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!

Work required to increase Earth/Moon separation

  1. Apr 8, 2013 #1

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    1. The problem statement, all variables and given/known data
    The Moon and Earth have masses of 7.4 × 1022 kg and 6.0 × 1024 kg, respectively, and the distance between their centers is 3.8 × 108 m. The amount of work required to pull them further apart so that their separation increases by a factor of 1.6 is most nearly


    2. Relevant equations
    Etot = -GMm/(2r)
    W=ΔE


    3. The attempt at a solution
    ΔE=(-GMm/2)(1/(1.6*3.8 x 108) - (1/3.8 x 108)) = 1.44 x 1028 J

    Answer key says 2.9 x 1028 J
    That's twice my answer. Is it possible that the person who made the answer key forgot to factor out the /2. Or should I be considering only ΔU? I did a similar problem from this chapter using ΔEtot and got the right answer. What am I missing?

    Thanks!
     
  2. jcsd
  3. Apr 8, 2013 #2

    mukundpa

    User Avatar
    Homework Helper

    Is the moon still in a stable circular orbit? I think if the motion of the moon is not considered than we have to take only change in potential energy.
     
  4. Apr 8, 2013 #3

    gneill

    User Avatar

    Staff: Mentor

    What energies does your Etot include?

    Why don't you do a check of the result by performing the work integration (integral of F.dr) for the increase in separation?
     
  5. Apr 8, 2013 #4

    BruceW

    User Avatar
    Homework Helper

    Why is the 2 here? The potential energy of the system is just -GMm/r
     
  6. Apr 8, 2013 #5

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    That formula is for total energy, not PE
    U=-GMm/r
    KE=0.5mv2, v=sqrt(GM/r)
    KE=0.5 GMm/r
    KE=GMm/(2r)
    U=-2GM/(2r) --add the 2's so U and KE have a common denominator
    KE+U=(GMm/(2r)) - (2GMm/(2r))
    KE+U=-GMm/(2r) = Etot

    I think mukundpa solved it. The problem did not state that the Moon achieved a circular orbit at the new distance. So with its velocity constant, all that changed is the PE. Doing the problem that way, I get the answer the key is looking for.

    Thanks everyone!
     
  7. Apr 8, 2013 #6

    BruceW

    User Avatar
    Homework Helper

    another way to interpret it, is that they wanted the work done, given that the kinetic energy of the moon and earth do not change.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted