Is Kinetic Energy Equal to Negative Potential Energy in Circular Orbital Motion?

[oldspice1212]

Hey, so I have a question about motions of planets and their energy basically.

When we have a circular orbit, why is it that the kinetic energy is just the opposite of potential energy? (Assuming it's a closed orbit)

Like if we have U = something, than the kinetic energy T = -1/2U? This would be saying the kinetic energy doesn't change for a circular orbit but the potential energy does, and than I would think this would be a parabolic orbit as energy would then equal to 0 and epsilon (eccentricity) is equal to 1.

I hope that made sense, I'm having trouble understanding such motion.

 

[Simon Bridge]

When we have a circular orbit, why is it that the kinetic energy is just the opposite of potential energy? (Assuming it's a closed orbit)

Have you followed the derivation?
http://www.pha.jhu.edu/~broholm/l24/node1.html

Like if we have U = something, than the kinetic energy T = -1/2U? This would be saying the kinetic energy doesn't change for a circular orbit but the potential energy does...

No - if U changes, the T will also change. If U does not change, then neither does T.
Note: that should be T=-(1/2)U

 

[oldspice1212]

Interesting, because recently I did a problem, for which the kinetic energy remained the same and the potential energy had changed, so that is where most of the confusion comes from.

 

[Simon Bridge]

It is possible T to remain the same and for U to change - this happens, for eg, when you lift an object at a constant speed - I'm not saying that cannot happen. I am saying that the relation T=-(1/2)U does not indicate that either U or T will change or remain the same. Instead it tells you the relationship between U and T for a circular orbit.
See the link in post #2.