# Kepler's Third Law and Motion of Two Point Masses

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1. Sep 18, 2016

### zxcvbnm

I'm trying to work through some equations in the paper 'Gravitational Radiation and the Motion of Two Point Masses' (Peters, 1964) but I can't get out the right values

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

For a binary star system with each mass = 1 solar mass, the equations give the results:
Period ~ 4.5 days
Lifetime for decay ~ 3x1012 years

2. Relevant equations
T = a4/4B
B = (64/5)G3m1m2(m1+m2)/c5

3. The attempt at a solution
Using solar mass = 1.989 x 1030kg,
G = 6.67408 x 10-11 m3kg-1s-2
c = 3 x 108 ms-1

I get B = (1/(3 x 108)5)(64/5)(6.67408 x 10-11)3 x 2 x 1.989 x1030 x 103 = 6.229 x 10-42

T = (10 x (695700 x 1000))4/4B = ~1.12 x 1051 s, which isn't at all near 3 x 1012 years :( Help?

2. Sep 18, 2016

### zxcvbnm

Oh, also regarding Kepler's third law. My lecture notes give it as

(G/4pi2)(m1 + m2) t2=a3

where t is orbital period in years, masses are in solar units, and a is in au. This formula also isn't working for me yet?