Distance between the centres 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

Options

A. `2/3sqrt((Gm)/a)`
B. `3/2sqrt((5Gm)/a)`
C. `2/3sqrt((5Gm)/a)`
D. `3/2sqrt((Gm)/a)`

Answer

B

Explanation

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.

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 16 M vanishes i.e., a minimum KE or velocity has to be imparted to m from surface of 16 M such that it is just able to overcome the gravitational pull of 16M. By law of conservation of energy
`rArr 1/2mv^2+[-(G(16M)M)/(2a)-(GMm)/(8a)]+[-(GMm)/(2a)-(G(16M)m)/(8a)]=0`
`rArr 1/2mv^2-(GMm)/(8a)(45)rArr v=3/2sqrt((5GM)/a)`