91Ó°ÊÓ

The given data can be listed below,

• The mass of each star is,$M$.
• The radius of each star is, $R$.

According to the law of conservation of energy, the total energy i.e. kinetic energy, gravitational potential energy, and other work of an isolated system is constant.

a)

The relation of gravitational force is expressed as,

$F_g=G\frac{m_1{m}_2}{r^2}$

Here $F_g$is the gravitational force, $G$is the gravitational constant, $m_1$is the mass of star one, $m_2$is the mass of star second and $r$is the distance between stars.

In given below figure, the two stars separated by a distance$2R$ and the mass of each star is$M$ . Then, the gravitational force is expressed as,

$\begin{array}{c}{F}_g=G\frac{M×M}{{(R+R)}^2}\\ =G\frac{M^2}{{\left(2R\right)}^2}\\ =\frac{G{M}^2}{4R^2}\end{array}$

Hence the gravitational force of one star on the other is,$\frac{G{M}^2}{4R^2}$.

b)

The relation of force with mass and acceleration is expressed as,

$F_g=M{a}_{rad}$ ...(i)

Here$M$is the mass of moving particle and $a_{rad}$is the acceleration and is always directed toward the center of the circle.

The relation of acceleration in terms of mass and radius is expressed as,

$a_{rad}=\frac{v^2}{R}$

Here$v$is the speed of moving particle in a circular orbit and$R$is radius of orbit.

Substitute the value of $F_g$and $a_{rad}$in equation (i). Then it is expressed as,

Hence the speed of each star in circular orbit is,$\sqrt{\frac{GM}{4R}}$.

And the relation of period of the orbit is in terms of radius of orbit and speed of moving particle in a circular orbit is expressed as,

$T=\frac{2\mathrm{Ï€}r}{v}$

Here$T$is the period, the time for one revolution and$r$is the radius of circular orbits. But in this problem stars moving in $R$radius of circular orbits, so it is expressed as,

$T=\frac{2\mathrm{Ï€}R}{v}$ ...(ii)

Substitute the value of $v$in the equation (ii).

$\begin{array}{c}T=\frac{2\mathrm{Ï€}R}{\sqrt{\frac{GM}{4R}}}\\ =4\mathrm{Ï€}\sqrt{\frac{R^3}{GM}}\end{array}$

Hence the period of orbit is, $4\mathrm{Ï€}\sqrt{\frac{R^3}{GM}}$.

c)

Apply the conservation of the energy to the system of two stars. It is expressed as,

$K_1+U_1+W_o=K_2+U_2$

Here $K_1$, $K_2$and $U_1$, $U_2$are the kinetic and gravitational potential energy, and $W_0$is the work required to separate the stars. Separate to infinity implies, $K_2=0$and $U_2=0$. Then it is expressed as,

$\begin{array}{c}{K}_1+U_1+W_o=0\\ W_o=−(K_1+U_1)\end{array}$ ...(iii)

The kinetic and potential energy at initially is expressed as,

$K_1=\frac{1}{2}M{v}^2+\frac{1}{2}M{v}^2$.

$K_1=M{v}^2$

...(iv)

Substitute the value of in the equation (iv).

$\begin{array}{c}{K}_1=M{\left(\sqrt{\frac{GM}{4R}}\right)}^2\\ K_1=\frac{G{M}^2}{4R}\end{array}$

And potential energy at initially is expressed as,

$U_1=−G\frac{m_1{m}_2}{r}$

Because mass of each star is$M$ and radius of orbits is $R$. So it is expressed as,

$\begin{array}{c}{U}_1=−G\frac{M×M}{R+R}\\ =−\frac{G{M}^2}{2R}\end{array}$

Substitute the value of and in the equation (iii).

$\begin{array}{c}{W}_o=−\left(\frac{G{M}^2}{4R}−\frac{G{M}^2}{2R}\right)\\ =\frac{G{M}^2}{4R}\end{array}$

Hence the work required to separate the two stars to infinity is, $\frac{G{M}^2}{4R}$.