Derivation of Gravitational Potential

  1. Hi,

    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.

    Thanks,
     
  2. jcsd
  3. jtbell

    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. [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. I can use integrals like these, to an extent.

    Unfortunately not.
     
  6. [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:
    [tex]\frac{-Gm_{1}m_{2}}{R}[/tex]

    Pretty sure the math is correct, someone might be able to fix any physics errors I have.
     
    Last edited: Feb 18, 2013
  7. 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.

    BiP
     
  8. Quick question (for my knowledge), why are the limits of intergration from R to infinity?
     
  9. 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.

    BiP
     
  10. 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
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?