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Gravitational Self Potential Energy

  1. Jan 20, 2016 #1
    Note: I know this question has been asked before, but I wasn't allowed to ask my question on that thread

    1. The problem statement, all variables and given/known data

    The gravitational self potential energy of a solid ball of mass density ρ and radius R is E. What is the gravitational self potential energy of a ball of mass density ρ and radius 2R?

    2. Relevant equations
    $$PE = -\frac{GMM}{R}$$

    3. The attempt at a solution
    My attempt was dimensional analysis, because I had no other idea on how to approach this. The energy was somehow going to be related to G, radius, density, and other constants. PE has units $$kg \frac{m^2}{s^2}$$, density $$\frac{kg}{m^3}$$, and G in terms of $$\frac{m^3}{s^2}$$. I tried using E = $$k G^a \rho^b R^c$$, but I couldn't eliminate enough masses from the equation.
  2. jcsd
  3. Jan 20, 2016 #2


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    These are not the units for G.

    Otherwise, your approach should work.

    See if it works out once you have the correct units for G. If not, post the details of your attempt.
  4. Jan 20, 2016 #3

    I will try that, but are there other ways to solve this question? I kinda think my method is a bit sloppy.
  5. Jan 20, 2016 #4


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    Yes. Look at the equation you wrote for the PE of two point masses. This equation tells you how to dimensionally get PE from G, M and R. Thus, you need to think about relating M to density and R so that you can see how PE is related to G, ρ, and R.
  6. Jan 20, 2016 #5
    Uh huh, so lemme see if I understand you:

    The mass will be 8 times the original mass of the ball due to unvarying density, and radius is doubled. Since $$PE = \frac{-GMM}{R}$$, this equals $$\frac{8^2}{2}$$ = 32 times? Is that correct?
  7. Jan 20, 2016 #6


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    Yes, that's correct. The actual formula for the gravitational PE of a solid sphere will have some numerical factor out front, but it must be proportional to GM2/R.
  8. Jan 20, 2016 #7
    Thank you for helping
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