If an orbiting artificial satellite were to slow down, what would happen?

Asked by: Christopher Alec Maquiling


Artificial satellites slow down all the time, and uniformly, they are no longer able to resist the pull of gravity and fall back to Earth. Satellites like the Space Shuttle do it intentionally and can control their descent. The Mir space station was a controlled crash in the South Pacific, and countless other satellites have been accidentally lost. The most common cause of what is known as orbital decay is the atmosphere. As you can imagine, there is no solid "border" to the atmosphere. Gravity holds most of it near the Earth, but out near the border of space there are very few particles, but certainly enough to slow down a satellite that ventures too close to the Earth. Satellites carry fuel to help maintain their orbits, but when that fuel runs out... sooner or later they will fall back to the ground. Hopefully it can be in a controlled manner. If you have any more questions, feel free to ask us, or hit up NASA.
Answered by: Frank DiBonaventuro, B.S., Physics grad, The Citadel, Air Force officer

It would fall to the planet's surface. Here's why: Imagine a car driving around in a perfect circle. This is a centripetal acceleration described by a = v^2 / r where v is the velocity of the car and r is the distance of its revolution. This equation of acceleration also applies to the motion of the satellite as it moves around the planet. (Here we are not considering the planet's rotational motion.) Now if we want to consider centripetal force, we multiply the centripetal acceleration by the mass of the satellite; this gives: F = m a Next consider that the weight of the artificial satellite is given by the equation: W = m g Now g is the acceleration due to gravity. In the physics classroom its value is usually given at sea level, but in reality g is a variable: closer to the center of mass it is strong, farther away it is weak. For the artificial satellite to orbit the planet without falling in, F and W must be equal. So then: F = W
m a = m g
a = g
v^2 / r = g Now if the satellite slows down, v decreases in value so that: v^2 / r < g This means that the acceleration due to gravity is greater than the centripetal acceleration. Gravity wins the fight and pulls the satellite to the planet's surface.
Answered by: Frederick Herrmann, B.A.