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MHT CET · Physics · Gravitation

The acceleration due to gravity on moon is \(\frac{1^{\text {th }}}{6}\) times the acceleration due to gravity on earth. If the ratio of the density of earth ' \(e_e\) ' to the density of moon ' \(e_m\) ' is \(\frac{5}{3}\) then the radius of moon ' \(R_m\) ' in terms of the radius of earth ' \(R_e\) ' is

  1. A \(\left(\frac{3}{18}\right) \mathrm{R}_{\mathrm{e}}\)
  2. B \(\left(\frac{1}{2 \sqrt{3}}\right) \mathrm{R}_{\mathrm{e}}\)
  3. C \(\left(\frac{5}{18}\right) \mathrm{R}_{\mathrm{e}}\)
  4. D \(\left(\frac{7}{6}\right) \mathrm{R}_{\mathrm{e}}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\left(\frac{5}{18}\right) \mathrm{R}_{\mathrm{e}}\)

Step-by-step Solution

Detailed explanation

We know \(g=\frac{\mathrm{GM}}{\mathrm{R}^2}\)
And the relation between mass and density of the planet is given by
\(\begin{aligned} & M=\rho\left(\frac{4}{3} \pi R^3\right) \\ & \therefore g=\left(\frac{4 \pi G}{3}\right) \rho R=K \rho R\end{aligned}\)
Given, \(\mathrm{g}_{\mathrm{m}}=\frac{1}{6} \mathrm{~g}_{\mathrm{e}}\) and \(\mathrm{e}_{\mathrm{m}}=\frac{3}{5} \mathrm{e}_{\mathrm{e}}\)
\(\begin{aligned} & \Rightarrow \frac{g_m}{g_e}=\frac{e_m R_m}{e_e R_e}=\frac{1}{6} \\ & \Rightarrow \frac{3 R_m}{5 R_e}=\frac{1}{6} \\ & \Rightarrow R_m=\frac{5}{18} R_e\end{aligned}\)
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