# Homework Help: 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.