# Is the gravitational potential of a planet in orbit always equal to minus the squared velocity?

Say a planet (mass $m$) is orbiting a star (mass $M$) in a perfect circle, so it is in circular motion.

$F=ma$ and the gravitational force between two masses $F=\frac{GMm}{r^2}$ so

$\frac{GMm}{r^2}=ma$

$\frac{GM}{r^2}=a$

And in circular motion $a=\frac{v^2}{r}$ so

$\frac{GM}{r^2}=\frac{v^2}{r}$

$\frac{GM}{r}=v^2$

And gravitational potential $V=-\frac{GM}{r}$

So $v^2=-V$

Is there a (qualitative/less math/wordy) reason why this is the case? (or have I got this wrong?) and is this limited to the specific case of perfect circular motion?

It basically states that in a bound system, the average of the potential energy $V$ is related to the average of the kinetic energy $T$ like $$\frac{\langle V \rangle}{2} = - \langle T\rangle$$ where the angular brackets indicate a time average.
Note that the formula above is specific only to a potential that goes as $1/r$. As the above Wikipedia link points out, the more general formula is given as $$\frac{n\langle V \rangle}{2} = \langle T\rangle$$ for any potential that goes as $r^n$.
• Note that stable orbits only exist for $n=1$ (though bound states can obviously still exist) – Jerry Schirmer Jul 29 '13 at 19:42
• Meant to say $n=-1$ and $n=2$ – Jerry Schirmer Jul 29 '13 at 19:48
Gravitational potential energy $U$ is measured in joules and is proportional to the particle mass $m$. When you multiply your $v^2 = -\frac{GM}{r}$ by $m$, you can obtain that the particle kinetic energy is 1/2 of the potential energy $U$. It is OK for a bound state. See the virial theorem.