Spring compression and escape speed

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data
    A package of mass 5 kg sits at the equator of an airless asteroid of mass 6.3*10^5 kg and radius 48 m, which is spinning so that a point on the equator is moving with speed 2 m/s. We want to launch the package in such a way that it will never come back, and when it is very far from the asteroid it will be traveling with speed 227 m/s. We have a large and powerful spring whose stiffness is 3.0*10^5 N/m. How much must we compress the spring?

    2. Relevant equations
    Kf = Ui
    (1/2) * m * v^2 = (1/2) * ks * s^2
    v = sqrt[ (2 * G * M) / ri ]

    3. The attempt at a solution
    I've listed out the variables I'll be using in this equation:
    m = 5 kg
    M = 6.3e5 kg
    k = 3e5 N/m
    r = 48m

    I'll get straight to the point, plugging in the variables was easy up to the point where I had to figure out what v was, where I find the escape speed. I used the equation I stated above ( v = sqrt[ (2 * G * M) / ri ] ) to find out the escape velocity needed to figure out how much compression is needed on the spring, and got 5.403e-6.

    However, it seems to be a wrong answer, but I have no idea why, but I have a feeling it has something to do with my velocity (which I got a value of 1.3235e-3 m/s), but again I don't know what else to do with it.
  2. jcsd
  3. Feb 22, 2009 #2


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    Homework Helper

    I don't think it is useful to use that v= formula when you are asked to provide not only the energy for escape velocity but also an additional speed of 227 m/s. Better to compute the total energy needed - the GmM/r plus the 1/2m*227^2. The spring energy must provide this, so you can now calculate the compression.
  4. Feb 23, 2009 #3
    Thanks for your response, Delphi, I used the equation provided to find the total energy needed to launch it, but I came up with yet again, a wrong answer. Here's how I got my answer:

    Find the total energy needed:
    GmM/r + (1/2) * m * 227^2 = { [ (6.673e-11) * 5kg * 6.3e5kg ] / 48 } + [ (1/2) * 5kg * 227^2 ]
    = 1.28823e5 m/s

    Plug that value in to the kf = Ui equation:
    (1/2) * m * v^2 = (1/2) * ks * s^2 = sqrt{ [ (5kg) * (1.28823e5)^2 ] / 3e5 }
    = 5.259e2 m/s

    Again, it's not correct.. Worse, I'm having trouble grasping the concept, which is what I'm really worried about.

    Thanks for the help though, really. I've been scratching my head about this for a while now.
  5. Feb 23, 2009 #4


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    1.28823e5 m/s should have units of Joules.
    And be subbed into E = 1/2k*x^2 in place of E.
    I get x = about 1 meter.
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