Orbital Period

1. Nov 2, 2013

woaini

1. The problem statement, all variables and given/known data

The three planets (v1, v2 and v3) in the diagram all have similar mass and are in a line equally spaced so that v1 and v3 are orbiting around v2 synchronously. If the mass of each of the planets are M and the radius of the orbit is R, what is the orbital period?

2. Relevant equations

T=$\frac{2*pi*r}{v}$

v=$\sqrt{GM/r}$

Fg=GMm/r^2

3. The attempt at a solution

I am pretty sure you just need to set two equations equal to each other and solve for the variable T, but I am unsure which two equations this is. It would be appreciated if somebody could explain all this to me. Thank you.

2. Nov 2, 2013

rude man

One equation.
Equate the centripetal force of v1 to the gravitational pull exerted on v1 by v2 and v3.

3. Nov 2, 2013

woaini

$\frac{Mv^2}{R}$=$\frac{2*G*M*M}{R^2}$

v=$\sqrt{2GM}$

T=$\frac{2*pi*R}{v}$=$\frac{2*pi*R}{sqrt(2GM)}$

4. Nov 2, 2013

rude man

You don't have the correct expression for the total gravitationl attraction of v2 and v3 on v1.

I also suggest changing mv^2/R to m(w^2)R.

5. Nov 2, 2013

woaini

How would I find the total gravitational attraction? Since they both have the same radius and masses, I assume that the gravitational forces would be the same and therefore have two times that.

6. Nov 2, 2013

rude man

Look at the picture. Is the distance from v1 to v2 the same as the distance of v1 to v3?

7. Nov 2, 2013

woaini

$\frac{Mv^2}{R}$=$\frac{G*M*M}{R^2}$+$\frac{G*M*M}{2R^2}$

v=$\sqrt{3GM}$

T=$\frac{2*pi*r}{\sqrt{3GM}}$

8. Nov 2, 2013

rude man

Still not right. re-examine the second term on the right.

9. Nov 2, 2013

woaini

$\frac{Mv^2}{R}$=$\frac{G*M*M}{R^2}$+$\frac{G*M*M}{2R^2}$

v=$\sqrt{5/4*GM}$

T=$\frac{2*pi*r}{\sqrt{5/4*GM}}$

10. Nov 2, 2013

rude man

What is the distance between v1 and v3? Ergo, what is the grav. attraction between them?

11. Nov 2, 2013

woaini

Isn't the distance 2R?

Alright, I figured out I simplified to v wrong resulting in a wrong answer.

Last edited: Nov 2, 2013
12. Nov 2, 2013

tms

Yes, but when you write it, you should write $(2R)^2$, not $2R^2$.

13. Nov 2, 2013

woaini

Yes, so it's 4R^2

14. Nov 3, 2013

So it is.