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Compression of a Spring on an Asteroid?

  1. Dec 10, 2006 #1
    1. The problem statement, all variables and given/known data
    A package with mass m sits on an airless asteroid of mass M and radius R. We want to launch a package straight up in such a way that its speed drops to zero when it is a distance 4R away from the center of the asteroid, where it's picked up by a waiting ship before it can fall back down. We have a powerful spring whose stiffness is Ks. How much must we compress the spring?

    2. Relevant equations
    I'm not quite sure what equations to use, except I'm almost positive we're going to use U(spring)= 1/2*Ks*s^2, where s is the abs value of the stretch/compression.
    Maybe also use:
    F(grav)=-G((M*m)/(|r21|^2))*rhat, where |r21| is the magnitude of the difference of locations between these two objects

    3. The attempt at a solution

    I know that the answer is: stretch=sqrt((3*G*m*M)/(2*R*Ks))
    I just can't figure out how to get there. I assumed that the way we got the "3" on the top was due to the actual distance between them was 3R (since 1R was the actual radius of the asteroid.

    please help!
  2. jcsd
  3. Dec 11, 2006 #2


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    You are on the right track with your equations. Consider conservation of enegy.
  4. Dec 11, 2006 #3
    If i chose the system where everything is in the system, there is no work external. SO, Ugrav=-Uspring, b/c Delta K and Delta mc^2 go to zero.

    but the answer is sqrt((3GmM)/(2Rks)) whereas my answer gives me:
    sqrt((2GmM)/(3Rks)). what am i missing?
  5. Dec 11, 2006 #4


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    I think that you are simply making a manipulation error.

    So we have conservation of energy;

    Ugi + Uspring = Ugf

    [tex]-\frac{GM}{R} + \frac{1}{2}k_{s}\cdot s^2 = -\frac{GM}{4R}[/tex]

    Can you now go from here?
  6. Dec 11, 2006 #5
    but doesn't that answer give me a 3 on the bottom and a 2 on the top? The answer has a 3 on the top and a 2 on the bottom. Am I doing my algebraic manipulation wrong?
  7. Dec 11, 2006 #6


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    Sorry, yes it does. I was reading your answers the wrong way round. From what I can see you have done nothing wrong, in my opinion your answer is correct. Perhaps, someone else can see the error?
  8. Dec 12, 2006 #7

    Doc Al

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    You must be if you're getting the answer you got.

    Consider this:
    [tex]-a + x = -\frac{a}{4}[/tex]

    That gives you:
    [tex]x = \frac{3a}{4}[/tex]

    Apply this manipulation and you'll get the expected answer.
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