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B What does gravitational potential mean ?

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  1. Oct 28, 2016 #1
    Gravitational potential energy between 2 objects is ##-{G\times M\times m\over R}.##
    My question is, does the value of this equation of this equation mean total gravitational potential energy of both objects ?
    i.e, say the gravitational potential energy of object with mass ##M## is ##p## and, that of object with ##m## is ##P##.
    So, ##p + P = -{G\times M\times m\over R}## or ##p + P = -{2G\times M\times m\over R}## ??
     
    Last edited: Oct 28, 2016
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  3. Oct 28, 2016 #2

    andrewkirk

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    That is the gravitational potential energy of ##m## with respect to the gravitational field of ##M##. It is also the gravitational potential energy of ##M## with respect to the gravitational field of ##m##. The opening sentence of your post is not correct because it should not contain the word energy.

    The gravitational potential is a slightly different concept and involves only one object. It is not energy, but needs to be multiplied by the mass of another object to give a potential energy.

    The potential of the gravitational field of ##M## at distance ##R## from its centre of mass is ##-\frac{GM}{R}##. We call that number the 'gravitational potential at distance ##R## from ##M##'.

    The potential of the gravitational field of ##m## at distance ##R## from its centre of mass is ##-\frac{Gm}{R}##. We call that number the 'gravitational potential at distance ##R## from ##m##'.
     
  4. Oct 28, 2016 #3

    God i am really sorry, i meant potential energy not just potential. That is the reason why you should do physics at 2 am morning. Sorry i will edit my question.
     
  5. Oct 29, 2016 #4

    vanhees71

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    No, you are right. What you've given is indeed the potential of the gravitational force on a test mass ##m## outside a spherically symmetric mass distribution at rest around the origin of your reference frame with total mass ##M##. The potential indeed is
    $$V(r)=-\frac{G M m}{r},$$
    and the force is
    $$\vec{F}=-\vec{\nabla} V=-V' \vec{\nabla} r=-\frac{G M m}{r^2} \frac{\vec{x}}{r}.$$
     
  6. Oct 29, 2016 #5
    should this read 'the potential energy of mass m at a distance r from mass M is $$V(r)=-\frac{G M m}{r},$$'
     
  7. Oct 30, 2016 #6

    vanhees71

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    Both is correct: The potential of a vector field is a scalar field, whose gradient gives the vector field (supposed there exists a potential for the vector field). If a force has a potential that does not explicitly depend on time, then energy is conserved, i.e., for a particle subject to this force you have
    $$E=\frac{m}{2} \vec{v}^2+V(\vec{x})=\text{const}.$$
    In this context ##V(\vec{x})## is called "potential energy" of the particle.
     
  8. Oct 30, 2016 #7

    PeroK

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    It depends on whether you take ##M >> m## and assume that ##M## does not move. In that case, ##m## has all the PE of the system, which is indeed:

    ##V(r) = -\frac{GMm}{r}##

    But, if you take the case where ##M## is not so large and both masses move, then the above is the total PE of the system and, for example, if ##m = M## then each mass has half the PE.
     
  9. Oct 30, 2016 #8

    PeroK

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    You need to be careful, as these are fields valid for a "test mass" and, if you have two masses, ##M## and ##m##, they cannot be test masses for each other simultaneously.

    The OP's question arose from a problem where ##m = M##, and so neither was a test mass for the other.
     
  10. Oct 30, 2016 #9

    vanhees71

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    If the masses are close, then you should rather interpret the potential as an interaction potential,
    $$V(\vec{x}_1,\vec{x}_2)=-\frac{GmM}{|\vec{x}_1-\vec{x}_2|}.$$
    Then it's convenient to introduce center-of-mass and relative coordinates
    $$\vec{R}=\frac{m \vec{x}_1+M \vec{x}_2}{M+m}, \quad \vec{r}=\vec{x}_1-\vec{x}_2.$$
    Then the center of mass moves with constant velocity, and the equation of motion for the relative coordinates is that of a particle with mass
    $$\mu=\frac{mM}{m+M},$$
    the reduced mass, and force potential
    $$V(\vec{r})=-\frac{GmM}{|\vec{r}|}.$$
     
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