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Precession of the perihelion of Mecury

  1. Sep 2, 2003 #1
    My problem involves the precession of the perihelion of Mecury
    F sub g = - (GMm)/r^2 * (1 + a/r) where a << r
    I proved in previous parts d^2r/dt^2 - r*(dtheta/dt)^2 =
    -GM/r^2 * (1 + a/r) [eqn 1] and r* d^2theta/dt^2 + 2*dr/dt*dtheta/dt = 0
    I also used u(theta) = 1/r(t) to turn eqn 1 to d^2u/dtheta^2 +
    u(1-GMa/l^2) = GM/l^2 where l = L/m.
    Where I'm stuck is showing the solution is u(theta) = u(sub 0) *
    (1+ ecos(n(theta - theta (sub 0))) where e and theta (sub 0) are constants of integration and u(sub 0) and n are in terms of a, G, M and l. I've tried many times but can't get it to work out.
    Last edited by a moderator: Feb 5, 2013
  2. jcsd
  3. Sep 3, 2003 #2


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    Your equation is d^2u/dtheta^2 + u(1-GMa/l^2) = GM/l^2 which we can write as d^u/dtheta^2+ Au= B by setting A= 1-GMa/l^2 and B= GM/l^2.

    The general solution to the differential equation d^u/dtheta^2+ Au= 0 is u= C1 cos(sqrt(A)theta)+ C2 sin(sqrt(A)theta). The constant solution u= B/A satisfies d^u/dtheta^2+ Au= B since the second derivative is 0. Adding those,

    u= C1 cos(sqrt(A)theta)+ C2 sin(sqrt(A)theta)+ B/A is the general solution to the differential equation. You should be able to write that in terms of your constants.

    You may need to use the facts that
    cos(theta- theta0)= cos(theta0) cos(theta)+ sin(theta0)sin(theta) and
    sin(theta- theta0)= cos(theta0) sin(theta)- sin(theta0)cos(theta) to put it in the form you want.
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