Question

A double star consists of two stars having masses $M$ and $2M$. The distance between their centres is equal to $r$. They revolve under their mutual gravitational interaction. Then, which of the following statements are not correct?

• A. Heavier star revolves in orbit of radius $r/3$.
• B. Kinetic energy of the heavier star is twice that of the other star.
• C. lighter star revolves in orbit of radius $\frac{2r}{5}$
• D. all above the correct .

Answer 1

Answer: (A), (C)

The correct options are
A Heavier star revolves in orbit of radius $r/3$.
C lighter star revolves in orbit of radius $\frac{2r}{5}$

The centre of mass of the double star system remains stationary and both the stars revolve around in circular orbits which are concentric with the centre of mass.

The distance of centre of mass from the heavier star is equal to $\left(\frac{Mr+2M.0}{M+2M}\right)=\frac{r}{3}$

Hence, the heavier star revolves in a circle of radius $r/3$ while the lighter star in a circle of radius $2r/3$. Hence, option (a) is incorrect.

To calculate period of revolution of a double star system, concept of reduced mass can be used.

The reduced mass of the system is equal to

$\frac{\left(M\right)\left(2M\right)}{(M+2M)}=\frac{2M}{3}$

Hence, the period of revolution will be equal to $\frac{2\pi }{\sqrt{2\frac{GM}{3}}}{r}^{\frac{3}{2}}$

Where r is distance between two stars. Hence, option (b) is correct.

Kinetic energy of a star will be equal to $1/2m{v}^2$ where v is speed of the star which is equal to (radius of its circular orbit) $\times \omega$.

Hence, KE of heavier star is $E_1=\frac{1}{2}\left(2M\right)(\frac{r}{3}\omega {)}^2$ and that of lighter star, $E_2=\frac{1}{2}M(\frac{2r}{3}\omega {)}^2$.

It means, KE of the lighter star is twice that of the heavier star.

Hence, option (c) is incorrect.