1. Not finding help here? Sign up for a free 30min 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!

Difference between potential energy on Earth and in orbit

  1. Nov 30, 2007 #1
    1. The problem statement, all variables and given/known data

    What is the difference between the potential on the surface of the Earth (R,M) and an object in orbit (r) about the earth.

    2. Relevant equations



    3. The attempt at a solution

    potential on earth = -GM/R
    potential of object in orbit = -GM/r

    diffrence = GM/R-GM/r

    answer should be : GM/R-3GM/2r --- I don't understand where the 3/2 came from.

    Thanks
     
  2. jcsd
  3. Nov 30, 2007 #2
    Perhaps the 3/2 comes from the potential energy due to the Sun?
     
  4. Nov 30, 2007 #3

    Astronuc

    User Avatar

    Staff: Mentor

    The question should be - "where does the 1/2 GM/r come from?".

    For an object "in orbit" it must be revolving, i.e. it has to have kinetic energy in addition to GPE, otherwise it would simply fall into the Mass providing the gravitational force.

    For an object to be in orbit, mg = mv2/r, means g = v2/r or v2 = gr, and that object has kinetic energy 1/2 mv2 or 1/2 mgr.

    g = GM/r2, so

    KE = 1/2 m (GM/r2) r = 1/2 GMm/r

    Also, check the form of the GPE

    http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html#ui
     
  5. Nov 30, 2007 #4
    But they're not talking about the total energy, just the difference between the potential on the south pole of earth and from distance r from the center of Earth.
     
  6. Nov 30, 2007 #5

    Astronuc

    User Avatar

    Staff: Mentor

    Where does the South Pole come into this. At the polls the effect of gravity is greater because one does not have the rotation of the earth to offset some of the gravitational force.
     
  7. Nov 30, 2007 #6
    Hello,

    The two are the same. Take a look at the problem statement again:

    The "in orbit" is key. Gravitational potential is defined as the work required to move a mass from one point to another. Here, you want to move the object to a point on the surface of the earth. Since the object is in orbit, ie moving, you can't just drop it to the surface of the earth; there's going to be work required to slow it down to a stop, and this needs to be accounted for. Hence the kinetic term.

    Hope this helps.
     
  8. Dec 1, 2007 #7
    I still don't know how they got the 3/2. Let me give you the exact question:

    A satellite is in circular orbit of radius r about the Earth (Radius R, mass M). Astandard clock C on the satellite is compared with an identical clock C0 at the south pole on Earth. Show that the ratio of the rate of the orbiting clock to that of the clock on earth is approximately:

    1+(GM/Rc^2)-(3GM/2rc^2).

    Note that the orbiting clock is faster only if r > 3/2 R, ir if r-R>3184 km.

    --------------

    The formula to find the rate is : 1+delta (potential)/c^2
    so I have to find the diffrence between the potential
     
  9. Dec 1, 2007 #8
    Hello,

    I'm not sure this is still an introductory physics problem :-)

    Here's my take:

    According to GR, the further into the gravitational well you get, the slower the clocks run. So if it were simply a matter of position in the well (gravitational potential), the earth clock would -always- run more slowly (its deeper).

    Now the hint states (and the answer shows) that the orbiting clock is faster only if [itex] r > \frac{3}{2}R[/itex], so there is an orbit where the clock is actually slower, counter to GR.

    The formula you quote for the rate, that has to do with the potential, I suspect is only useful in finding the dilation due to the gravitational field. You still would need to apply the SR time dilations due to differing velocities, and I expect this will produce your 3/2 factor. If you re-read the problem, it simply asks for the the rate of the orbiting clock to the clock on earth, and doesn't specify which effects to ignore and which to include-- so you should probably address both of them.

    So, again, I think ignoring the kinetic part is sinking you. I invite someone else to comment, as I'm not as up on this stuff as some others are.
     
  10. Dec 4, 2007 #9
    Thanks so much, but the formula is the I mensioned is correct according to my book because that's what it finds, the rate. I'll try posting the problem in advanced physics and see if anybody has an idea, because my instructor will probably put it in the final exam.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?