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!

Total Work done by gravity

  1. Oct 9, 2013 #1
    So I was thinking about the Universal Law of Gravitation, and the force of an object depends on the radial distance from Earth for simplicity. As an object travels to the center of the earth the Fg increases as the radial distance decreases.F=(G*m*m*1)/(r^2). Knowing that the Force is not constant, the work done by gravity would be ∫F*dr from lower limit of r to upper limit of 0. Now, if you put the F in terms of r you get ∫[(G*m*m)/(r^2)]*dr from r to 0. As I tried to solve for Work, I arrived at 1/0 and now im stuck. Was all this valid or just nonsense? If you think about this in vector form 3 dimensions, it is more understandable to me at least. Here is a picture of my work. Help please.

    Attached Files:

    • Work.jpg
      File size:
      18.7 KB
  2. jcsd
  3. Oct 9, 2013 #2
    Think about what you mean by r being 0. Are you really taking the limit of integration as the center of Earth? I would think you would want to substitute 0 for the radius of Earth. Or even better, just use r_initial and r_final as your limits, it gives you a more general answer. Then just plug-n-chug for your solution.
  4. Oct 9, 2013 #3
    oh your right haha. 0 radial distance would be like super imposed on earth hahaa
    they would occupy the same space right? haha ty.
  5. Oct 9, 2013 #4


    User Avatar

    Staff: Mentor

    No, it doesn't. When the radial distance is less than the earth's radius, only the mass inside the radial distance "counts" in calculating the gravitational force. Google "shell theorem" for details.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook