Question

Distance between the centers of two stars is 10a. The masses of these stars are M and 16M and
their radii a and 2a respectively. A body of mass m is fired straight from the surface of the larger
star towards the smaller star. The minimum initial speed for the body to reach the surface of
smaller star is

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

Answer 1

The correct option is B $\frac{3}{2}\sqrt{\frac{5GM}{a}}$

First we have to find a point where the resultant field due to both is zero.
Let the point P be at a distance x from centre of bigger star.
$\Rightarrow \frac{G\left(16M\right)}{x^2}=\frac{GM}{(10a-x{)}^2}\Rightarrow x=8a\left(from{0}_1\right)$

i.e., once the body reaches P, the gravitational pull of attraction due to M takes the lead to make m move towards it automatically as the gravitational pull of attraction due to 16M vanishes i.e., a minimum KE or velocity has to be imparted to m from surface of 16M such
that it is just able to overcome the gravitational pull of 16M. By law of conservation of energy
$\frac{1}{2}m{v}^2+[-\frac{G\left(16M\right)}{2a}-\frac{GMm}{8a}]=0+[-\frac{GMm}{2a}-\frac{G\left(16M\right)m}{8a}]\Rightarrow \frac{1}{2}m{v}^2-\frac{GMm}{8a}\left(45\right)\Rightarrow v=\frac{3}{2}\sqrt{\frac{5GM}{a}}$