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Gravitational Energy and Kepler's Law

  1. Feb 1, 2010 #1
    1. In a double-star system, two stars of mass 6.0 *10^30 kg each rotate about the system's center of mass at a radius of 2.0 * 10^11 m.

    (a) What is their common angular speed?

    (b) If a meteoroid passes through the system's center of mass perpendicular to their orbital plane, what minimum speed must it have at the center of mass if it is to escape to "infinity" from the two-star system?

    2. Relevant equations

    U(gravitational potential energy) = -GMm/R

    3. The attempt at a solution

    I tried to do F=ma and a=v^2/r so m(v^2/r)=GMm/r^2

    any help would be greatly appreciated. I need this done by midnight. thanks so much
    Last edited: Feb 1, 2010
  2. jcsd
  3. Feb 1, 2010 #2


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    Homework Helper

    Welcome to PF!

    In all circular orbit problems, begin with the fact that the centripetal force is the gravitational force: Fc = Fg
    Put in the two detailed formulas for the forces (with the mass and radius in them). Then you can solve for the velocity or period easily. And one more step to get the angular velocity.
  4. Feb 2, 2010 #3
    isn't that what I did in my attempt at a solution? the only other thing I realize is that the force of gravity is 2r not r. so m(v^2/r)=GMm/(2r^2). Do I need two separate equations for both stars?
  5. Feb 2, 2010 #4


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    Homework Helper

    Sorry, jetsfan - I failed to see that you already had done what I suggested!
    Now I don't understand why you are stalled.
    You have v² = GM/(4r) so v = 2.24 x 10^4 m/s.
    From there you can get ω = v/r in a moment . . . and you're done.
    Oh, is it the wrong answer?
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