Question

A system of binary stars of masses $m_{A}and{m}_B$ are moving in circular orbits of radii $r_{A}and{r}_B$ respectively. If $T_{A}and{T}_B$ are the time periods of masses $m_{A}and{m}_B$ respectively, then

• A. $\frac{T_A}{T_B}={\left(\frac{r_A}{r_B}\right)}^{3/2}$
• B. $T_A>T_B(if{r}_A>r_B)$
• C. $T_A>T_B(if{m}_A>m_B)$
• D. $T_A=T_B$

Answer 1

The correct option is D $T_A=T_B$

Binary stars of masses $m_A$ and $m_B$ are moving in circular orbit. of radius $r_A$ and $r_B$ respectively.

Now, the time periods are $T_A$ and $T_B$ respectively.

(Refer to Image)

At that moment, figure shows that two binary stars of respective mass $M_1$ and $M_2$ rotating in circular orbits of radius $R_1$ and $R_2$ respectively.

The common centre of both orbits is also centre of mass of both the stars.

From the definition of centre of mass, we have

$M_1\times R_1=M_2{R}_2⟶\left(1\right)$

Force acting on both the stars, i.e. gravitational force and centripetal forces are balanced

or, $\frac{M_1{M}_2}{R_2}=\frac{M_1{V}_1^2}{R_1}=\frac{M_2{V}_2^2}{R_2}⟶\left(2\right)$

where, $R+R_1+R_2$. Let $T_1$ and $T_2$ are time periods of revolution of respective stars.

Let us substitute,

$V_1=\frac{2\Pi R_1}{T_1}$ and $V_2=\frac{2\Pi R_2}{T_2}$

In equation (2) and after simplification we get

$\frac{M_1{R}_1}{T_1^2}=\frac{M_2{R}_2}{T_2^2}⟶\left(3\right)$

Using eqauation (1) in equation (3) hence we get, $T_1=T_2$ .