Question

For a binary star system of mass 'M' and '2M' separated by distance 'd'.

• A. Period of revolution of stars is $\frac{2\pi d^{\frac{3}{2}}}{\sqrt{3GM}}$
• B. Radius of circular path of star of mass 'M' is $\frac{d}{3}$
• C. Speed of revolution of star of mass 2M is $\sqrt{\frac{GM}{3d}}$
• D. Both stars can move in different plane

Answer 1

Answer: (A), (C)

The correct options are
A Period of revolution of stars is $\frac{2\pi d^{\frac{3}{2}}}{\sqrt{3GM}}$
C Speed of revolution of star of mass 2M is $\sqrt{\frac{GM}{3d}}$


$F=\frac{G2M^2}{d^2}$
'c' is the COM of the two stars
The distance of '2M' from the COM should be half that of the distance from 'M'.

$\therefore r_1=\frac{d}{3}$ and $r_2=\frac{2d}{3}$

By equating the forces to centripetal force,
$F=\frac{2M{v}_1^2}{r_1}\Rightarrow \frac{G2M^2}{d^2}=\frac{2M{v}_1^2}{\frac{d}{3}}$

$v_1=\sqrt{\frac{GM}{3d}}$

$T=\frac{2\pi r_1}{v_1}=\frac{2\pi d\sqrt{3d}}{3\times \sqrt{GM}}\Rightarrow T=\frac{2\pi d^{\frac{3}{2}}}{\sqrt{3GM}}$

$\overrightarrow{v_1}$ is the speed of $M_1$ and $\overrightarrow{v_2}$ is the speed of $M_2$.
By conservation of momentum,
$M_1\overrightarrow{v_1}+M_2\overrightarrow{v_2}=0$
This means $v_1$ and $v_2$ are along the same line, so they are also in the same plane.