Shortest possible period of revolution of two identical gravitating solid spheres

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Homework Statement

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.

Homework Equations

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?

Thanks in advance

 

[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
r_d= distance between centre of spheres
r_r= distance between a spheres centre and the midpoint between the spheres
clearly r_d=2r_r
Your equation for T was wrong and it is more useful to use
$$v=\frac{2\pi r_r}{T}$$

 

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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

 

[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?

 

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

 

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alright I'm so close now...i realize 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

 

[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
$$v^2=\frac{4\pi^2 r^2}{T^2}$$

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

 

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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?

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

 

[apelling]

Are you using

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

 

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No, I'm using

Your equation for T was wrong and it is more useful to use
$$v=\frac{2\pi r_r}{T}$$

But same answer if I use the squared version

 

[apelling]

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

 

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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.

 

[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

 

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