Question

Question: A certain triple-star system consists of two stars, each of mass m , revolving in the same circular orbit of radius raround a central star of mass M (Fig. 13-54).The two orbiting stars are always at opposite ends of a diameter of the orbit. Derive an expression for the period of revolution of the stars.

Short answer

Answer

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

Step-by-step solution

Identification of given data

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

Significance of Newton’s law of universal gravitation

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:

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

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

Determining the expression for the period of revolution of the stars

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,

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

According to the expression of orbital velocity of star,

$$\begin{gathered} v=\frac{2\pi r}{T} \\ {v}^2=\frac{4{\pi }^2{r}^2}{T^2} \end{gathered}$$

Equation (i) becomes

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

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