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kamiar · Jan 26, 2009, 02:53 PM

I have been looking for the answer to this problem for month now and I am looking forward to see if anyone can solve it...

ebaines · Jan 26, 2009, 03:32 PM

This is a formula for the period of the orbit for a two-body problem, right?

Start with two formulas:

1. The gravitational force between the two bodies is
$ F = \frac {GM_1 M_2 } {R^2} $

where R = distance between the two bodies.

2. The centripetal acceleration of body M_2 in orbit about the center of gravity of the two bodies is
$ a = R_2 \omega ^2 $

where $R_2$ = the distance from the center of gravity of the two bodies to body 2, and $\omega$ is the rotational velocity of body 2 about the center of gravity. Hence the force required to keep body 2 in orbit is:

$ F = M_2 R_2 \omega ^2 $

Now set these two forces equal to each other, and combine with the fact that the distance R_2 is related to R by:

$ R_2 = \frac {M_1} {M_1 + M_2} \cdot R $

The last step is to convert from $\omega$ to P (the period) by using:

$ P = \frac { 2 \pi } {\omega} $

Post back if you have questions.

kamiar · Jan 26, 2009, 04:09 PM

How do you get the answer in terms of m1,m2, R, G??

ebaines · Jan 27, 2009, 07:46 AM

First, take a look at the attached drawing, and note that there are three values of R:

R = distance between the two bodies
$R_1$ = distance from Body 1 to the center of gravity of the two-body system.
$R_2$ = distance from Body to the CG

Remember that the two bodies orbit about the center of gravity.

Set the gravitational force and the centripetal force equal and pull omega out to the left side:

$ \frac {G M_1 M_2} {R^2} = M_2 R_2 \omega ^2 \\ \omega ^2 = \frac {G M_1} {R^2 R_2} $

Then sub in
$ R_2 = \frac {M_1} {(M_1 + M_2)} \cdot R $

to get:
$ \omega^2 = \frac {G M_1} {R^2 \frac {M_1} {M_1 + M_2} R} \\ \omega ^2 = \frac{G (M_1 + M_2)} {R^3} $
Replace $\omega$ with:
$ \omega = \frac {2 \pi } P $
and rearrange:
$ P^2 = \frac {4 \pi ^2 R^3} {G(M_1 + M_2)} $

Voilà!