Question

A planet of mass $M$ and a star of mass $2M$ are revolving around their common $\text{COM}$ under influence of mutual gravitational pull. Distance between them is $r$. Find the total mechanical energy of the system.

• A. $\frac{-3G{M}^2}{r}$
• B. $\frac{-2G{M}^2}{r}$
• C. $\frac{-G{M}^2}{2r}$
• D. $\frac{-G{M}^2}{r}$

Answer 1

The correct option is D $\frac{-G{M}^2}{r}$


The two planets revolve about their $\text{COM}$. From figure,

$r_1+r_2=r...\left(1\right)$

From the centre of mass formula,

$r_1=\frac{2Mr}{2M+M}=\frac{2r}{3}$

Using equation $\left(1\right)$,

$\Rightarrow r_2=r-\frac{2r}{3}=\frac{r}{3}$

Since, the mutual gravitational force provides the centrifugal force to planet and the star to rotate about the $\text{COM}$,

$F_{\text{centrifugal force}}=F_{\text{gravitational force}}$

$\Rightarrow M{r}_1{\omega }^2=2M{r}_2{\omega }^2=\frac{G\left(M\right)\left(2M\right)}{r^2}$

Where, $\omega$ is angular speed.

$\Rightarrow {\omega }^2=\frac{2GM}{r_1{r}^2}$

$\Rightarrow {\omega }^2=\frac{2GM}{\frac{2r}{3}\times r^2}=\frac{3GM}{r^3}$

Total mechanical energy :

$E=U+K$

$\Rightarrow E=-\frac{G\left(M\right)\left(2M\right)}{r}+\frac{1}{2}M{r}_1^2{\omega }^2+\frac{1}{2}\left(2M\right)r{{}_2}^2{\omega }^2$

Substituting the values,

$\Rightarrow E=\frac{-2G{M}^2}{r}+\frac{1}{2}\left(\frac{3G{M}^2}{r^3}\right){\left(\frac{2r}{3}\right)}^2+\frac{1}{2}\left(\frac{6G{M}^2}{r^3}\right){\left(\frac{r}{3}\right)}^2$

$\Rightarrow E=\frac{-2G{M}^2}{r}+\frac{2}{3}\frac{G{M}^2}{r}+\frac{1}{3}\frac{G{M}^2}{r}$

$\therefore E=\frac{-G{M}^2}{r}$

Hence, option (d) is the correct answer.

Let's think: Why Total energy of the system is negative in the above given problem?