Kepler's 3d Law: Stable?
Most recent answer: 05/01/2014
- david (age 21)
kristiansund. norway
The basic connection between orbital size and speed is easy to get for circular orbits from Newton's gravity. Since the inward acceleration due to gravity goes as 1/r2 and the inward acceleration of a circular orbit of speed v and radius r is v2/r, we get that v2 is proportional to 1/r so v is proportional to 1/r1/2.
You raise a great question about stability. What happens if something gives the satellite a little push in the direction it's going? You have more velocity at that moment, so it sounds like maybe you should end up in a smaller orbit, if our v-r relationship holds. It's more complicated than that because now the satellite will go in an elliptical orbit, with both the speed and distance changing over the course of the orbit. So let's explore what happens.
It's helpful to talk about the kinetic energy, mv2/2 and the gravitational potential energy -GmM/r, where m is the satellite mass and M is the mass of the planet. The total energy mv2/2 - GmM/r doesn't change as v and r change in the elliptical orbit. You can derive a virial relationship that the time-average over the orbit has
Ave(mv2/2)=-Ave(GmM/2r). The gravitational potential energy is on average twice as big as the kinetic energy, but negative. So the total energy is negative, and just minus the average kinetic energy. When you speed up the satellite you make the total energy a little less negative. That means that even though you increased the kinetic energy the new orbit has less kinetic energy on average than the old one. The satellite spends a lot of time farther away, moving more slowly.
So the orbits aren't unstable. When you give the satellite a little push, one way or another, it shifts to a new elliptical orbit. Only a push so big as to make the total energy positive would make the satellite escape from orbitting the planet.
Mike W.
(published on 05/01/2014)