Distance between the centres of two stars is `10a`. The masses of these stars are `M` and `16 M` and their radii a and `2a` respectively. A body of mass `m` is fired straight from the surface of the larger star towards the surface of the smaller star. What should be its minimum initial speed to reach the surface of the smaller star? Obtain the expression in terms of `G`, `M` and a.

Explanation

Let `P` be the point on the line joining the centre of the two plantes s.t the net field at it is zero
Then `(GM)/(r^(2)) (G. 16M)/((10a -r)^(2)) =0 rArr (10 a-r)^(2) = 16r^(2)`
`rArr 10a-r = 4r rArr r =2a`
Potential at point P `v_(P) = (-GM)/(r) - (G.16M)/((10a -r)) = (-GM)/(2a) - (2GM)/(a) = (-5GM)/(2a)`
Now if the particle projected from the larger planet has enough energy to cross this point it will reach the smaller planet For this the `K.E` imparted to the body must be just enough to raise its total mechnical energy to a value which is equal to `P.E` at point `P`
`(1)/(2)mv^(2) - (G (16M)m)/(2a) (-GMm)/(8a) =mv_(P)`
or `(v^(2))/(2) (-8GM)/(a) (-GM)/(8a) = (-5GMm)/(2a)`
or `v^(2) = (45GM)/(4a)` or `v_(min) = (3)/(4) sqrt((5GM)/(a))`
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