Question

Suppose two planets (spherical in shape) of radii \(R\) and \(2R\), but mass \(M\) and \(9~M\) respectively have a centre to centre separation \(8~R\) as shown in the figure. A satellite of mass \('m'\) is projected from the surface of the planet of mass \('M'\) directly towards the centre of the second planet. The minimum speed \('v'\) required for the satellite to reach the surface of the second planet is \(\sqrt{\dfrac{a}{7}\dfrac{GM}{R}}\) then the value of \('a'\) is _____.

[Given : The two planets are fixed in their position]



8R 

Answer 1

Let net force experienced by the mass (m) at point P be zero.

$\frac{GMm}{x^2}=\frac{G9Mm}{{(8R-x)}^2}$

$8R-x=3x⟹4x=8R$

$x=2R$

So, we have to give at least enough velocity to reach point (P), because after that bigger planet will attract mass (m) towards itself automatically.

Now, applying conservation of energy between projection and point $\text{P}$ and consider velocity at ( P) is zero.

$\frac{1}{2}m{v}^2-\frac{GMm}{R}-\frac{G9Mm}{7R}=0-\frac{GMm}{2R}-\frac{G9Mm}{6R}$

$\frac{1}{2}m{v}^2-\frac{16GMm}{7R}=\frac{-12GMm}{6R}$

$\frac{1}{2}m{v}^2=\frac{-12GMm}{6R}+\frac{16GMm}{7R}$

$\frac{1}{2}m{v}^2=\frac{-84GMm+96GMm}{42R}$

$\frac{1}{2}m{v}^2=\frac{12GMm}{42R}$

$v=\sqrt{\frac{4GM}{7R}}$

the value of ('a') is __

$\sqrt{\frac{4GM}{7R}}$

___.