# B What does gravitational potential mean ?

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1. Oct 28, 2016

### Buffu

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
2. Oct 28, 2016

### andrewkirk

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 $m$'.

3. Oct 28, 2016

### Buffu

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.

4. Oct 29, 2016

### vanhees71

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
$$\vec{F}=-\vec{\nabla} V=-V' \vec{\nabla} r=-\frac{G M m}{r^2} \frac{\vec{x}}{r}.$$

5. Oct 29, 2016

### lychette

should this read 'the potential energy of mass m at a distance r from mass M is $$V(r)=-\frac{G M m}{r},$$'

6. Oct 30, 2016

### vanhees71

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.

7. Oct 30, 2016

### PeroK

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.

8. Oct 30, 2016

### PeroK

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.

9. Oct 30, 2016

### vanhees71

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
$$V(\vec{r})=-\frac{GmM}{|\vec{r}|}.$$