Question

Two planets A and B have the same average density. Their radii $R_A$ and $R_B$ are such that $R_A : R_B = 3 : 1$. If $g_A $ and $g_B$ are the acceleration due to gravity at the surfaces of the planets, the $g_A : g_{B}$ equals

• A. 3:01
• B. 1:03
• C. 9:01
• D. 1:09

Answer 1

The Correct Option is A

Given, $\frac{R_{A}}{R_{B}}=\frac{3}{1}$ and $\rho_{A}=\rho_{B}$ $\because$ Average density, $\rho=\frac{3 g}{4 \pi R G}$ $\therefore$ From E (ii), $\Rightarrow \quad \frac{3 g_{A}}{4 \pi R_{A} G}=\frac{3 g_{B}}{4 \pi R_{B} G}$ $\Rightarrow \frac{g_{A}}{g_{B}}=\frac{R_{A}}{R_{B}}$ From E (i), $\frac{g_{A}}{g_{B}}=\frac{3}{1}$