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Derivation of Gravitational Potential

  1. Feb 18, 2013 #1

    The derivation of the Gravitational Potential formula, as I understand, is:

    [itex] W = Fd [/itex] (1)

    [itex] W = G \frac{M_1m_2}{r^2}d [/itex] (2) Substituting the Gravitational Force formula

    [itex] W = - \int_R^∞G \frac{M_1m_2}{r^2} \, dr [/itex] (3) Integrating within the boundaries of the initial distance (R) and Infinity

    Which allows us to arrive at:

    [itex] E_p = - \frac{GM_2m_1}{R}[/itex] (4)

    However, what I don't understand is how we are able to proceed from step 3 to step 4.
    What method must be used in order to proceed as such?

    My proficiency with Calculus is still in the works.

  2. jcsd
  3. Feb 18, 2013 #2


    User Avatar

    Staff: Mentor

    Do you know how to do integrals like this one?

    $$\int_a^b {x^n dx}$$

    If so, here's a hint: ##\frac{1}{r^2} = r^{-2}##.

    If no, then you'd best develop your calculus up to that point.
  4. Feb 18, 2013 #3
    [tex]\lim_{t \to \infty} \int_{R}^{t} \frac{Gm_{1}m{2}}{r^{2}} dr[/tex]

    Do you know how to solve that?
  5. Feb 18, 2013 #4
    I can use integrals like these, to an extent.

    Unfortunately not.
  6. Feb 18, 2013 #5
    [tex]\lim_{t \to \infty} -\int_{R}^{t} \frac{Gm_{1}m{2}}{r^{2}} dr[/tex]
    If you're taking an integral with respect to r and G, m1, and m2 are all constants, then what happens to the integral:
    [tex]\lim_{t \to \infty} -Gm_{1}m_{2}\int_{R}^{t} \frac{1}{r^{2}}dr[/tex]
    Now take the fact that 1/r^2 = r^-2
    So the integral is then solvable:
    [tex]\lim_{t \to \infty} (-Gm_{1}m_{2} \frac{-1}{r})|_{t}^{R}[/tex]
    So then this becomes:
    [tex]\lim_{t \to \infty} (\frac{-Gm_{1}m_{2}}{R}-\frac{-Gm_{1}m_{2}}{t})[/tex]
    Finally take the limit, anything over infinity tends to 0.
    So you end up with:

    Pretty sure the math is correct, someone might be able to fix any physics errors I have.
    Last edited: Feb 18, 2013
  7. Feb 18, 2013 #6
    Also, note that potential at infinity is conventionally defined to be 0. The convention, being what it is, isn't derivable mathematically, so you need to use it as a given in solving the problem when evaluating potential at infinity in your integral.

  8. Feb 18, 2013 #7
    Quick question (for my knowledge), why are the limits of intergration from R to infinity?
  9. Feb 18, 2013 #8
    The gravitationalpotential is defined as the work done by gravity to bring an object from infinity to a distance R from an object along a straight line, hence the limits of integration. A theorem from vector calculus shows that it does not matter what path the object travels, so the definition can be adjusted to an "arbitrary path from infinity to a distance R" from an object. But that is an offshoot of vector calculus.

  10. Feb 21, 2013 #9
    I think I figured it out, please correct me if I'm wrong.

    [itex]W = Fd[/itex]

    [itex]W = G\frac{M_1m_2}{r^2}d[/itex]

    [itex]W =- \int_R^∞ G\frac{M_1m_2}{r^2}\,dr[/itex]

    [itex]W = - G M_1m_2 \int_R^∞ r^{-2}\,dr [/itex] (Initially I was uncertain about pulling [itex]GM_1m_2[/itex] out)

    [itex]W = - G M_1m_2 [\frac{1}{r}]^R_∞[/itex]

    Am I right in assuming that this is the reason why [itex]-G\frac{M_1m_2}{∞}[/itex] produces zero?

    [itex] W = [-G \frac{M_1m_2}{r} - -G \frac{-GM_1m_2}{∞}][/itex]

    Which leads to

    [itex] E_p = -G\frac{M_1m_2}{r} [/itex]

    Thanks for the help
    Last edited: Feb 21, 2013
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