Does everything in the same Earth orbit have the same velocity, or is the radius of the orbit a function of mass and velocity?
Short answer, yes. The orbit is independent of the mass of the orbiting object (planetary-magnitude masses excepted.)
So for circular orbits, all objects in the same orbit will have the same speed relative to Earth (strictly speaking, “velocity” includes the direction of travel.)
Objects in different circular orbits of the same radius (e.g. equatorial and pole-to-pole) will also have the same speed as each other.
Objects in elliptical orbits have a greater speed closer to Earth and a smaller speed further away, but the speeds are still independent of the mass of the orbiting object.
Not really. The gravitational attraction between two objects is what would determine he necessary speed to maintain an orbit. For sufficiently small masses in orbit around a large one the effect is negligible, but for larger masses (say, the moon) I don’t doubt that mass becomes a relevant factor.
Think about it. If you have a mass in orbit around the earth at a certain speed, and you slowly increase its mass, the force the earth’s drag on it increases. To counter this force, you must increase the speed.
Just for more clarity, the atraction between two masses in Newtonian systems would be [sup]Gm[sub]1[/sub]m[sub]2[/sub][/sup]/[sub]d[sup]2[/sup][/sub].
To maintain an orbit, the orbitting mass must move tangent to the sphere it orbits in the same amount as it falls due to gravitational attraction.
So, if we either increase the mass or decrease the distance then we must increase the speed to maintain orbit.
Heh. You’ve forgotten the other equation!
If m[sub]1[/sub] is the mass of the Earth, and m[sub]2[/sub] is the mass of the orbiting object then for circular motion, force towards the centre of motion is:
F=m[sub]2[/sub]v[sup]2[/sup]/d
where v is the orbital speed and d is the orbital radius. Equating the two:
m[sub]2[/sub]v[sup]2[/sup]/d=Gm[sub]1[/sub]m[sub]2[/sub]/d[sup]2[/sup]
which reduces to:
v[sup]2[/sup]=Gm[sub]1[/sub]/d
So the mass of the orbiting object cancels, and orbital velocity is just a function of the mass of the Earth, which is constant, and the orbital radius d.
Or, without the math: me and my twin are in identical orbit side by side. We each have mass M. We link hands. We now form a single object of mass 2M. The force acting on us has doubled! Do we drop plummeting to the ground? No, because the force required to keep an object of mass 2M moving in the same circle has also doubled.
The situation becomes more complicated for satellites of very large mass. In reality, both objects orbit around their common centre of mass. For artificial satellites, this is so close to the centre of the earth as to make no difference. However, when the mass becomes significant in comparison to Earth’s, this is no longer a valid assumption.
Good show, I hadn’t quite thought about it like that. Well, I’ve said it before and I’ll say it again… you learn something every day on the SDMB! 