newtonian mechanics - 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?