Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Two separate "gravitational potential energy" equations?

  1. Jul 12, 2018 #1
    I am hoping to get a deeper understanding of the difference between two different gravitational potential energy equations, the first of which is given by

    U = mgh

    and the second given by

    U = (Gm1m2)/r

    I first assumed that in a system consisting of the Earth and, say, a tennis ball, these two equations would yield the same result for the potential energy of the tennis ball as it was held 1 meter above the ground. It another second or two to realize that the results of these two equations are extremely different. This confuses me, since both equations describe "gravitational potential energy."

    What is the explanation behind the difference in these two equations? What type of scenarios do each describe? Is it the fact that we are not considering the Earth itself as an object in the first equation? Is the second equation specifically describing gravitational force between two objects? Any insight on this would be very much appreciated!
     
  2. jcsd
  3. Jul 12, 2018 #2

    Dale

    Staff: Mentor

    Take the second equation, do a coordinate transformation from r to h, then do a Taylor series expansion about h=0. What do you get?
     
  4. Jul 12, 2018 #3

    jtbell

    User Avatar

    Staff: Mentor

    If Dale's hint doesn't ring a bell, show us exactly how you set up the calculation for the two formulas, for an object 1 m above the ground. That is, show us the numbers that you used.
     
  5. Jul 13, 2018 #4
    I forgot that g = GM/r2, so I understand a little better how the equations are related. As for my calculations:

    Ball: m = 1 kg
    h = 1 m

    With equation 1: U = mgh = (1 kg)(9.8 m/s2)(1 m) = 9.8 J

    With equation 2: -Gm1m2/r = - (G)(1 kg)(ME kg)/(RE + 1 m) = 6.26E7 J,

    where ME = mass of Earth = 5.97E24 kg, and RE = radius of Earth = 6.37E6 J

    Is the reason for this large value of equation 2 because it is the potential energy of the ball relative to the center of the Earth?

    Also, I'm not sure I fully understand Dale's hint. Is it that U = mgh is a linear approximation of gravitational potential energy?
     
  6. Jul 13, 2018 #5
    With equation 2 that is the GPE relative to an unbound mass at infinite distance, what happens if you consider the difference between the GPE at the surface and 1 m above the surface?
    i.e. What do you get if you do (G)(m)(ME)/(RE + 1) - (G)(m)(ME)( RE)?
     
  7. Jul 13, 2018 #6
    correct ..! i got it
     
  8. Jul 13, 2018 #7

    Dale

    Staff: Mentor

    It is, but you really should work through it because there is a very important point to come out of the math. Don’t just think about it, actually work it out and post your result. Use ##r=R+h## for the coordinate transformation to start, where R is the radius of the earth. Then do the Taylor series expansion about h=0.

    You dropped a minus sign.
     
    Last edited: Jul 13, 2018
  9. Jul 13, 2018 #8

    jtbell

    User Avatar

    Staff: Mentor

    He did do that:
    Now do it again, with ##r = R##, to get the potential energy at ground level, and compare this to what equation 1 gives for the potential energy at ground level. This should tell you an important difference between the two equations.

    Next, find the difference between the potential energies at ground level and at a height of 1 m. First use equation 1, which should be pretty trivial. Then use equation 2, which might need some effort to get a useful answer.
     
  10. Jul 13, 2018 #9
    As Vagn and jtbell say, I see now, using equation 2, that by calculating the change in potential energy of the ball initially at ground level and at a final position of 1 m above the ground, the result is the same as that given by equation 1, that is:

    Uf = -Gm1ME/(RE + 1)
    Ui = -Gm1ME/(RE)
    Uf - Ui = 9.8 J

    As for the Taylor series:

    f(h) = -Gm1ME/(r + h), r = R

    which can be rearranged as (-Gm1ME/R)*(1+(h/R))-1

    so f(h) = (-Gm1ME/(R) ∑ (-1)n(h/R)n

    The first few terms of the series would be (-Gm1ME/R)[1 - (h/R) + (h/R)2 - (h/R)3 + ...]

    (GME/R2) = g, so (-Gm1ME/R2) = -m1g

    therefore, the linear term is m1gh, which is consistent with equation 1. I'm hoping I did this correctly...

    I see that as the order increases, for h << R, the terms of the series become increasingly insignificant.

    What is the reason for ignoring the first term of the series, the 1? I know sometimes the first term of a series is considered insignificant; is that the case here?
     
    Last edited: Jul 13, 2018
  11. Jul 13, 2018 #10

    jbriggs444

    User Avatar
    Science Advisor

    Suppose you evaluate the series at one height, then at another height and take the difference? What is 1 minus 1?
     
  12. Jul 13, 2018 #11
    Ah, yes. I'm always forgetting to evaluate final - initial. Thanks for that.
     
  13. Jul 13, 2018 #12

    jtbell

    User Avatar

    Staff: Mentor

    Yep, only the difference in PE between two locations is physically significant.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted