A planet of mass $m_1$ revolves round the sun of mass $m_2$. The distance between the sun and the planet is $r$. Considering the motion of the sun find the total energy of the system assuming the orbits to be circular.

Both the planet and the sun revolve around their centre of mass with same angular velocity (say $\omega$ )

$r=r_1+r_2 ...\left(i\right)$

$m_1{r}_1{\omega }^2=m_2{r}_2{\omega }^2=\frac{G{m}_1{m}_2}{r^2} ...\left(ii\right)$

Solving Eqs. (i) and (ii), we get

        $r_1=r\left(\frac{m_2}{m_1+m_2}\right)$

        $r_2=r\left(\frac{m_1}{m_1+m_2}\right)$

and  ${\omega }^2=\frac{G\left(m_1+m_2\right)}{r^3}$

Now, total energy of the system is

   $E=PE+KE\text{ or }E=-\frac{G{m}_1{m}_2}{r}+\frac{1}{2}{m}_1{r}_1^2{\omega }^2+\frac{1}{2}{m}_2{r}_2^2{\omega }^2$

Substituting the values of $r_1, r_2$ and ${\omega }^2$, we get

   $E=-\frac{G{m}_1{m}_2}{2r}$