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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

  1. A \(\frac{11.2}{24} \mathrm{kms}^{-1}\)
  2. B \(11.2 \times \sqrt{24} \mathrm{kms}^{-1}\)
  3. C \(\frac{11.2}{\sqrt{24}} \mathrm{kms}^{-1}\)
  4. D \(11.2 \times 24 \mathrm{kms}^{-1}\)
Verified Solution

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, \(\frac{v_{m}}{v_{e}}=\sqrt{\frac{2 \times g_{m} \times R_{m}}{2 \times g_{e} \times R_{e}}}\)…