Question

A double star consists of two stars having masses $m$ $2$ $m$ separated by a distance $r$. Of the following the correct option is

• A. Radius of circular path traced by $2$ $m$ is $\frac{2r}{3}$.
• B. Kinetic energy of heavier star is double than that of lighter star.
• C. Time period of revolution of both are not same.
• D. Angular momentum of lighter star is more

Answer 1

The correct option is D Angular momentum of lighter star is more

Both the stars in double star system revolve in circular orbits which are concentric with the centre of mass.

So, The distance of centre of mass from the heavier star= $\frac{mr+2m\times 0}{m+2m}=\frac{r}{3}$

So, the heavier star revolves in a circle of radius r/3 while the lighter star in a circle of radius 2r/3. So, option (a) is not correct.

Since center of mass of double star system is stationary, so it must be necessary that time period of both the stars should be same, so $\omega$ is also same for both star. So option (3) is also incorrect.

Now, Kinetic energy of heavier star= $\frac{1}{2}m{v}^2=\frac{1}{2}2m{\left(\omega \frac{r}{3}\right)}^2=\frac{1}{9}m{\omega }^2{r}^2$

Kinetic energy of lighter star= $\frac{1}{2}m{v}^2=\frac{1}{2}m{\left(\omega \frac{2r}{3}\right)}^2=\frac{2}{9}m{\omega }^2{r}^2$

So kinetic energy of lighter star more, so option(2) is also incorrect.

Now, angular momentum of heavier star = $I\omega$ =$2m{\left(\frac{r}{3}\right)}^2\omega =\frac{2}{9}m\omega r^2$

angular momentum of lighter star = $I\omega$ =$m{\left(\frac{2r}{3}\right)}^2\omega =\frac{4}{9}m\omega r^2$

So, angular momentum of lighter star is more, so soption (D) is correct.