Question

In the above question, a projectile of mass m is projected from the surface of the sphere of mass M directly towards the center of the second sphere. The minimum speed v of the projectile so that it reaches the surface of the second sphere is :

• A. $\sqrt{\frac{3GM}{5R}}$
• B. $3\sqrt{\frac{GM}{5R}}$
• C. $\sqrt{\frac{GM}{5R}}$
• D. $\sqrt{\frac{2GM}{5R}}$

Answer 1

The correct option is A $\sqrt{\frac{3GM}{5R}}$

the position of balancing point is towards the lighter body

the mechanical energy at M = kinetic energy + Potential energy

$E_m=\frac{1}{2}m{v}^2-\frac{GMm}{R}-\frac{G\left(4M\right)m}{R}$

energy at the balancing point $E_B=-\frac{GMm}{2R}-\frac{G\left(4M\right)m}{4R}$

from law of conservation of energy total energy is constant

$\frac{1}{2}m{v}^2-\frac{GMm}{R}-\frac{G\left(4M\right)m}{R}=-\frac{GMm}{2R}-\frac{G\left(4M\right)m}{4R}$

by simplifying the above equation we get $v=\sqrt{\frac{3GM}{5R}}$