Shortest possible period of revolution of two identical gravitating solid spheres

1. Jan 16, 2013

autodidude

1. The problem statement, all variables and given/known data
From K&K's 'Intro to Mechanics'

Find the shortest possible period of revolution of two identical gravitating solid spheres which are in circular orbit in free space about a point midway between them.

2. Relevant equations

3. The attempt at a solution

So I figured the gravitational force exerted on each sphere by the other would be

$$F=\frac{2mg}{r^2}$$

according to Newton's law of gravitation (m being each sphere's mass). This force would be providing the centripetal acceleration that's keeping them going in a circle so the angular velocity can't exceed a certain value and this is related to the period of revolution.

$$F_c=\frac{2mG]{r^2}=m\frac{v^2}{r}$$
∴$$(\frac{2G}{r})^{1/2}=v$$

So plugging that into $$T=\frac{\omega}{2\pi}$$ gives me $$T=(\frac{G}{2\pi^2r^3})^{1/2}$$

Is this correct? If not, am I at least on the right track?

2. Jan 16, 2013

apelling

Your final equation suggests that as r tends to infinity then T tends to zero. Does that sound reasonable?
Your method is essentially right but you have got errors in the details of your equations.

Applying Newton's law of gravity gives a force of $$F=-\frac{Gm^2}{r_d^2}$$
Which should be equated to centripetal force (as you rightly did)
$$F=-\frac{mv^2}{r_r^2}$$
Notice I have two different distances
rd= distance between centre of spheres
rr= distance between a spheres centre and the midpoint between the spheres
clearly rd=2rr
Your equation for T was wrong and it is more useful to use
$$v=\frac{2\pi r_r}{T}$$

3. Jan 16, 2013

autodidude

thanks apelling, redoing it i got:

$$T=(\frac{4\pi^2r^3}{G})^{1/2}$$

but when I looked at the units, I seem to have a mass unit in there

4. Jan 17, 2013

apelling

You should get

$$T=(\frac{16\pi^2r^3}{Gm})^{1/2}$$

With r being the distance between the centre of a sphere and the midpoint about which the spheres rotate.

Now I suspect you need to replace the mass m with density x volume of a sphere. This introduces another distance: the radius of the sphere. For minimum time period the orbital radius r should be a minimum. What is the smallest it can be?

5. Jan 20, 2013

autodidude

Oh, thanks...back to work then! Wow, you gotta be really careful when doing these problems in this book!

6. Jan 22, 2013

autodidude

alright i'm so close now...i realise the mistake i made before was adding the masses rather than squaring. my answer now is like yours except i have a 4 in place of your 16

7. Jan 22, 2013

apelling

let r= distance between centre of sphere and midpoint between spheres.
then distance between centres of spheres=2r

We then get

$$F=-\frac{Gm^2}{4r^2}=-\frac{mv^2}{r}$$
which simplifies to
$$\frac{Gm}{4r}=v^2$$
Subbing in
$$v^2=\frac{4\pi^2 r^2}{T^2}$$

$$T=(\frac{16\pi^2r^3}{Gm})^{1/2}$$

8. Jan 24, 2013

autodidude

Redoing it once again, I now have an 8. I've been taking r to be the distance between the two bodies and so have r/2 for the centripetal force side....but shouldn't the same answer pop out?

$$\frac{Gm^2}{r^2}=\frac{2mv^2}{r}$$
$$(\frac{Gm}{2r})^{1/2}=v$$
$$T=(\frac{8\pi^2r^3}{Gm})^{1/2}$$

9. Jan 24, 2013

apelling

Are you using

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

10. Jan 24, 2013

autodidude

No, I'm using

But same answer if I use the squared version

11. Jan 25, 2013

apelling

You should be using r/2 in this equation to be consistent with your definition of r.

12. Jan 25, 2013

autodidude

Hmmm...I've just redone it a few times now and I'm now getting a 2 (have tried another formula as well, T=2pi/w, which gives the same answer). I'll keep checking though...must've made another error somewhere.

13. Jan 28, 2013

apelling

The 2 is correct. I got 16 because I used r=1/2 separation of masses. When cubed in the final equation this leads to a factor of 8 difference. 2 x 8 =16

14. Jan 31, 2013

autodidude

Ah yeah...thanks a lot for your help apelling!