Two hypothetical planets of masses m_1 and m_2 are at rest when...

Two hypothetical planets of masses

\mathrm{m}_1

and

\mathrm{m}_2

are at rest when they are infinite distance apart. Because of the gravitational force they move towards each other along the line joining their centres. What is their speed when their separation is '

d

'? (Speed of

\mathrm{m}_1

\mathrm{v}_1

and that of

\mathrm{m}_2

\mathrm{v}_2

)

Options

A $\text{ } v_1=v_2$
B $\begin{aligned}&\mathrm{v}}_{1}= \mathrm{m}}_{2} \sqrt{\left.\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right.}\right)}\\ &\mathrm{v}}_{2}= \mathrm{m}}_{1} \sqrt{\left.\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right.}\right)} \end{aligned}$
C $\begin{aligned}&\mathrm{v}}_{1}= \mathrm{m}}_{1} \sqrt{\left.\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right.}\right)}\\ &\mathrm{v}}_{2}= \mathrm{m}}_{2} \sqrt{\left.\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right.}\right)} \end{aligned}$
D $\begin{aligned}&\mathrm{v}}_{1}= \mathrm{m}}_{2} \sqrt{\frac{2 \mathrm{G}}{\mathrm{m}_1}}\\ &\mathrm{v}}_{2}= \mathrm{m}}_{2} \sqrt{\frac{2 \mathrm{G}}{\mathrm{m}_2}} \end{aligned}$

Correct answer

B. $\begin{aligned}&\mathrm{v}}_{1}= \mathrm{m}}_{2} \sqrt{\left.\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right.}\right)}\\ &\mathrm{v}}_{2}= \mathrm{m}}_{1} \sqrt{\left.\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right.}\right)} \end{aligned}$

Step-by-step solution

We choose reference point, infinity, where total energy of the system is zero. So, initial energy of the system

=0

Final energy

=\frac{1}{2} \mathrm{~m}_1 \mathrm{v}_1^2+\frac{1}{2} \mathrm{~m}_2 \mathrm{v}_2^2-\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}}

From conservation of energy, Initial energy = Final energy

\therefore 0=\frac{1}{2} \mathrm{~m}_1 \mathrm{v}_1{ }^2+\frac{1}{2} \mathrm{~m}_2 \mathrm{v}_2{ }^2-\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}}

\frac{1}{2} \mathrm{~m}_1 \mathrm{v}_1^2+\frac{1}{2} \mathrm{~m}_1 \mathrm{v}_2^2=\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}} \ldots(1)

By conservation of linear momentum

\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2=0

\frac{\mathrm{v}_1}{\mathrm{v}_2}=-\frac{\mathrm{m}_2}{\mathrm{~m}_1} \Rightarrow \mathrm{v}_2=-\frac{\mathrm{m}_1}{\mathrm{~m}_2}

\mathrm{v}_1

Putting value of

\mathrm{v}_2

in equation (1), we get

\mathrm{m}_1 \mathrm{v}_1^2+\mathrm{m}_2\left(-\frac{\mathrm{m}_1 \mathrm{v}_1}{\mathrm{~m}_2}\right)^2=\frac{2 \mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}}

\begin{aligned} &\frac{m_1 m_2 v_1^2+m_1^2 v_1^2}{m_2}=\frac{2 \mathrm{Gm}_1 m_2}{\mathrm{~d}} \\ &\mathrm{v}_1=\sqrt{\frac{2 \mathrm{Gm}_2^2}{\mathrm{~d}\left(\mathrm{~m}_1+\mathrm{m}_2\right)}}=\mathrm{m}_2 \sqrt{\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right)}} \\ &\text{ Similarly } \mathrm{v}_2=-\mathrm{m}_1 \sqrt{\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right)}} \end{aligned}

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