Shortest possible period of revolution of two identical gravitating solid spheres

In summary, K&K says that to solve for the period of revolution of two identical spheres which are in circular orbit around a point midway between them, one must use Newton's law of gravity and the equation v=\frac{2\pi r_r}{T}. They also mention that when solving this equation, it is important to remember to use the correct units (masses vs. density or volume of a sphere). Finally, they mention that the smallest orbital radius r a sphere can have is 2r.
  • #1
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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

[tex]F=\frac{2mg}{r^2}[/tex]

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.

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

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


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

Thanks in advance
 
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  • #2
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 [tex]F=-\frac{Gm^2}{r_d^2}[/tex]
Which should be equated to centripetal force (as you rightly did)
[tex]F=-\frac{mv^2}{r_r^2}[/tex]
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
[tex]v=\frac{2\pi r_r}{T}[/tex]
 
  • #3
thanks apelling, redoing it i got:

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

but when I looked at the units, I seem to have a mass unit in there
 
  • #4
You should get

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

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
Oh, thanks...back to work then! Wow, you got to be really careful when doing these problems in this book!
 
  • #6
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
 
  • #7
let r= distance between centre of sphere and midpoint between spheres.
then distance between centres of spheres=2r

We then get

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

leads to
[tex]T=(\frac{16\pi^2r^3}{Gm})^{1/2}[/tex]
 
  • #8
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?

[tex]\frac{Gm^2}{r^2}=\frac{2mv^2}{r}[/tex]
[tex](\frac{Gm}{2r})^{1/2}=v[/tex]
[tex]T=(\frac{8\pi^2r^3}{Gm})^{1/2}[/tex]
 
  • #9
Are you using

[tex]v^2=\frac{\pi^2 r^2}{T^2}[/tex]
 
  • #10
No, I'm using

apelling said:
Your equation for T was wrong and it is more useful to use
[tex]v=\frac{2\pi r_r}{T}[/tex]

But same answer if I use the squared version
 
  • #11
You should be using r/2 in this equation to be consistent with your definition of r.
 
  • #12
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
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
Ah yeah...thanks a lot for your help apelling!
 

Related to Shortest possible period of revolution of two identical gravitating solid spheres

1. What is the definition of "shortest possible period of revolution"?

The shortest possible period of revolution refers to the amount of time it takes for two identical gravitating solid spheres to complete one full rotation around each other.

2. What factors affect the shortest possible period of revolution?

The shortest possible period of revolution is affected by the masses of the two spheres, their distance from each other, and the strength of their gravitational pull.

3. Can the shortest possible period of revolution be calculated?

Yes, the shortest possible period of revolution can be calculated using the formula T = 2π√(r^3/GM), where T is the period, r is the distance between the spheres, G is the gravitational constant, and M is the combined mass of the two spheres.

4. How does the shortest possible period of revolution differ from the average period of revolution?

The shortest possible period of revolution is the minimum amount of time it takes for the spheres to complete one full rotation, while the average period of revolution takes into account multiple rotations and is affected by the initial velocity and direction of the spheres.

5. Is the shortest possible period of revolution affected by external forces?

Yes, external forces such as gravitational pull from other objects or changes in the distance between the spheres can affect the shortest possible period of revolution.

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