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The form of gravitational potential energy

  1. Mar 8, 2013 #1
    http://www.scholarpedia.org/article/Celestial_mechanics#Newton.E2.80.99s_Celestial_Mechanics In this source, the gravitational potential energy is given as [itex] \frac{-MmG}{r}-\frac{mmG}{r}[/itex], seeming to imply that the [itex]\frac{MmG}{r}[/itex] result only applies to a body, mass [itex]m[/itex], in a gravitational potential, not a two-body sytem. Why is this, or is it an error?

    I would have thought, given that the force is [itex]-\frac{mMG}{r^2}[/itex], the potential energy is [itex]-\int (-\frac{mMG}{r^2}) =-\frac{mMG}{r}[/itex]: certainly not the result in the source.
  2. jcsd
  3. Mar 8, 2013 #2
    I suspect the radius r in the Force and energy equations in Scholarpedia are slightly different.

    Often because M is so much greater than m, I have elsewhere see the approximation M + m about equal to M...which matches your proposed integration result.

    The equation here


    for escape velocity shows - GMm/r [as you suspected] for the potential energy BUT that 'r' is from the surface of earth...and my old physics book has the same -GMm/r when the sun is considered to be at rest....

    and I just watched Leonard Susskind derive the same potential energy in lecture 2 of his Cosmology series.

    In other formulations, the center of mass of a system of an orbiting sun and planet about each other [sun is not stationary] is based on mr =mR.....where such a center of mass is stationary....

    So you are on the right track, I think, but exactly what was assumed and approximated in 'Scholarpedia' is not clear to me.
  4. Mar 9, 2013 #3
    Since the r they use is the relative position vector between the Sun and the planet, the acceleration of this quantity must be the sum of the accelerations of the two bodies (since they both act along this vector to change its length, each attracting the other). So, the force on the planet is [itex]-\frac{G(M+m)m}{r^2}[/itex] (the M coming from the acceleration of the planet by the Sun and the m in parenthesis coming from the acceleration of the Sun by the planet) yielding a potential of [itex]-\frac{G(M+m)m}{r}[/itex]. The total potential energy uses μ = Mm/(M+m) in place of the m outside of the parenthesis (since the Sun also has potential energy due to the gravity of the planet). [itex]-\frac{G(M+m)\mu}{r}=-\frac{G(M+m)Mm}{(M+m)r}=-\frac{GMm}{r}[/itex]. The total energy of the Sun-planet system is: [itex]E=\frac{1}{2}\mu{}v^2-\frac{GMm}{r}[/itex] (where [itex]r=|\vec{r}|[/itex] and [itex]v=\dot{\vec{r}}[/itex]). The energy equation on the scholarpedia site is just for the energy of the planet.
  5. Mar 9, 2013 #4
    Thank you, that's an excellent answer, clearing all my previous problems.

    By the way, in the model where the sun isn't stationary (i.e.[itex]F=-\frac{G(M+m)m}{r^2}[/itex]), is it a simplification to assume that the sun is stationary, and exerts a force of [itex]-\frac{G(M+m)m}{r^2}[/itex] on the planet: does this pedagogical simplification actually reproduce the same results as reality, or must one use the frame of reference where the sun and Earth are moving?
    Last edited: Mar 9, 2013
  6. Mar 9, 2013 #5
    Yes, you get the same results as reality. Of course this simplification does not work as perfectly in systems with more than two bodies.
  7. Mar 13, 2013 #6
    Thank you, Isometric Pion. It seems counter-intuitive that treating the COM of the sun (i.e. a non-inertial reference-frame (the sun's accelerating)) as the origin of the co-ordinate system yields the same result as a inertial reference frame, but that's probably just me. Thanks again.
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