Gravitational Potential of Ring Legendre

1. The problem statement, all variables and given/known data

So I’ve figured out most of this problem but there is a bit that erks me:

For a thin uniform ring of rad "a" and mass M, the observation point is in the x-z plane (phi = 0). r = origin to observation point, r’ = position vector in plane at distance a from origin, and R the vector from r to r’ (r-r’).

Now I’ve the expression for potential but it says that

R = sqrt (r^2 + a^2 – 2ra sin θ cos ø

I can’t really work out where this comes from (the angle part).

Anyway I don’t really need to know that as much, after that I have to find the monopole, dipole, and quadrapole terms. I’ve got the first two but the integral for the quadrapole is out of my reach.

I’ve got the integral

phi from 0 to 2*pi of -G(M/2*pi)/((3/8)*[(a/r)^2 – 2(a/r)*cos theta sin phi]^2)

2. Relevant equations

Thought it might be cosine rule, R = r – r’

3. The attempt at a solution

I have already broken it down to the Legendre form and can get the integral for the mono and di, but this quad I am lost on.

Leave a Reply

Name *
Email *