# 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.

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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.

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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! –  user30768 Oct 9 '13 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.

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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. –  user30768 Oct 9 '13 at 20:41