1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    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


    User Avatar
    Science Advisor

    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

    User Avatar
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook