_2}\right)$

$\Delta E = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 1000}{2} \times \left(\frac{1}{7.0 \times 10^6} - \frac{1}{1.0 \times 10^7}\right)$

$\Delta E = 1.99 \times 10^{17} \times (1.429 \times 10^{-7} - 1.000 \times 10^{-7})$

$\Delta E = 1.99 \times 10^{17} \times 4.29 \times 10^{-8} = 8.54 \times 10^9$ J.

About $8.5$ GJ of work is required to raise the satellite to the higher orbit.

Try it: Orbital energy calculator - get $K$, $U$, $E$ at two radii and the energy required to transfer between them.

Common traps

Using $U = m g h$. That is only valid near Earth's surface. For orbital problems, always use $U = -G M m / r$.

Forgetting the negative sign in $E$. Total mechanical energy of a bound orbit is negative by convention (zero at infinity).

Assuming faster means more energy. At a higher orbit, kinetic energy is lower but total energy is higher. Speed alone is not a measure of total energy in gravity wells.

Treating $\Delta K$ and $\Delta U$ as equal in magnitude. For circular-to-circular transfers, $\Delta U = -2 \Delta K$, so $\Delta E = \Delta K + \Delta U = -\Delta K$.

Forgetting that $E = 0$ corresponds to escape. Any positive total energy means the satellite is no longer bound.

In one sentence

The total mechanical energy of a satellite in a circular orbit is $E = -G M m / (2 r)$, with $K$ and $U$ in the fixed ratio $-2K = U$, so raising the orbit increases $E$ (rocket does positive work), increases $U$, and decreases $K$.