From the first sesquipage of these revision notes:
http://cp4space.files.wordpress.com/2013/04/kepler.pdf you can derive the orbit equation, namely: r = (L^2/GMm^2) / (1 + epsilon cos(theta)) Now consider the perihelion and aphelion to get simultaneous equations for epsilon. At these two points, the velocity is entirely tangential so is given by: v = L/mr and therefore the kinetic energy is: T = L^2/2mr^2 and therefore: E = L^2/2mr^2 - GMm/r Those are sufficiently many equations to derive your result by routine algebraic manipulation. Sincerely, Adam P. Goucher http://cp4space.wordpress.com
----- Original Message ----- From: Mike Stay Sent: 11/30/13 11:45 PM To: math-fun Subject: [math-fun] Eccentricity of hyperbolic path of deflected light
I'm trying to follow the derivation in http://arxiv.org/pdf/physics/0508030v4.pdf, but I have no idea where equation 5 comes from: ------ For a light particle with mass m, total energy E and angular momentum L (with respect to the Sun's center) and where the Sun has mass M, the eccentricity epsilon is
epsilon = sqrt(1 + (2 E L^2)/(G^2 m^3 M^2)) (5) ------
I could probably get some of that by dimensional analysis, but certainly not all of it. How would one derive this? -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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