How does one derive the total energy in an elliptic orbit: [tex]E= - \frac{GMm}{2a}[/tex] where a is the semi-major axis? I did manage to get the result for the special case of circular orbit, as [tex]v = \sqrt{\frac{GM}{R}}[/tex] But the problem is that I can't figure out a way to express v in an elliptic orbit. If at all possible, give hints (that is: not a direct answer), as I'd rather try it myself first :).
Not really, no. I can't get the given equation: [tex]v=\sqrt{2\mu\left({1\over{r}}-{1\over{2a}}\right)}[/tex] without assuming the result (total energy) I'm trying to get. EDIT: Either I am getting paranoid or someone replied, but deleted his/her message :).
Isn't that the equation I wrote in my second message? I can get it by assuming [tex]E= - \frac{GMm}{2a}[/tex] but that's the equation I want to prove.
Sorry for the confusion. That was me, then I thouight that maybe that was too much of hint - that you didn't really want !
The speed can be resolved into two orthogonal components - radial and tangential. Since [itex]v^2 = v_t^2 + v_r^2[/itex] (Pythagoras) the total kinetic energy is simply the sum of the tangential and radial kinetic energies. Use that, together with the fact that angular momentum and total energy is constant, to derive the relationship between radius, speed and total energy. [tex]E(r) = KE + PE = \frac{1}{2}mv^2 - \frac{GMm}{r} = \frac{1}{2}mv_t^2 + \frac{1}{2}mv_r^2 - \frac{GMm}{r}[/tex] Use the fact that the radial KE (middle term) is 0 when r is maximum or minimum (ie. when r = a or r=b) AM