Total mechanical energy of two neutron stars orbiting a common center ($E=U+K$) [closed]

Question

Suppose we have two neutron stars of mass $M$, orbiting a common centre at a distance $R$. (i.e. the two neutron stars are $2R$ apart from each other, and $R$ apart from the centre). I've attempted to calculate the total mechanical energy, though I'm not sure which one is the correct answer.

Attempt:

$F=\frac{GMm}{r^2}=\frac{GM^2}{(2R)^2}=\frac{GM^2}{4R^2}$ (Newton's law of universal gravitation).

Consider one neutron star,

$F = Ma = \frac{Mv^2}{R}=\frac{GM^2}{4R^2}$ (centripetal force)

Therefore, $v=\sqrt{\frac{GM}{4R}}$ for one of the neutron stars

Hence, the Kinetic energy of one neutron star is

$K=\frac{GM^2}{8R}$,

making the total kinetic energy of the system:

$$K_{tot}=2K=\frac{GM^2}{4R}$$

Now, this is the part where I'm stuck.

We know that the potential energy of an orbit is given by $U=-\frac{GMm}{r}$, but in this case does $r=2R$ (i.e. the distance between the two neutron stars), or does $r=R$, the distance between a neutron star and its centre?

So depending on which expression for $U$ is correct, we would have either $$U=-\frac{GM^2}{R}$$ or $$U=-\frac{GM^2}{2R}$$

So our total mechanical energy $E=K_{tot}+U$ will be either $$E=-\frac{GM^2}{4R}$$ or $$E=-\frac{3GM^2}{4R}$$

By the way, if it helps, $U$ is defined as the mutual potential energy between the two neutron stars.

Answer 1

The centripetal acceleration is $Mv^2/R$ as the masses move in a circle of radius R.. The attraction is $GM^2/4R^2$. Equating the two gives $Mv^2 = GM^2/4R$, which is also the total kinetic energy. Add to that the potential energy of $-GM^2/2R$ and you arrive at $E=-GM^2/4R$.