JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2021
- Hard
- Numerical
Suppose two planets (spherical in shape) of radii \({R}\) and \(2 {R}\), 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 planets is \(\sqrt{\frac{a}{7} \frac{G M}{R}}\) then the value of \('a'\) is \(....\) [Given: The two planets are fixed in their position]

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Answer & Solution
Correct Answer
(A) \(4\)
Step-by-step Solution
Detailed explanation
Assume that at a distance x from the planet of mass M, the net gravitational field becomes zero.
\(\frac{G M}{x^{2}}=\frac{G 9 M}{(8 R-x)^{2}}\)
\(8 R-x=3 x\)
\(x=2 R\)
Apply conservation of energy and consider velocity at \(P\) is zero.
\(\frac{1}{2} m v^{2}-\frac{G M m}{R}-\frac{G 9 M m}{7 R}=0-\frac{G M m}{2 R}-\frac{G 9 M m}{6 R}\)
\(\frac{1}{2} v^{2}=\frac{2 G M}{7 R} \Rightarrow v=\sqrt{\frac{4}{7} \frac{G M}{R}}\)
\(a=4\)
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