Double star mutual orbit calculations

  • Thread starter charlie05
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  • #1
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Homework Statement


A tight pair of stars orbiting with circular paths at a distance a, the weight of the stars is the same: M1=M2=M
What period T and frequency f is the pair circulating? Explain the results generally with G, M, a.

Homework Equations


Kepler's law M1+M2=a^3/T^2

The Attempt at a Solution


T^2=a^3/2M
f=1/T
 

Answers and Replies

  • #2
BvU
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Hi,

Is there a question in all this ?

Explain the results generally with G, M, a.
Don't see a G in your answer.
 
  • #3
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yes, that's just the problem....

maybe: T^2=4πa^3/2GM......?

but is there ,, a ,, distance of the stars? rather it should be semiaxis...but when the paths are circular, it is the distance of the center of the stars....
 
  • #4
stockzahn
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yes, that's just the problem....

maybe: T^2=4πa^3/2GM......?

but is there ,, a ,, distance of the stars? rather it should be semiaxis...but when the paths are circular, it is the distance of the center of the stars....
I'm not familiar with this kind of calculations, but I'd try to find the force equilibrium between gravitation and centrifugal force ...
 
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  • #5
BvU
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  • #6
haruspex
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yes, that's just the problem....

maybe: T^2=4πa^3/2GM......?

but is there ,, a ,, distance of the stars? rather it should be semiaxis...but when the paths are circular, it is the distance of the center of the stars....
Be careful. Is that equation for one small body orbiting a much larger one? If so, there may be an assumption that the radius of orbit equals the distance between the mass centres. Does that apply here?
 
  • #7
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here are the two stars of the same weight... so distance in the formula must be 1/2a?
 
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  • #8
ehild
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here are the two stars of the same weight... so distance in the formula must be 1/2a?
No, you need to use the distance between the centers as the radius of the orbit. The real radius is a/2, but the formula comes from the solution of the "two-body problem", that the motion of two interacting bodies can be treated as the motion of a single body of mass μ=(m1m2)/(m1+m2) around a central object of mass (m1+m2), and at distance equal to the distance between the bodies.
https://en.wikipedia.org/wiki/Two-body_problem
For circular orbit, the solution is very easy: The angular velocity is the same for both bodies, and the centripetal force is equal to the force of gravitational interaction.
If the masses are m1 and m2, the radii (from the CoM ) are r1 and r2, then r1+r2=a, and
##m_1r_1\omega^2=G\frac {m_1m_2}{(r_1+r_2)^2}##
##m_2r_2\omega^2=G\frac {m_1m_2}{(r_1+r_2)^2}##
divide the first equation by m1, the second one by m2, and add the equations together, you get
##a\omega^2=G\frac{m_1+m_2}{a^2}## which is the same equation that would hold for a single mass orbiting around a cental object of mass m1+m2 on a circular orbit of radius a
 
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  • #9
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ω^2=2πf....m1+m2=2M...f^2=GM/2π^2a^3....is it right?
 
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  • #10
ehild
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ω^2=2πf....m1+m2=2M...f^2=GM/()^2a^3....is it right?
You had some mistakes.
 
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  • #11
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yes, sorry, so: ω=2πf....m1+m2=2 M...f^2=G2M/4π^2 a^3........f^2=GM/2π^2 a^3
 
  • #12
SammyS
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Homework Statement


A tight pair of stars orbiting with circular paths at a distance a, the weight of the stars is the same:
:confused:

If this is the literal wording of the problem I'm flabbergasted.

Weight being the force due to gravity: If the two stars can be considered to be isolated, then of course they have the same weight, no matter what is the ratio of their masses. But I suppose what was meant was that they have the same mass.
 
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  • #13
ehild
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yes, sorry, so: ω=2πf....m1+m2=2 M...f^2=G2M/4π^2 a^3........f^2=GM/2π^2 a^3
You still miss parentheses. Do you divide or multiply by π^2 a^3?
 
  • #14
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:confused:

If this is the literal wording of the problem I'm flabbergasted.

Weight being the force due to gravity: If the two stars can be considered to be isolated, then of course they have the same weight, no matter what is the ratio of their masses. But I suppose what was meant was that they have ahe same mass.
yes, bad translation, sorry.....
 

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