binary

THE PHYSICS ofBINARY STARS

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Gravitation|Astronomy|Topics

A simple binary star system may be modelled on the following assumptions:

stars are perfect spheres.
stars act as though their own mass acts at their own centre of gravity.
centre of mass of system (G) lies on a line joining their centres.
stars move in nearly circular orbits about their common centre of mass (G).
the time period of orbit of each star about G is the same.

Here is a diagram to illustrate the situation:

We now develop a simple analysis of such a system.

.
CENTRE OF MASS POSITION
For this we take 'Moments of Mass' about the line XY:

M ( r + R ) = ( M + m ) r

This leads immediately to:

r= M( r +R) ( m + M )

This gives us the position of G in terms
of the masses of the stars and the distance between their centres.

ANALYSIS OF FORCES
This is developed using Newton's second law for the motion of m.

F = - mv²
r

Circular motion of m about G

and using Newton's Gravitational Law:

Gravitational attraction over r + R

F = - GmM
(r+R)²

Equating these gives:

mv² = GMm
r ( r + R )²

-->

v² = GMr
( r + R )²

This may now be combined with our centre of mass result:

v² = GM . M( r + R )
( r + R )² ( M + m )

-->

v² = GM²
(r + R)(M + m)

We now use the standard circular motion results:

v = rw

w =2p
T

v² = 4p²r²
T²

Using the last two results and the position of centre of mass result we now find :

4p²( r + R )²M² = GM²
T²( M + m )² (R + r)(M + m)

This gives us a formula for the time period of rotation of the binary star system in terms of the masses of the stars, the distance between their centres and the Universal Gravitational Constant, G:

Time Period of Binary Star System

Questions:
1. Fill in the missing algebra steps that lead to this result.
2.Show that an equivalent formula for the time period is:

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