Kepler's Third Law and Motion of Two Point Masses

In summary, the equation for the decay time of a binary star system with 1.989 x 1030 kg masses is not accurate, and needs to be changed to Newton's version of Kepler III.
  • #1
zxcvbnm
8
0
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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  • #2
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
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
To clarify, trying to solve for T and t :)
 
  • #5
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
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.
 
  • #7
"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
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
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!
 

1. What is Kepler's Third Law?

Kepler's Third Law, also known as the Harmonic Law, states that the square of a planet's orbital period is directly proportional to the cube of its semi-major axis. This means that the farther a planet is from the Sun, the longer its orbital period will be.

2. How is Kepler's Third Law related to the motion of two point masses?

Kepler's Third Law can be applied to any two objects that are orbiting each other, as long as their masses are much smaller than the masses of the objects they are orbiting. This law helps us understand and predict the motion of these two objects, such as the Earth and the Moon.

3. What is the mathematical formula for Kepler's Third Law?

The mathematical formula for Kepler's Third Law is T2 = (4π2/G(M1 + M2))a3, where T is the orbital period, G is the gravitational constant, M1 and M2 are the masses of the two objects, and a is the semi-major axis of their orbit.

4. How did Kepler discover his Third Law?

Johannes Kepler discovered his Third Law after years of studying the motion of planets and their orbits around the Sun. He used the data collected by his mentor, Tycho Brahe, and the laws of planetary motion established by his predecessors, Copernicus and Galileo, to formulate his Third Law.

5. Can Kepler's Third Law be applied to objects other than planets?

Yes, Kepler's Third Law can be applied to any two objects that are orbiting each other in space, as long as their masses are much smaller than the masses of the objects they are orbiting. This can include stars, moons, asteroids, and other celestial bodies.

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