Asteroid orbiting each other

  • Thread starter CornerCase
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  • #1

Homework Statement


Consider only two asteroids, both of uniform mass ma, r=10km radius, separated by s=10m. What are the equations for velocity and orbital period for a stable circular orbit?

Homework Equations


If the asteroids are considered point masses then the distance between them, R, is 2r+s.
Planetary orbital period equation: T = [tex]\stackrel{2\piR^{\stackrel{3}{2}}}{\sqrt{Gm_{a}}}[/tex]
Planetary orbital velocity equation: v = [tex]\sqrt{\stackrel{Gm_{a}}{R}}[/tex]
The barycenter is always .5R
Sorry, I don't know how to put the division bar in between yet

The Attempt at a Solution


The problem is that the planetary equations are based upon m1 >> m2. The barycenter is always within m1 (and wobble occurs). But with m1 = m2 they each orbit a common center. Is the answer to assume point masses and use the above equations assuming R=2r+s? Then calculate T and v but know that they circle at a point halfway between the two?
 

Answers and Replies

  • #2
Delphi51
Homework Helper
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Welcome to the Physics Forums.
Maybe you are more advanced at this than I am!
Why isn't it just
centripetal force = gravitational force
mv²/(R+5) = Gm²/(2R+10)²
 
  • #3
ehild
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1,912

The Attempt at a Solution


But with m1 = m2 they each orbit a common center. Is the answer to assume point masses and use the above equations assuming R=2r+s? Then calculate T and v but know that they circle at a point halfway between the two?

Yes, assume point masses orbiting round the centre of mass, both with the same angular velocity, and the distance between them is d=2r+s. Write up the equation for the centripetal force which is equal to the force of gravity between the masses.

ehild
 
  • #4
Thanks for letting me know that centripetal force = gravitational force is the way to go. Now, it would affect the stresses on the mass from the points closest to the center to those farther away, but its good to know that doesn't affect orbital calculations. And to think, AFTER I got the orbital calculations right I wanted to look at the centripetal force on various points to see how it adjusted the apparent gravity, but I didn't think to have it help me in the orbit. :redface:
 
  • #5
ehild
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I do not get you. The problem asks the velocity and orbital period of a stable circular orbit. Consider both asteroids as point masses d=2r+s distance apart. Both of them orbit around the same circle of radius R=r+s/2 with speed v (like a dumbbell). The force of gravity yields the centripetal force for both of them. Use the equation in Delphi's post

mav2/(r+s/2)=G ma2/(2r+s)2

to get v and T=2π(r+s/2)/v. You do not need to do tidal force calculations.

ehild
 
  • #6
Thank you. I was overcomplicating the idea. As the two asteroids spin, their (point-mass) centers are moving at v, but the points nearest each other (s apart) are moving <v and the points farthest out (.5s+2r) are moving >v. I didn't know if masses that close would modify the calculations.
 
Last edited:
  • #7
ehild
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If the asteroids are perfect spheres with uniform mass distribution then you can apply this simple calculation.

ehild
 

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