Total Energy of Binary System
We already know the following formulas (\(U\) being GPE)
\[U = -\frac{GMm}{r}\]
\[E_{tot} = KE_1+KE_2+U\]
\[F = \frac{GMm}{r^2}\]
We can get the velocity as:
\[F = \frac{mv^2}{r}\]
\[v = \sqrt{\frac{Fr}{m}}\]
Example
Now consider the following system:
let \(m_1 = 3m\) and \(m_2 = m\)
let the distance between the center of the two masses be \(d\)
the center of mass is \(\frac{d}{4}\) away from \(m_1\) and \(\frac{3d}{4}\) away from \(m_2\)
\[F = \frac{G(3m)m}{d^2} = \frac{G3m^2}{d^2}\]
using \(v_{3m} = \sqrt{\frac{Fr}{m}}\):
\[v_{3m} = \sqrt{\frac{G3m^2}{d^2}\cdot\frac{\frac{d}{4}}{3m}} = \sqrt{\frac{Gm}{4d}}\]
\[v_{m} = \sqrt{\frac{G3m^2}{d^2}\cdot\frac{\frac{3d}{4}}{m}} = \sqrt{\frac{9Gm}{4d}}\]
Hence,
\[KE_{3m} = \frac1{2}(3m)\frac{Gm}{4d}=\frac{3Gm^2}{8d}\]
\[KE_{m} = \frac1{2}m\frac{9Gm}{4d}=\frac{9Gm^2}{8d}\]
\[U = -\frac{G3m^2}d\]
Therefore:
\[E_{tot} = KE_{3m}+KE_m+U=-\frac3{2}\frac{Gm^2}d\]
Question
Now you can try to work this out youself for the general case where \(m_1=nm\) and \(m_2 = m\).
\[E_{tot} = -\frac{n}{2} \frac{Gm^2}{d}\]