COMEDK · Physics · 9. Gravitation
Assuming, \(g_{\text {(Moon) }}=\left(\frac{1}{6}\right) g_{\text {Earth }}\) and \(D_{\text {(Moon) }}=\left(\frac{1}{4}\right) D_{\text {Earth }}\), where \(g\) and \(D\) are the acceleration due to gravity and diameter respectively, the escape velocity from the Moon is
- A \(\frac{11.2}{24} \mathrm{kms}^{-1}\)
- B \(11.2 \times \sqrt{24} \mathrm{kms}^{-1}\)
- C \(\frac{11.2}{\sqrt{24}} \mathrm{kms}^{-1}\)
- D \(11.2 \times 24 \mathrm{kms}^{-1}\)
Answer & Solution
Correct Answer
(C) \(\frac{11.2}{\sqrt{24}} \mathrm{kms}^{-1}\)
Step-by-step Solution
Detailed explanation
As we know, escape velocity, \(v_{e}=\sqrt{2 g R}\) Let \(v_{m}\) be escape velocity from moon and \(v_{e}\) be the escape velocity from Earth. Then, \(\frac{v_{m}}{v_{e}}=\sqrt{\frac{2 \times g_{m} \times R_{m}}{2 \times g_{e} \times R_{e}}}\)…
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