1
$\begingroup$

I understand that as a satellite in low-Earth orbit experiences atmospheric drag, it is slowed down. Therefore it no longer has the velocity to maintain its orbit at that point, as per the orbital velocity formula.

However, what I don't understand is, how does it maintain an orbit as it descends? The orbital velocity increases as the orbital radius decreases, meaning it needs a higher linear orbital velocity, not a lower one, to maintain its orbit. The drag also increases as the satellite descends.

I'd have thought that it simply crashes straight down.

$\endgroup$
1
  • $\begingroup$ "as per the orbital velocity formula." Which formula would that be. I suspect you mean $v = \sqrt{GM/r}$. Note, however, that that expression applies only to circular orbits, and you can't analyze this problem in those terms alone. $\endgroup$ Jan 6 '16 at 18:01
1
$\begingroup$

Consider this an orbit of a satellite.

Angular momentum $\vec{L}=\vec{r} \times \vec{p}$ where $\vec{r}$ and $\vec{p}$ are vectors.

angmom

Angular momentum is a conserved quantity. In a satellite with a decaying orbit due to atmospheric drag , as $r$ becomes smaller by the loss of energy, conservation of angular momentum means that $p$ has to increase and therefor the velocity, since $\vec{p}=m \vec{v}$.

It cannot fall straight down because conservation of angular momentum would be violated.

$\endgroup$
2
  • $\begingroup$ Does the drag force exert a torque on the satellite? $\endgroup$
    – Farcher
    Jan 15 '18 at 13:55
  • $\begingroup$ @farcher a correct calculation would show a small diminution of the velocity due to the energy and momentum (and angular momentm transferred to the molecules of the atmosphere), but it is a correction to the first order effects described above. $\endgroup$
    – anna v
    Jan 15 '18 at 14:43
0
$\begingroup$

"The orbital velocity increases as the orbital radius decreases, meaning it needs a higher linear orbital velocity, not a lower one, to maintain its orbit."

This is true for different points of an elliptical orbit of constant energy. A decaying orbit has a decreasing energy, so it does not necessarily travel faster as it descends. I believe, in fact, that given the proper orbital parameters and drag coefficients and such, a satellite could theoretically fall from orbit to the surface at a constant speed.

At any rate, I see no clear reason to think a satellite would somehow lose all of its orbital velocity at once and fall straight down.

$\endgroup$
0
$\begingroup$

There are several concepts in play here. First of all, you're correct that for stable orbits, lower orbits have both greater orbital speed and a shorter orbital period (higher angular velocity). When a satellite's orbit decays it can be approximated as slowly passing through lower stable orbits.

"The orbital velocity increases as the orbital radius decreases, meaning it needs a higher linear orbital velocity, not a lower one, to maintain its orbit."

It would indeed need a greater velocity to maintain a lower orbit, but by definition of "decay" it is not maintaining that orbit, only passing through it, so to speak.

When the satellite begins to enter the atmosphere air drag slows it more quickly and the speed of decay increases. This is a positive feedback system: the lower the orbit, the greater the drag AND the greater the drag the faster the orbit decays. When the instantaneous velocity vector of a falling body is parallel to any line intercepting that body and the (much) larger body that it's falling towards, then the falling body will not achieve even one more "orbit". This means that it's no longer useful to think of the body as being in orbit; it is now merely "falling."

In principle, a (very) round planet with no atmosphere could have a satellite orbit it only several feet off the ground!

In the case of the earth's gravity/atmosphere a satellite in decaying orbit loses its orbit status long before it gets near the ground (and generally before a person standing on the ground could see it well enough to estimate it's angle of approach.

$\endgroup$
0
$\begingroup$

Gravitationally bound systems (such as satellites in orbit) have the property that when energy is removed from the system, the average speed actually increases. Thinking again of the satellite, but temporarily ignoring atmospheric drag for a bit, think of the satellite in an initially circular orbit, then something happens to it and slows it down. It will now be in a elliptical orbit, and its average speed over the course of the orbit will actually be greater than it had on the initially circular orbit. (The math behind this is the virial theorem.) The apogee speed is lower than what the satellite had before, but the perigee speed is greater, and it ends up that the speed averaged over the orbit is also greater.

Bringing atmospheric drag back into it, at the beginning, when the atmospheric drag is small, this slow sapping of orbital energy brings the satellite into a lower, elliptical orbit with higher average speed than its original orbit. As drag continues to act on the satellite, though, and it is brought lower and lower into the atmosphere, eventually the atmosphere will be thick enough that the velocity gains from falling closer to Earth are lost to the ever-increasing atmospheric drag. The satellite will then burn and crash.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.