MHT CET · Physics · Gravitation
The density of a new planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of earth. If \(R\) is the radius of earth, then radius of the planet would be
- A \( 4 \mathrm{R}\)
- B \(\mathrm{R} / 2\)
- C \(\frac{R}{4}\)
- D \(2 \mathrm{R}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{R} / 2\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Given, } \rho_{\mathrm{p}}=2 \rho_{\mathrm{e}}, \mathrm{g}_{\mathrm{p}}=\mathrm{g}_{\mathrm{e}} \\ & \mathrm{g}=\frac{4}{3} \pi \rho \mathrm{GR}\end{aligned}\)
\(\begin{array}{ll}\therefore & \frac{R_p}{R_e}=\left(\frac{g_p}{g_e}\right)\left(\frac{\rho_e}{\rho_p}\right)=(1) \times\left(\frac{1}{2}\right) \\ \therefore & R_p=\frac{R_e}{2}=\frac{R}{2}\end{array}\)
\(\begin{array}{ll}\therefore & \frac{R_p}{R_e}=\left(\frac{g_p}{g_e}\right)\left(\frac{\rho_e}{\rho_p}\right)=(1) \times\left(\frac{1}{2}\right) \\ \therefore & R_p=\frac{R_e}{2}=\frac{R}{2}\end{array}\)
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