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Is $-\frac{GMm}{R}$ just an approximation? I believe that it is since we assume that one of the mass is at rest when deriving it.

Qmechanic
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Jin
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    It's an approximation to the general relativistic gravitational potential. But staying in the regime of Newtonian gravity, it's not an approximation, it's true by linear superposition. – Daddy Kropotkin Dec 07 '21 at 10:42
  • How do you expect neither mass being at rest to change the result? I'd be interested to read a motivation of $-GMm/R$ in which an at-rest assumption occurs. – J.G. Dec 07 '21 at 10:45
  • @J.G. I am reading Physics Volume 1, Halliday Resnick Krane where they assumed $M$ to be at rest. – Jin Dec 07 '21 at 10:46
  • Are you talking about two point-like bodies? Because when written in this form ( $M$ and $m$) it may also imply something, e.g., the Earth, by a point. – Roger V. Dec 07 '21 at 11:00
  • @J.G. Page 308, "A particle of mass M, which we assume to be at rest at the origin, exerts a gravitational force on m". Note that it's on Physics 1 and not on 'Fundamentals of Physics' which is another book by the same author. – Jin Dec 07 '21 at 11:02
  • @RogerVadim By $M$ and $m$, I meant two point particles with mass $M$ and $m$ respectively. (However, it's not restricted to only particles by Shell theorem) – Jin Dec 07 '21 at 11:04
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    It is worth specifying this in your question - to avoid misinterpretation. – Roger V. Dec 07 '21 at 11:07
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    Please see this related question: https://physics.stackexchange.com/q/3534/123208 – PM 2Ring Dec 07 '21 at 11:39
  • @PM2Ring Interesting question. I think it has the same "flavour" as this one. – Jin Dec 07 '21 at 12:02

1 Answers1

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So when we use this Potential we tend to use it in cases whereon mass is much larger than another so any movement from gravity is negligible, like with us on earth we are pulling back on the earth as the earth pulls on us so we give it a minuscule change in momentum so in our reference frame negligible movement so can just be seen as being at rest. This applies to all masses in a Newtonian framework no matter the velocity.

Baxwell bolt
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  • Exactly, this is what I think. We can apply it only for problems where one mass is way larger than the other one. I noticed this when trying to solve the problems directly using the work-energy theorem (doing the integration). That means that it's just an approximation right? – Jin Dec 07 '21 at 10:51
  • No this applies for all masses at all velocities because this is a just a snap shot in time so the speed of the respective particles doesn't matter – Baxwell bolt Dec 07 '21 at 10:52
  • The proof assumed that M is at rest, so indeed it works for all masses. However when applying it to solve some problems, I think we need to assume $M >> m$, when solving for the speed. – Jin Dec 07 '21 at 10:55
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    when solving for the speed yes it makes life a lot easier to state that having a much larger mass, due to the fact we can just use an equipotential (the same potential so basically orbiting ) but the potential formula you stated in your question applies to all masses. – Baxwell bolt Dec 07 '21 at 10:58