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Escape Velocity of a spherical asteroid

  1. Nov 16, 2005 #1
    Consider a spherical asteroid with a radius of 5 km and a mass of 8.65x10^15 kg.
    (a) What is the acceleration of gravity on the surface of this asteroid?
    ANSWER: ___ m/s2
    (b) Suppose the asteroid spins about an axis through its center, like the Earth, with an angular speed . What is the greatest value can have before loose rocks on the asteroid's equator begin to fly off the surface?
    ANSWER: ___ rad/s

    For (a) which I got right.... I used the equation g=(G*M)/R^2
    g=(6.67e-11*8.65e15)/(5000^2)=.023078 m/s2

    For (b) I used the equation Wf^2= Wi^2 + 2*a*d and i substituted the numbers in to get... Wf^2 = 0 + 2(.023037*2pi)(31415.9m*2pi) and I got it to be 95.45 rad/s and for some reason this is not right... am I even using the right equation or is there another equation i can use to get the answer such as maybe... F= (G*M*m)/R^2 ....
  2. jcsd
  3. Nov 16, 2005 #2


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    I've got no idea what you've done in (b), but have you taken gravity into account at all?

    I'd use centrifugal force.
  4. Nov 16, 2005 #3
    Consider that, in circular motion:


    If a > g, then what would happen?

    Does this help you?

  5. Nov 16, 2005 #4
    so you are saying a =w^2 * r..... so w = sqroot(a/r) so w= sqroot(.023037/5000) = .0021465 ?????
  6. Nov 16, 2005 #5
    already tried that method

    i already tried that method greg and i got the wrong answer...im not liking this topic at all lol:surprised
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