Artificial satellites don't orbit the earth forever. Eventually the Earth's atmosphere, thin as it may be up there, will bring them down.

But did you know the linear speed of a satellite in a near circular orbit will increase because of the air drag?

The satellite will experience an acceleration forward along its path, and the accelerations's magnitude will be the same as if the air drag were turned around and were pushing the satellite along. How can that be?

Hello I don't really understand this question and hope someone can help me solve it.

  • $\begingroup$ Can you provide a reference for what you are claiming? $\endgroup$ – NeutronStar Feb 16 '15 at 18:49
  • $\begingroup$ I just found this question on a site: physics.ohio-state.edu/~furnstah/courses/261questions2.html (It's the last question) $\endgroup$ – mariam.. Feb 16 '15 at 18:51
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    $\begingroup$ The only mechanism I can think of to explain this is that the drag initially slows the satellite causing it to drop to a lower orbit and thus accelerate...? $\endgroup$ – lemon Feb 16 '15 at 19:41

As said by lemon, the reason is that the satellite drops to a lower orbit. This can be seen in a very simple way.


The orbit can be approximated with a circular motion with a very slowly decaying radius. This implies we can write down equations for circular orbital motion:

$$F = \frac{GMm}{r^2} = m\frac{v^2}{r} \Rightarrow$$ $$v = \sqrt{\frac{GM}{r}}$$

For a decaying $r$, our satellite's velocity $v$ increases.


If the orbital velocity is decreased (by drag), the satellite's circular motion is disturbed. It can't maintain anymore it's constant altitude and will start falling. The gravitational field will do work on the satellite as it drops and give it additional radius-directed (actually anti-parallel to radius) velocity. If the satellite moves on, the radius-directed velocity transforms to radius-perpendicular velocity. This will in time again convert to radius-directed velocity (now really parallel, not anti-parallel to radius), however the contributions from many such impacts (as drag is continuous) will cancel each other's created vertical velocity out. In contrary, the contributions to horizontal velocity do not cancel each other (only drag partially cancels it, however as seen in the quantitative part, the effect due to falling is stronger).


It may be easier to consider a change in orbit caused by a pair of separate short thruster burns, to split the process into its various parts.

Consider a satellite in a perfectly circular orbit around the Earth. Its speed is constant, say $v_1$.

To go to a lower orbit, the satellite fires a thruster that opposes its motion. The satellite slows down, to a smaller velocity, ($v_2$) and is no longer in its original circular orbit.

Instead, it is now at the high point of a new elliptical orbit. As it continues in its orbit, it comes closer to the earth and speeds up. If it were left alone for a complete orbit, it would reach its maximum speed , $v_3$, a half orbit later at perigee, then climb back up until it reaches the high point of its new elliptical orbit again, with the same $v_2$.

Instead, the satellite fires its thruster again at perigee, opposing its motion and slowing down again from $v_3$ to $v_4$. With proper planning, this last velocity change leaves the satellite with the correct $v_4$ for a new lower circular orbit; this is a larger velocity than $v_1$!

The point is that the velocity increase during the coast phase is larger that the two combined velocity decreases in the two burn phases.

The same thing happens in reverse. Transferring a satellite from low earth orbit to geosynchronous orbit requires two burns. Both speed the satellite up, and the velocity in geosynch is lower than in LEO.

In the case of air friction, you have infinitely many "burns" slowing the satellite. It makes the math a bit more complex, but the prinicple is the same...


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