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Hooke's Law on the moon

  1. Dec 5, 2007 #1
    1. The problem statement, all variables and given/known data

    How would F(g) and (delta) X change if the Spring experiment was done on the moon where the gravitational acceleration is six times smaller than on earth?

    2. Relevant equations
    F(g) = (delta) mg
    F(s) = k(delta) X

    3. The attempt at a solution

    I would think that F(g) and (delta) X would be 6 times less on the moon because there would be less gravitational pull on the spring. So if F = 68600 g cm/sec^2 and X is 26.1 cm on earth, the F = 11433 g cm/sec^2and X = 4.35 cm on the moon.

    Please help.I think I'm getting the concept confused. Thanks!
    Last edited: Dec 5, 2007
  2. jcsd
  3. Dec 5, 2007 #2
    I don't think delta X would change, since that is a problem of conservation of energy assuming the system is in the horisontal position.
  4. Dec 5, 2007 #3


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    Dearly Missed

    You are indeed right, for the hanging spring.

    The only (problem-relevant) change in this lunar setup with respect to a tellar one is the change in the value of local g, the mass and the spring constant remaining the same.

    Since lunar g is one sixth of tellar g, the lunar delta will be one sixt of the delta on Earth.
  5. Dec 5, 2007 #4


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    Yes, obviously if g on the moon is "6 times smaller than on earth" (1/6 the value) and the mass remains the same, then F(g) is 1/6 what it is on earth. (Actually it works the other way- because F(g) is 1/6 what it is on earth, g is 1/6.)
    On earth, F= mg. Dividing both sides of that equation by 6, (F/6)= m(g/6).

    But the "spring" question is not a clear. If the spring is lying on a flat surface, gravity plays no part. Are you assuming that a weight is hanging from the spring and gravitational force is stretching it, then the same thing happens. If F is the gravitational force on earth and F= kX, then dividing both sides by 6, (F/6)= k(X/6). Because, on the moon, the gravitational force if F/6, we also have the "stretch" equal to X/6.
  6. Dec 5, 2007 #5
    I believe that it is in a vertical position since the weight (mass) is hanging on the end of a spring and the spring is attached to the ceiling of a room.
    Last edited: Dec 5, 2007
  7. Dec 5, 2007 #6
    Thank you Halls of Ivy.

    I'm still unclear about X. The formula on the moon is (F/6) = K (X/6).. Is this correct?
  8. Dec 7, 2007 #7

    Shooting Star

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    I'm unclear about you problem. Use basic concepts.

    When you hang a mass m on earth, and the extension is x, then force = mg = kx.

    When you hang the same mass m on moon, and the extension is x2, then m(g/6) = kx2, which gives, x2 = x/6.
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