Question

A pair of stars rotates about a common centre of mass. One of the stars has a mass M and the other m. Their centres are a distance d apart, d being large compared to the size of either star. Derive an expression for the period of revolution of the stars about their common centre of mass. Compare their angular momentum and kinetic energies.
A. $\sqrt {\dfrac{{G\left( {M + m} \right)}}{{{d^3}}}} ,\dfrac{{2m}}{M}$
B. $\sqrt {\dfrac{{G\left( {M + m} \right)}}{{{d^3}}}} ,\dfrac{{4m}}{M}$
C. $\sqrt {\dfrac{{G\left( {M + m} \right)}}{{{d^3}}}} ,\dfrac{{3m}}{M}$
D. $\sqrt {\dfrac{{G\left( {M + m} \right)}}{{{d^3}}}} ,\dfrac{m}{M}$

Answer 1

Hint: The distance of centre of mass from a given mass is obtained by the product of mass and total distance which is again divided by total mass of the system. The necessary force required for rotation is provided by the gravitational force.

Complete step by step answer:
A pair of stars is rotating about a common centre of mass and mass of one star is M and the other is m. Their centres are d distance apart. Let ${d_1}$ be the distance of centre of mass from M and ${d_2}$ be the distance of centre of mass from m.
${d_1} = \left( {\dfrac{m}{{M + m}}} \right)d$ and
$\Rightarrow{d_2} = \left( {\dfrac{M}{{M + m}}} \right)d$ i.e.,
$\Rightarrow\dfrac{{{d_1}}}{{{d_2}}} = \dfrac{m}{M}$

The gravitational force between the stars provides the necessary centripetal force which helps them to rotate about a common centre of mass. Hence,
\[\dfrac{{GMm}}{{{d^2}}} = M{\omega _1}^2{d_1}\] and
$\Rightarrow\dfrac{{GMm}}{{{d^2}}} = m{\omega _2}^2{d_2}$
$\Rightarrow{\omega _1} = \sqrt {\dfrac{{Gm}}{{{d^2}{d_1}}}} $ and
$\Rightarrow{\omega _2} = \sqrt {\dfrac{{GM}}{{{d^2}{d_2}}}} $
Now, we substitute the value of ${d_1}$ and ${d_2}$ in the above equations,
${\omega _1} = \sqrt {\dfrac{{Gm}}{{{d^2}\left( {\dfrac{{md}}{{M + m}}} \right)}}} $ and
$\Rightarrow{\omega _2} = \sqrt {\dfrac{{GM}}{{{d^2}\left( {\dfrac{{Md}}{{M + m}}} \right)}}} $
$\Rightarrow{\omega _1} = \sqrt {\dfrac{{G\left( {M + m} \right)}}{{{d^3}}}}$ and
$\Rightarrow{\omega _2} = \sqrt {\dfrac{{G\left( {M + m} \right)}}{{{d^3}}}} $
$\left[ {\therefore {\omega _1} = {\omega _2}} \right]$
Thus, time period of revolution =$\dfrac{{2\pi }}{\omega } = \dfrac{{2\pi }}{{\sqrt {\dfrac{{G\left( {M + m} \right)}}{{{d^3}}}} }}$

Angular momentum(L) = linear momentum×distance
${L_1}$(star of mass M) = $Mv{d_1}$ and
$\Rightarrow{L_2}$ (star of mass m) = $mv{d_2}$ [linear momentum = mv]
$\Rightarrow {L_1} = M\left( {{\omega _1}{d_1}} \right){d_1}$ and
$\Rightarrow{L_2} = m\left( {{\omega _2}{d_2}} \right){d_2}$
$\Rightarrow {L_1} = M{\omega _1}{d_1}^2$ and
$\Rightarrow{L_2} = m{\omega _2}{d_2}^2$
$\Rightarrow \dfrac{{L_1}}{{L_2}} = \dfrac{{M{\omega _1}{d_1}^2}}{{m{\omega _2}{d_2}^2}}$
$\Rightarrow \dfrac{{L_1}}{{L_2}} = \dfrac{M}{m}{\left( {\dfrac{m}{M}} \right)^2} \\
\Rightarrow \dfrac{{L_1}}{{L_2}} = \dfrac{m}{M}$ [as ${\omega _1} = {\omega _2}]$

Kinetic energy = $\dfrac{1}{2}m{v^2}$
${K_1}$(star of mass M) = $\dfrac{1}{2}M{\left( {{\omega _1}{d_1}} \right)^2}$ and
$\Rightarrow{K_2}$(star of mass m)= $\dfrac{1}{2}m{\left( {{\omega _2}{d_2}} \right)^2}$
$\Rightarrow{K_1} = \dfrac{1}{2}M{\omega _1}^2{d_1}^2$,
$\Rightarrow{K_2} = \dfrac{1}{2}m{\omega _2}^2{d_2}^2$
$\Rightarrow \dfrac{{K_1}}{{K_2}} = \dfrac{{\dfrac{1}{2}M{\omega_1}^2{d_1}^2}}{{\dfrac{1}{2}m{\omega _2}^2{d_2}^2}} \\
\Rightarrow \dfrac{{K_1}}{{K_2}} = \dfrac{M}{m}{\left( {\dfrac{{{d_1}}}{{{d_2}}}} \right)^2}\left[ {{\omega _1} = {\omega _2}} \right] \\
\therefore \dfrac{{K_1}}{{K_2}} = \dfrac{M}{m}{\left( {\dfrac{m}{M}} \right)^2} \Rightarrow \dfrac{m}{M}
$

So, the correct answer is “Option D”.

Note: The centre of mass of a two particle system lies in between them on the line joining the two particles i.e., stars. The ${d_1}$ is the distance from mass M and ${d_2}$ is the distance from mass m.Centre of mass of a body or system of a particle is defined as, a point at which whole of the mass of the body or all the masses of a system of particle appeared to be concentrated.When we are studying the dynamics of the motion of the system of a particle as a whole, then we need not bother about the dynamics of individual particles of the system. But only focus on the dynamic of a unique point corresponding to that system.