Two stars of masses $m_{1}$ and $m_{2}$ distance $r$ apart revolve about their centre of mass. The period of revolution is:

2 π \sqrt{\frac{r^{3}}{2 G (m_{1} + m_{2})}}

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2 π \sqrt{\frac{r^{3} (m_{1} + m_{2})}{2 G (m_{1} m_{2})}}

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2 π \sqrt{\frac{{2 r}^{3}}{G (m_{1} + m_{2})}}

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2 π \sqrt{\frac{r^{3}}{G (m_{1} + m_{2})}}

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Solution

The correct option is D $2 π \sqrt{\frac{r^{3}}{G (m_{1} + m_{2})}}$
Let the distances of the stars from the center of mass be $r_{1}$ and $r_{2}$ .

We can write that,
$(r_{1} + r_{2}) = r$ ,

The centeripetal force for revolution about center is provided by the gravitational force of attraction.
$\frac{G m_{1} m_{2}}{r^{2}} = m_{1} r_{1} ω^{2}$

and $\frac{G m_{1} m_{2}}{r^{2}} = m_{2} r_{2} ω^{2}$ .

So, the angular velocities can be written as:
$ω^{2} = \frac{G m_{2}}{r^{2} r_{1}} = \frac{G m_{1}}{r^{2} r_{2}}$

For whole system:
$ω = \frac{G (m_{1} + m_{2})}{r^{2} (r_{1} + r_{2})} = \frac{G (m_{1} + m_{2})}{r^{3}}$ .

From this one can find the expression of the period of revolution
$T = \frac{2 π}{ω} = 2 π \sqrt{\frac{r^{3}}{G (m_{1} + m_{2})}}$

So, option $(D)$ is correct.

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Two stars of masses m1 and m2 distance r apart revolve about their centre of mass. The period of revolution is:

Two stars of masses $m_{1}$ and $m_{2}$ distance $r$ apart revolve about their centre of mass. The period of revolution is: