Kepler's Third Law and Motion of Two Point Masses

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SUMMARY

This discussion focuses on the application of Kepler's Third Law to a binary star system consisting of two solar mass stars. The user attempts to calculate the orbital period and decay time using equations from the paper "Gravitational Radiation and the Motion of Two Point Masses" by Peters (1964). Key results include a calculated separation of approximately 10 solar radii, an expected orbital period of 4.5 days, and a lifetime for decay of about 3 x 1012 years. The user struggles with the calculations, particularly in deriving the correct values for T and t using the formulas provided.

PREREQUISITES
  • Understanding of Kepler's Third Law and its application to binary star systems
  • Familiarity with gravitational physics and the constants involved (G, c)
  • Ability to perform unit conversions (solar mass to kg, solar radius to meters)
  • Knowledge of orbital mechanics and the significance of orbital period
NEXT STEPS
  • Review the derivation of Kepler's Third Law for binary systems
  • Learn about gravitational radiation effects on binary star systems
  • Study the implications of mass transfer in binary star evolution
  • Explore numerical methods for solving orbital mechanics problems
USEFUL FOR

Astronomy students, astrophysicists, and anyone studying binary star dynamics and gravitational interactions will benefit from this discussion.

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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. Homework Statement

For a binary star system with each mass = 1 solar mass, the equations give the results:
Separation ~ 10 solar radii
Period ~ 4.5 days
Lifetime for decay ~ 3x1012 years

Homework Equations


T = a4/4B
B = (64/5)G3m1m2(m1+m2)/c5

The Attempt at a Solution


Using solar mass = 1.989 x 1030kg,
solar radius = 695700 km
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?
 
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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.
 
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.
 
To clarify, trying to solve for T and t :)
 
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.
 
Simon Bridge said:
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.
Sorry, I'll try my best to break it down in a way that seems clear!
To start with, Kepler's law problem.

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.
 
"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.
 
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.
 
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!
 

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