Question

The density of a newly discovered 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 the earth. If the radius of the earth is $R$, the radius of the planet would be

• A. 2R
• B. 4R
• C. $\frac{1}{4} R$
• D. $\frac{1}{2} R$

Answer 1

The Correct Option is D

The correct answer is D:\(\frac{1}{2}R\)
The formula for acceleration due to gravity is \((g = \frac{GM}{R^2})\), where \((G)\) is the gravitational constant, \((M)\) is the planet's mass, and \((R) \) is the planet's radius. According to the question, the acceleration due to gravity is the same on both planets. Therefore, we can set up the equation: \((\frac{GM_p}{R_p^2} = \frac{GM_e}{R_e^2}).\)
Simplifying this equation leads to: \((G \times \frac{\frac{4}{3} \pi R_p^3 \rho_p}{R_p^2} = G \times \frac{\frac{4}{3} \pi R_e^3 \rho_e}{R_e^2})\). Further simplification gives: \((R_p \rho_p = R_e \rho_e).\)
Given the information that the density of planet \((p)\) is twice the density of planet \((e)\), i.e., \((\rho_p = 2 \rho_e)\), we can substitute this into the previous equation: \((R_p \times 2 \rho_e = R_e \rho_e)\). This leads to the conclusion: \((R_p = \frac{R_e}{2} = \frac{R}{2})\).