Question

In a double star system one of mass ${m_1}$and another of mass ${m_2}$ with a separation $d$ rotate about their common center of mass. Then rate of sweeps of area of star of mass ${m_1}$ to the star of mass ${m_2}$ about their common center of mass is:
$A.\,\,\dfrac{{{m_1}}}{{{m_2}}}\\\ B.\,\,\dfrac{{{m_2}}}{{{m_1}}}\\\ C.\,\,\dfrac{{{m_1}^2}}{{{m_2}^2}}\\\ D.\,\,\dfrac{{{m_2}^2}}{{{m_1}^2}}$

Answer 1

In the question we have to consider and apply the concept of the Kepler’s Law of gravitation. Also. We have to apply the concept of center of mass for a system of a number of bodies. With these two we have to calculate the rate of area swept by the star ${m_1}$ to the star ${m_2}$.

Complete step by step answer:
First, we will make some assumptions so that we can apply the relations and formulas.
Let us consider the Center of mass of the system of both stars to be at the distance of $r$ from mass ${m_1}$ and then let us take the distance between the masses to be $d$. Here we will consider the mass ${m_1}$as the origin.
Therefore, with these assumptions, we can calculate the distance $r$ using the formula of the Center of mass:
$r = \dfrac{{{m_1} \times {d_1} + {m_2} \times {d_2}}}{{{m_1} + {m_2}}}\\\ \implies r = \dfrac{{{m_1} \times 0 + {m_2} \times d}}{{{m_1} + {m_2}}}\\\ \implies r = \dfrac{{{m_2}d}}{{{m_1} + {m_2}}}$
Therefore, we now have the distance of the center of mass from the star of mass ${m_1}$.
Now the distance of the centre of mass from the star of mass ${m_2}$ will be: $d - r$.
If we put the value of r which we calculate earlier, we get the distance of the center of mass from the star of mass ${m_2}$
$d - r = d - \dfrac{{{m_2}d}}{{{m_1} + {m_2}}}\\\ \implies d - r = \dfrac{{{m_1}d + {m_2}d - {m_2}d}}{{{m_1} + {m_2}}}\\\ \implies d - r = \dfrac{{{m_1}d}}{{{m_1} + {m_2}}}$
Hence now we have the distance of the center of mass from both the stars.

Now we have to see the fact that the centripetal force required for the planetary motion is provided by the gravitational force acting between the two stars, therefore if we equate the two forces, we get the following relation:
gravitational force = centripetal force
$\dfrac{{G{m_1}{m_2}}}{{{d^2}}} = \dfrac{{{m_1}{v_1}^2}}{r}$
Here we have considered the force on the star ${m_1}$, and ${v_1}$is the velocity of the same star and G is the gravitational constant.
Making ${v_1}$ the subject and putting the value of r in the above relation we get:
$\dfrac{{G{m_1}{m_2}}}{{{d^2}}} = \dfrac{{{m_1}v_1^2}}{{\left( {\dfrac{{{m_2}d}}{{{m_1} + {m_2}}}} \right)}}\\\ \implies v_1^2 = \dfrac{{G{m_2}^2d}}{{{d^2}\left( {{m_1} + {m_2}} \right)}}\\\ \implies {v_1} = \sqrt {\dfrac{{G{m_2}^2}}{{d\left( {{m_1} + {m_2}} \right)}}} \\\ \implies {v_1} = {m_2}\sqrt {\dfrac{G}{{d\left( {{m_1} + {m_2}} \right)}}}$
The velocity of the star of mass ${m_1}$ is known to us.
Similarly, for the star of mass ${m_2}$:
${v_2} = {m_1}\sqrt {\dfrac{G}{{d\left( {{m_1} + {m_2}} \right)}}}$
Now that we have the velocities of both the stars, we will take the ratio of the both velocities:
$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{{m_2}\sqrt {\dfrac{G}{{d\left( {{m_1} + {m_2}} \right)}}} }}{{{m_1}\sqrt {\dfrac{G}{{d\left( {{m_1} + {m_2}} \right)}}} }}\\\ \implies \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{{m_2}}}{{{m_1}}}$
Now we have the relation between the velocities of both the stars.
We will now come to the area swept by the stars,
The area swept by the area vector of the stars is given by:
$A = \dfrac{1}{2}{r^2}\omega$, where r is the radius and $\omega$ is the angular velocity.
The ratio of area swept by both the stars will be:
$\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{\dfrac{1}{2}r_1^2{\omega _1}}}{{\dfrac{1}{2}r_2^2{\omega _2}}}$
The value of ${r_2} = d - r$ and ${r_1} = r$ putting these values:
$\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{\dfrac{1}{2}{r^2}{\omega _1}}}{{\dfrac{1}{2}{{\left( {d - r} \right)}^2}{\omega _2}}}\\\ \implies \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{{r^2}{\omega _1}}}{{{{\left( {d - r} \right)}^2}{\omega _2}}}$

Now that we have calculated the ratio of the area swept by the stars, we will use the following relation:
$v = r\omega$
Therefore, applying the above relation for both stars in the ratio of area swept:
$\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{{r^2}{\omega _1}}}{{{{\left( {d - r} \right)}^2}{\omega _2}}}\\\ \implies \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{r \times r{\omega _1}}}{{\left( {d - r} \right) \times \left( {d - r} \right){\omega _2}}}\\\ \implies \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{r \times {v_1}}}{{\left( {d - r} \right) \times {v_2}}}$
In the above we will put the values of $r$ and $d - r$ which we calculated in the beginning:
$\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{r \times {v_1}}}{{\left( {d - r} \right) \times {v_2}}}\\\ \implies \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{\dfrac{{{m_2}d}}{{{m_1} + {m_2}}} \times {v_1}}}{{\dfrac{{{m_1}d}}{{{m_1} + {m_2}}} \times {v_2}}}\\\ \implies \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{\left( {\dfrac{d}{{{m_1} + {m_2}}}} \right)}}{{\left( {\dfrac{d}{{{m_1} + {m_2}}}} \right)}}\dfrac{{{m_2} \times {v_1}}}{{{m_1} \times {v_2}}}\\\ \implies \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{{m_2} \times {v_1}}}{{{m_1} \times {v_2}}}$

Now we will put the value of the ratio of the velocities which we calculated earlier:
$\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{{m_2}}}{{{m_1}}} \times \dfrac{{{v_1}}}{{{v_2}}}\\\ \implies \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{{m_2}}}{{{m_1}}} \times \dfrac{{{m_2}}}{{{m_1}}}\\\ \therefore \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{{m_2}^2}}{{{m_1}^2}}$

So, the correct answer is “Option D”.

Note:
These types of questions require very deep knowledge of a topic, otherwise deriving such an equation cannot be done. These questions test multiple concepts of a students and having a grasp of the concept helps in doing such long derivations