Question

Two stars of masses $m_1$ and $m_2$ are in mutual interaction and revolving in orbits of radii $r_1$ and $r_2$ respectively. The time period of revolution for this system will be,

• A. $2\mathrm{\pi }\sqrt{\frac{{\left({\mathrm{r}}_1-{\mathrm{r}}_2\right)}^3}{\mathrm{G}\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}}$
• B. $2\mathrm{\pi }\sqrt{\frac{{\left({\mathrm{r}}_1+{\mathrm{r}}_2\right)}^3}{\mathrm{G}\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}}$
• C. $2\mathrm{\pi }\sqrt{\frac{{\left({\mathrm{r}}_1-{\mathrm{r}}_2\right)}^3}{\mathrm{G}\left({\mathrm{m}}_1-{\mathrm{m}}_2\right)}}$
• D. $2\mathrm{\pi }\sqrt{\frac{{\left({\mathrm{r}}_1+{\mathrm{r}}_2\right)}^3}{\mathrm{G}\left({\mathrm{m}}_1-{\mathrm{m}}_2\right)}}$

Answer 1

The correct option is B

$2\mathrm{\pi }\sqrt{\frac{{\left({\mathrm{r}}_1+{\mathrm{r}}_2\right)}^3}{\mathrm{G}\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}}$

Step 1: Given

Mass of first star= $m_1$

Mass of second star= $m_2$

Radius of first star= $r_1$

Radius of second star= $r_2$

Step 2: Formula used

Gravitational force between two objects is given by $F_G=\frac{G{m}_1{m}_2}{r^2}$, where $G$ is gravitational constant, $m_1$ and $m_2$ are masses of the objects and $r$ is the distance between them.

Centripetal force on an object is given by $F_C=mr{\omega }^2$, where $m$ is the mass of the object, $r$ is the radius of orbit and $\omega$ is the angular velocity.

Time period is given by $T=\frac{2\pi }{\omega }$, where $\omega$ is the angular velocity.

Step 3: Find an expression for angular velocity

Equate the gravitational force and centripetal force acting on the first mass, since the gravitational force between masses provides the necessary centripetal force. The distance between the masses will be equal to the sum of their radii.

$$\begin{gathered} F_C=F_G \\ \Rightarrow m_1{r}_1{\omega }^2=\frac{G{m}_1{m}_2}{{\left(r_1+r_2\right)}^2} \\ \Rightarrow m_1\frac{m_2\left(r_1+r_2\right)}{m_1+m_2}{\omega }^2=\frac{G{m}_1{m}_2}{{\left(r_1+r_2\right)}^2}\left[\because r_1=\frac{m_2\left(r_1+r_2\right)}{m_1+m_2}\right] \\ \Rightarrow {\omega }^2=\frac{G{m}_1{m}_2}{{\left(r_1+r_2\right)}^2}\times \frac{m_1+m_2}{m_1{m}_2\left(r_1+r_2\right)} \\ \Rightarrow \omega =\sqrt{\frac{G\left(m_1+m_2\right)}{{\left(r_1+r_2\right)}^3}} \end{gathered}$$

Step 4: Calculate the time period of revolution using the formula

$\begin{array}{rcl}T& =& \frac{2\pi }{\omega }\\ & =& \frac{2\pi }{\sqrt{\frac{G\left(m_1+m_2\right)}{{\left(r_1+r_2\right)}^3}}}\\ & =& 2\pi \sqrt{\frac{{\left(r_1+r_2\right)}^3}{G\left(m_1+m_2\right)}}\end{array}$

Hence, option B is correct.