AlphaNumeric said:
Assume a circular orbit for simplicity, then work out the kinetic and potential energy of an object of mass m orbiting a distance r from a static object of mass M (M>>m for simplicity). Then consider it in a circular orbit at a distance R with R<r, and then work out the difference. In the absense of any other effects, the loss of energy will be due to gravitational radiation, to a decent approximation I'd imagine.
[itex]v_{circular,r}=\sqrt{GM/r}[/itex]
[itex]v_{circular,R}=\sqrt{GM/R}[/itex]
[itex]KE_r=.5mv_{circular,r}^2=.5mGM/r[/itex]
[itex]KE_R=.5mv_{circular,R}^2=.5mGM/R[/itex]
[itex]KE_R-KE_r=.5mGM\left(1/R-1/r)[/itex]
That was simple enough.
According to what you say, wouldn't gravitational radiation also be emitted equal to half the change in gravitational potential such that the there is:
[itex]\Delta KE=.5mGM\left(1/R-1/r)[/itex]
[itex]\Delta gravitational\ radiation=.5mGM\left(1/R-1/r)[/itex]
But what about heat?
For non-circular orbits:
I know (from other sources) that for a collapsing gravitating object, half of the change in the gravitational binding energy will lead to kinetic energy inside the object and the other half will leave as electromagnetic radiation (per the virial theorem). How do I distinguish and understand the value of binding energy in comparison to the value of gravitational potential energy? Is there a clear path from [itex]GM^2/r[/itex] to [itex]GMm/r[/itex] ? Now this a bit tricky (I think). For pairs of objects originally at infinite distance from each other, the change in kinetic energy is [itex].5GMm/r_{in\ between}[/itex], but for forming stars, the gravitational contribution to kinetic energy (I guess) is [itex].5GM^2/r_{star}[/itex]... I know it's a lot of questions. But I hope you can answer some of them at least.