Question:
Two planets have masses M and 16 M and their radii are \(a\) and \(2a\), respectively. The separation between the centres of the planets is \(10a\). A body of mass m is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach at the surface of smaller planet, the minimum firing speed needed is :
Two planets have masses M and 16 M and their radii are \(a\) and \(2a\), respectively. The separation between the centres of the planets is \(10a\). A body of mass m is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach at the surface of smaller planet, the minimum firing speed needed is :
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Find the point on the line where the net gravitational field is zero; the body must just reach it, after which the smaller planet pulls it in.
Updated On: Nov 5, 2026
- \(2\sqrt{\frac{GM}{a}}\)
- \(\sqrt{\frac{GM^2}{ma}}\)
- \(\frac{3}{2}\sqrt{\frac{5GM}{a}}\)
- \(4\sqrt{\frac{GM}{a}}\)
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The Correct Option is C
Solution and Explanation
Let N be the neutral point at distance r from the centre of the larger planet (mass 16M); its distance from the smaller planet (mass M) is \(10a - r\).
Zero net field: \(\frac{G\cdot16M}{r^2} = \frac{GM}{(10a - r)^2}\), so \(\frac{4}{r} = \frac{1}{10a - r}\), giving \(r = 8a\). Thus N is \(8a\) from the larger and \(2a\) from the smaller planet.
Potential at N: \(V_N = -\frac{G(16M)}{8a} - \frac{GM}{2a} = -\frac{2GM}{a} - \frac{GM}{2a} = -\frac{5GM}{2a}\).
Potential at the surface of the larger planet (radius \(2a\), the smaller planet is \(8a\) away): \(V_S = -\frac{G(16M)}{2a} - \frac{GM}{8a} = -\frac{8GM}{a} - \frac{GM}{8a} = -\frac{65GM}{8a}\).
The body must just reach N with zero speed: \(\frac12 v^2 = V_N - V_S = \frac{GM}{a}\left(\frac{65}{8} - \frac52\right) = \frac{45GM}{8a}\).
\(v^2 = \frac{45GM}{4a}\), so \(v = \frac32\sqrt{\frac{5GM}{a}}\).
Answer: (3) \(\frac{3}{2}\sqrt{\frac{5GM}{a}}\)
Zero net field: \(\frac{G\cdot16M}{r^2} = \frac{GM}{(10a - r)^2}\), so \(\frac{4}{r} = \frac{1}{10a - r}\), giving \(r = 8a\). Thus N is \(8a\) from the larger and \(2a\) from the smaller planet.
Potential at N: \(V_N = -\frac{G(16M)}{8a} - \frac{GM}{2a} = -\frac{2GM}{a} - \frac{GM}{2a} = -\frac{5GM}{2a}\).
Potential at the surface of the larger planet (radius \(2a\), the smaller planet is \(8a\) away): \(V_S = -\frac{G(16M)}{2a} - \frac{GM}{8a} = -\frac{8GM}{a} - \frac{GM}{8a} = -\frac{65GM}{8a}\).
The body must just reach N with zero speed: \(\frac12 v^2 = V_N - V_S = \frac{GM}{a}\left(\frac{65}{8} - \frac52\right) = \frac{45GM}{8a}\).
\(v^2 = \frac{45GM}{4a}\), so \(v = \frac32\sqrt{\frac{5GM}{a}}\).
Answer: (3) \(\frac{3}{2}\sqrt{\frac{5GM}{a}}\)
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