Question

Two hypothetical planets of masses $m_1$ and $m_2$ 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 $m_1$ is $v_1$ and that of $m_2$ is $v_2$)

• A. $v_1=v_2$
• B. $v_1=m_2\sqrt{\frac{2G}{d(m_1+m_2)}},v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)}}$
• C. $v_1=m_1\sqrt{\frac{2G}{d(m_1+m_2)}},v_2=m_2\sqrt{\frac{2G}{d(m_1+m_2)}}$
• D. $v_1=m_2\sqrt{\frac{2G}{m_1}},v_2=m_2\sqrt{\frac{2G}{m_2}}$

Answer 1

Answer: (B)

$\text{Solution}:$

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

So, initial energy of the system = 0

Final energy $=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2-\frac{G{m}_1{m}_2}{d}$

From conservation of energy,Initial energy = Final energy

$\therefore 0=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_1{v}_2^2=\frac{G{m}_1{m}_2}{d}...\left(1\right)$

By conservation of linear momentum

$m_1{v}_1+m_1{v}_2=0$

or $\frac{v_1}{v_2}=\frac{m_2}{m_1}\Rightarrow v_2=\frac{m_1}{m_2}{v}_1$

Putting value of $v_2$ in equation , we get

$m_1{v}^2+m_2{(-\frac{m_1{v}_1}{m_2})}^2=\frac{2G{m}_1{m}_2}{d}$

$\frac{m_1{m}_2{v}_1^2+m_1^2{v}_1^2}{m_2}=\frac{2G{m}_1{m}_2}{d}$

$v_1=\sqrt{\frac{2G{m}_2^2}{d(m_1+m_2)}}=m_2\sqrt{\frac{2G}{d(m_1+m_2)}}$

Similarly $v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)}}$

$\text{Hence B is the correct option}$