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Asteroid Orbits

  1. Oct 16, 2006 #1
    Two small asteroids orbit each other in circular orbits. The distance between the centers
    of the two asteroids is 35 m. The mass of the larger asteroid is 3 x 10 6 kg and the mass of the smaller asteroid is 1.2 x 10 5 kg.
    a. )What is the magnitude of the gravitational force of interaction between the
    two asteroids?
    b.) What is the acceleration of each asteroid?
    c.) What is the period of the circular orbits of each of the two asteroids?
    Ignore the force of the sun acting on the asteroids for parts b & c.

    a)F= Gm1m2/r2=(6.667*10-11 Nm2/kg2)(3*106 kg)(1.2*105 kg)/(35 m)2=0.020 N

    b)I am try to find r, so I can insert into this equation:

    This is the equation i used to determine rE:
    I got, re=0.0009037 m

    That doesn't look right to me
    Any help?

    c)How do I solve for the period of circular orbits?
  2. jcsd
  3. Oct 16, 2006 #2


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    Are you trying to do this with reduced mass? If so, look here


    You don't have to do it this way, and since you know the force, that acceleration is easily found from Newton's second law. The period is a bit more difficult because the center of the circular orbit is the center of mass of the system. Each asteroid has its own acceleration, and its own distance from the center of motion. You can find the center of mass and use it do find the distances for each object, or you can convert to the equivalent reduced mass problem.
  4. Oct 16, 2006 #3
    Okay I got b figured out, but now I am just looking at c

    I know there is a Kepler's law, but not sure how to incorporate this.

    Any help?
  5. Oct 16, 2006 #4


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    There is a Kepler law, but you don't need it. You know the acceleration and you know the motion is circular. If you found the distance to the CM, that is the radius of the circle. You can find the CM if you need it. From the acceleration and the radius, you can find the speed (centripetal acceleration) and from the speed and the radius you can find the period. Both of course have the same period.
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