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

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1. Sep 18, 2016

### zxcvbnm

I posted this before but I think it was in the wrong, place, so sorry for the duplicate :O

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?

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?

2. Sep 18, 2016

### Simon Bridge

What are you trying to achieve? Do you have a problem statement for us?
Note - it can help to use the homework template to lay out your questions.

3. Sep 18, 2016

### zxcvbnm

I'm trying to use the separation and masses to calculate the decay time and orbital period that's given. I want to check that I'm using this equation correctly by doing it on this worked example before I use it for an assignment question, but I still can't get out the numbers that are given.

4. Sep 18, 2016

### zxcvbnm

To clarify, trying to solve for T and t :)

5. Sep 18, 2016

### Simon Bridge

OK - what do those letters stand for and for what?
In your problem statement you only have the results of equations ... you have not described the setup.
So far I see you have a binary star system or 1 solar mass stars ... and there is some separation given for some situation not given.

Please try to be more descriptive, and start from the physics of the situation.

6. Sep 18, 2016

### zxcvbnm

Sorry, I'll try my best to break it down in a way that seems clear!

I have two masses, of 1 solar mass each, orbiting at a distance of a = 10 solar radii (~10/215 AU, I believe).
From Kepler's third law, I get t = ((10/215)3 x 4 pi2/2G)1/2 = 5,455 years. The orbital period given in the paper is 4.5 days, though.

7. Sep 18, 2016

### zxcvbnm

"Consider the case of circularly orbiting binary stars, for which we neglect deformation, mass flow, and other radiation processes." so Kepler's third should apply here because it's a circular orbit, by my understanding.

8. Sep 18, 2016

### Staff: Mentor

Hmm. Newton's version of Kepler III for two masses should give:

$T = \frac{2 \pi}{\sqrt{\mu}} a^{3/2}~~~~~~~~~~$ where: $\mu = G(m_1 + m_2)$ and $a$ is the separation.

Plugging in the values in standard units (kg, m, etc.) for two solar mass stars separated by 10 solar radii I get T = 2.239 x 105 seconds, or about 2.59 days.

9. Sep 18, 2016

### zxcvbnm

Oh gosh, I have no idea what's gone on with my working then. I'll try redoing it while converting everything to standard units and see if that gets me to your answer. Thanks!