# Gravitational Potential Energy (GPE) to Kinetic Energy (KE) transfer in Satellite Orbits

I am stumped by a mundane A Level Physics question (teacher of physics here obviously a bit short of practice!). My colleagues and I are stumped and were wondering if any one could help us.

It concerns the change in KE of a satellite in orbit, if a satellite drops it’s orbit to that of a lower radius, it’s change in GPE is given by the expression $GMm(1/r_1-1/r_2)$. Due to the conservation of energy, I would have thought that the gain in KE would therefore be equal. But I keep seeing in past exam papers that the answer is HALF of this value! However I cannot seem to find a satisfying reason why anywhere.

Hints:

1. There is no conservation of mechanical energy between the initial and final periodic orbits, due to burning of the satellite's rocket engine(s).

2. Use e.g. the virial theorem, which says that the (time-averaged) kinetic energy is minus half the (time-averaged) potential energy in a periodic orbit.

• Thanks for this, again the Virial theorem is something I remembered vaguely from my year 2 structures in the universe course at uni and again it has helped! Thanks a lot! Commented Oct 9, 2013 at 20:46

The equation you present is correct for a satellite in an elliptical orbit, while the answer from past exams is certainly comparing the KE of circular orbits with different radius:

In the case of an elliptical orbit, the radius is varying at each instant (as the central body is in the focus of the ellipse and the trajectory is an ellipse) but the total mechanical energy is constant and given by

$E = KE + GPE = \frac{1}{2}mv^2-\frac{GMm}{r} = constant$.

Note that the potential energy is negative but still increases with altitude. So, comparing KE in the same orbit at different instants, when the satellite is at different positions (and So different distances $r_1$ and $r_2$ from the focus) just results in the expression you presented; at initial distance $r_1$ and final distance $r_2$, conservation of energy results in:

$E=constant = \frac{1}{2}mv_1^2-\frac{GMm}{r_1}=KE_1 + GPE_1 = KE_2+GPE_2=\frac{1}{2}mv_2^2-\frac{GMm}{r_2}$

and so

$\Delta KE=KE_2-KE_1=-\Delta GPE=GPE_1-GPE_2=GMm(\frac{1}{r_1}-\frac{1}{r_2})$.

The situation is different when comparing different circular orbits. For each circular orbit KE and GPE do not vary. In the case of circular orbits, the inertial force is just centripetal, and equating it to the gravitational force results that the speed is simply given by

$v=\sqrt{\frac{GM}{r}}$ (circular orbits only).

So, for two orbits with different radius $r_1$ and $r_2$ (and different total energies)we have:

$KE_1=\frac{1}{2}mv_1^2=\frac{1}{2}m(\sqrt{\frac{GM}{r_1}})^2$ and $KE_2=\frac{1}{2}mv_2^2=\frac{1}{2}m(\sqrt{\frac{GM}{r_2}})^2$

and finally

$\Delta KE=KE_2-KE_1=\frac{1}{2}GMm(\frac{1}{r_2}-\frac{1}{r_1})$

which is the "half" answer you mentioned. Note also the different sign of the answer compared to the case of a single elliptic orbit.

• This is brilliant, thank you very much! Your explanation really made sense and I can clearly see where the factor of a half comes from now. The ellipse/circular orbit discrepancy was one I didn't think I would see students encounter in A level physics and I would never of come to this conclusion independently. Once again, thank you. Commented Oct 9, 2013 at 20:41
• This suggests that the work done to get you from orbit 1 to orbit 2 and change speed from $v_1$ to $v_2$ is half the change in potential energy of the two orbits. How can I explain this? If I calculate the integral of force x displacement I still get $GMm (r_2^{-1}-r_1^{-1})$.. Commented Oct 14, 2018 at 18:10

You are correct that the difference in potential gravitational energy is converted into an increase of kinetic energy. However you calculated the difference in gravitational incorrectly, since you should subtract them from each other and not their radii. It might also help if you looked at specific orbital energy. Also keep in mind that the specific relative angular momentum is conserved.

• Specific orbital energy is something I hadn't thought about and again it helps, thanks very much! Commented Oct 9, 2013 at 20:44

For a satellite in a nonperturbed orbit, the change in gravitational potential energy does equal the change in kinetic energy.

Increasing the size of a circular orbit requires energy. Half of the energy is used to lift the satellite, while the other half is used to speed up the satellite.

For a satellite in a circular orbit perturbed by atmospheric drag, friction heats the satellite and the atmosphere, and decreases the orbital radius. Reducing the radius of an orbit releases gravitational potential energy. Half of the released energy goes to speeding up the satellite, the other half is lost as heat.