Two masses m $_{1}$ and m $_{2}$ are initially at rest and are separated by a very large distance. If the masses approach each other subsequently, due to gravitational attraction between them, their relative velocity of approach at a separation distance of $d$ is

\frac{2 G d}{(m_{1} + m_{2})}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{(m_{1} + m_{2}) G}{2 d}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{[(m_{1} + m_{2}) \frac{2 G}{d}]}_{1}^{1 / 2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(m_{1} + m_{2})^{1 / 2} 2 G d

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B ${[(m_{1} + m_{2}) \frac{2 G}{d}]}_{1}^{1 / 2}$

We use the energy balance

Initial potential energy equals final kinetic energy

$\frac{G m_{1} m_{2}}{d} = \frac{1}{2} m v_{1}^{2} + \frac{1}{2} m v_{2}^{2}$

also from the conservation of momentum we have

$m_{1} v_{1} = m_{2} v_{2}$

or

$v_{1} = \frac{m_{1} v_{1}}{m_{2}}$

Substituting this we get

$v_{1} = \sqrt{\frac{2 G m_{2}^{2}}{d (m_{1} + m_{2})}}$

Similarly we have

$v_{2} = \sqrt{\frac{2 G m_{1}^{2}}{d (m_{1} + m_{2})}}$

Now as velocities are in opposite direction their relative velocity is $v_{1} - (- v_{2}) = v_{1} + v_{2}$

or

$\sqrt{\frac{2 G (m_{1} + m_{2})}{d}}$

Suggest Corrections

Similar questions

M_{1}

M_{2}

d

M_{1}

M_{2}

d

m_{1}

m_{2}

m_{1}

m_{2}

m_{1}

m_{2}

r

Join BYJU'S Learning Program