When a satellite in a circular orbit around the earth enters the atmospheric region it encounters small air resistance to its motion. Then it's said its kinetic energy decreases. But I can't understand how they are comparing as the total energy decreases as well as potential energy decreases as radius decreases. Whether the answer is right or I am mistaking?


2 Answers 2


The potential energy doesn't increase as the radius decreases - instead it decreases. Two ways of seeing this.

The first way is to use $PE = mgh$. The higher up something is, the more potential energy it has.

The other way is to use $PE = -GMm/r$. As $r$ decreases, the magnitude of the potential energy increases; however because of the negative sign, the potential energy still decreases.

Since KE decreases and PE decreases, the total energy must decrease.

NB: the KE doesn't usually decrease; in fact it should increase as the PE decreases. However the total energy of the spacecraft decreases since the PE decreases by more than the KE increases (see virial theorem).

  • $\begingroup$ The kinetic energy of a satellite with an orbit decaying due to drag doesn't decrease until the satellite is on the verge of plunging deep into the atmosphere. $\endgroup$ Mar 14, 2018 at 14:26
  • $\begingroup$ @DavidHammen it doesn't, yeah, but if the question says it decreases ... will edit this into the answer anyway. $\endgroup$
    – Allure
    Mar 15, 2018 at 9:22
  • $\begingroup$ The question also asks "Whether the answer is right or I am mistaking?" The questioner is mistaken. $\endgroup$ Mar 15, 2018 at 9:56
  • $\begingroup$ Sorry actually I should have said that the answer given in the book is kinetic energy increases, which is wrong. Thanks by the way. $\endgroup$ Mar 18, 2018 at 19:34

Then it's said its kinetic energy decreases.

This happens only when the satellite's orbit has decayed to the extent that the satellite less than an orbit to live. What happens in general is that the satellite's total mechanical energy $\frac E m = \frac 1 2 v^2 - \frac \mu r$ decreases as the satellite encounters drag.

The question asks about a satellite in a circular orbit. It might help in understanding to see what happens to a satellite in an elliptical orbit. Suppose some satellite's perigee is hundreds of kilometers above the surface of the Earth and its apogee is thousands of kilometers above the surface of the Earth. This makes the satellite subject to drag only when it is near it's perigee point, and the drag force is a small perturbing force because perigee is hundreds of kilometers above the surface. The effect of the small drag force isn't immediately apparent. It becomes apparent half an orbit later: The satellite's apogee distance decreases.

Eventually drag circularizes the orbit, with the satellite orbiting circularly at a bit less than the initial perigee distance. Drag now acts continuously, but it remains a small perturbing force. The orbital velocity is a tiny bit less than it would be if drag wasn't present. The primary effect is to instead decrease the satellite's orbital radius.

Drag becomes significant (i.e., it has a visible effect on velocity) once the satellite's orbit is only a hundred kilometers or so above the surface. It's only at this point that the velocity decreases. This marks the satellite's last orbit.


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