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Astronaut on Asteroid

  1. Sep 13, 2007 #1
    Sorry guys, one more question for now. It is the only other one I couldn't really get on this homework assignment.

    1. The problem statement, all variables and given/known data
    An astronaut is about to explore an asteroid of average density [tex]\rho[/tex]=5000 kg/[tex]m^3[/tex]. He is worried that he may accidentally jump from its surface and float off into space. How big (what diameter) must it be before he can neglect this possibility? The astronaut knows that his mass, m, including space suit, is 91 kg. He also knows that on the earth he can raise his center of gravity 0.6 m by jumping with his space suit on. It seems reasonable to assume that the maximum energy output of his legs will be the same on the asteroid as on the earth. For simplicity assume M>>m, such that M does not move significantly during the action, and then check this assumption [How much does the asteroid move relative to the astronaut?].

    2. Relevant equations
    [tex]\rho = m/V[/tex]

    V = 4/3[tex]\pi[/tex][tex]r^3[/tex]

    3. The attempt at a solution
    I figure if I can get the mass of the asteroid I can calculate the radius, and hence, the volume. I am not sure what clue the 0.6 m jumping on earth is supposed to give me. Also, why is m of the astronaut given if we can neglect it, since M>>m?
  2. jcsd
  3. Sep 13, 2007 #2
    oops, I meant to say "calculate the radius, and hence the diameter". Also, I am not sure now why I included V, as it doesn't say anything about the asteroid being spherical.
  4. Sep 13, 2007 #3

    D H

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    Regarding the 0.6 m jump on the Earth: What does this translate to in terms of velocity? Do you know what escape velocity is?
  5. Sep 13, 2007 #4
    I'm not sure...I know escape velocity is [tex]\sqrt{2GM}/R[/tex]
  6. Sep 13, 2007 #5

    D H

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    The astronaut changes his velocity from 0 to some value to enable him to jump 0.6 m into the air on the Earth. He will get this same "delta-V" on the asteroid. (That's one of the givens.) So relate that to escape velocity.
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