For a satellite in a circular orbit at radius r

Gravitational Fields - OCR A-Level Physics

For a satellite in a circular orbit at radius r:
Kinetic energyThe capacity to do work. Measured in joules (J).The energyThe capacity to do work. Measured in joules (J). an object possesses due to its motion.: KE = $\frac{1}{2}$mv^2 = $\frac{GMm}{2r}$ (always positive).
Potential energyThe capacity to do work. Measured in joules (J).: PE = -$\frac{GMm}{r}$ (always negative).
Total energy: E = KE + PE = -$\frac{GMm}{2r}$ (always negative for a bound orbit).
The total energy is negative, meaning the satellite is bound to the central body.
Notice: |PE| = $2 \times$KE, and E = -KE = PE/2.
To move a satellite to a higher orbit, you must add energy. This increases both KE at the point of boost AND the orbital PE. The total energy becomes less negative (closer to zero).
If total energy reaches zero, the satellite has exactly escape velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.The minimum speed at which an object must be launched from the surface of a body to escape its gravitational field entirely (reach infinity with zero kinetic energyThe energy an object possesses due to its motion.). and follows a parabolic path.

Worked Example

A 500 kg satellite orbits Earth at a height of 400 km. Calculate its total energy. (M_E = $5.97 \times 10^{24}$ kg, R_E = $6.37 \times $$10^{6}$$ m)

Show Solution

$r = R_E + h = 6.37 \times 10^6 + 400 \times 10^3 = 6.77 \times 10^6 m.$

$E = -GMm/(2r).$

$E = -(6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 500) / (2 \times 6.77 \times 10^6).$

$E = -(1.99 \times 10^{17}) / (1.354 \times 10^7).$

$E = -1.47 \times 10^{10} J \approx -14.7 GJ.$

Answer

$$E \approx -14.7$ GJ$

Common Mistake MEDIUM

Wrong: Expecting a satellite in a higher orbit to have more kinetic energy.
Right: A satellite in a higher orbit has LESS kinetic energy (slower speed) but more (less negative) total energy. The energy input goes primarily into increasing GPE.