If satellite orbits the earth at low altitude, and residual atmosphere exerts tiny air drag on it?

Question

will orbital speed of the satellite

a) decrease
b) remain the same
c) decrease
d) all of the above

Answer 1

Like all your questions, Alex...it depends.

As the orbit decays and begins to enter into the atmosphere, drag forces d = 1/2 rho Cd A v^2 begin to come into play to slow that critter down. Rho, the air density is increasing with loss of altitude as the satellite slows. v is the tangential velocity, which is the direction of the satellite through the air; so v is the relative velocity of the air over the satellite.

The other forces acting on the satellite are the gravity (centripetal force) and the faux force, centrifugal, which is the equal, but opposite counterfore, but if and only if the orbit is not decaying. In which case, mg = mv^2/R; the weight (gravitational force) is offset by centrifugal force.

As v slows down, from the drag, g > v^2/R and f > mg - v^2/R until such time that R --> r where r < R, the original orbit altitude. At which time v --> V where V > v and f = mg - V^2/r = 0 and the satellite orbit no longer decays. But now, because Rho > rho and V > v, we have D = 1/2 Rho Cd A V^2 > 1/2 rho Cd A v^2 = d; where the drag D at r < R is greater than that d at R.

And the cycle begins once again, slowing and decaying with increasing drag; more slowing and decaying with more increasing drag. In the end, the slowing of the tangential velocity reaches a value where there is insufficient velocity to offset the weight of the satellite (i.e., the orbit takes it into the Earth rather than around it) and the satellite comes to a fiery end as the drag friction heats it up.

So there you have it, it slows, the orbit decays, and in doing that speeds up again. Understand, all this slowing, decaying, and speeding up is done in small differential increments. So the real solution to this question can be described through differential equations and their solution. (By the way, you've got two "decreases" in your selection.)

Oh, the answer is all of the above (assuming one of those decreases is an increases. That results because tangential velocity both increases and decreases; so it must pass through a v = constant phase however slight. That follows from the fact one cannot go from a deceleration to an acceleration without going through a = 0, which means dv/dt = 0 and v = constant or remains the same for that dt.

Answer 2

The satellite will slow down proportionally to the amount of air drag being put on it. In the case of high altitiude satellites the only force acting on the would be gravity and therefore they won't slow down. They do, however get afftected by bodies other than the earth so they still must have small thrusters for corrective action. If the drag on the satellite in your problem is small enough then the thrusters could possibly overcome it.

Answer 3

the drag acts against the motion of the satellite. So the satellite slows down.

Answer 4

increase.

As the drag pulls it into a lower orbit, it speeds up.