First slide

Gravitational potential and potential energy

Question

If a particle of mass z is projected with minimum velocity from the surface of a star with kinetic energy $K_{1} GMm / a$ and potential energy at surface of the star $K_{2} GMm / a$ towards the star of same mass m and radius a (K₁ and K₂ are constants) to reach the other star. Find the distance between the centre of the two stars

Moderate

A

$\frac{2 a}{(K_{2} - K_{1})}$
B

$\frac{4 a}{(K_{2} - K_{1})}$
C

$\frac{2 a}{(K_{1} - K_{2})}$
D

$\frac{a}{(K_{1} - K_{2})}$

Solution

Since the mass and radius of two stars are same, point P at which gravitational force will be equal, will be at equal distance (say r) from centre of each star. The particle should be projected to cross P, beyond which the particle will be attracted towards the other star.
KE = PE of body at P- P at surface of the star
$\frac{K_{1} GMm}{a} = [- \frac{GMm}{r} - \frac{GMm}{r}] - [- K_{2} \frac{GMm}{a}]$
on solving, $r = \frac{2 a}{(K_{2} - K_{1})} 2 r = \frac{4 a}{(K_{2} - K_{1})}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App