91Ó°ÊÓ

The triple star system, in which the central star has mass m and two stars of mass M are orbiting about diameter

Every particle in the universe is attracted to every other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance, according to Newton's Law of Universal Gravitation.

Formula:

$$\begin{gathered} F=\frac{GMm}{r^2} \\ {F}_c=\frac{m{v}^2}{r} \\ v=\frac{2\mathrm{Ï€}r}{T} \end{gathered}$$

Where,

F is the gravitational force

F_c is the centripetal force

G is the gravitational constant

M is the mass of earth

v is the speed of object

m is the mass of object

T is the time period of object

According to Newton’s law of gravitation, the net gravitational force acting on the mass due to the others is

$$\begin{gathered} F_{net}=\frac{GMm}{r^2}+\frac{Gmm}{{\left(2r\right)}^2} \\ {F}_{net}=\frac{Gm}{r^2}\left(M+\frac{m}{4}\right) \end{gathered}$$

This star of mass m is moving in an orbit; hence, the centripetal force can be provided by the gravitational force acting on the star. The expression for the centripetal force is

$F_c=\frac{m{v}^2}{r}$

The gravitational force can be balanced by the centripetal force. Therefore,

$\frac{Gm}{r^2}\left(M+\frac{m}{4}\right)=\frac{m{v}^2}{r}$ …(¾±)

According to the expression of orbital velocity of star,

$$\begin{gathered} v=\frac{2\mathrm{Ï€}r}{T} \\ {v}^2=\frac{4{\mathrm{Ï€}}^2{r}^2}{T^2} \end{gathered}$$

Equation (i) becomes

$$\begin{gathered} \frac{Gm}{r^2}\left(M+\frac{m}{4}\right)=\frac{m}{r}\frac{4{\mathrm{Ï€}}^2{r}^2}{T^2} \\ G\left(M+\frac{m}{4}\right)=\frac{4{\mathrm{Ï€}}^2{r}^3}{T^2} \\ T=\frac{2\mathrm{Ï€}{r}^{3/2}}{\sqrt{G\left(M+\frac{m}{4}\right)}} \end{gathered}$$

The expression for the period of revolution of the stars is$T=\frac{2\mathrm{Ï€}{r}^{3/2}}{\sqrt{G\left(M+\frac{m}{4}\right)}}$