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Two spherical planets A and B have the same mass but their densities are in a ratio 8:1. For these planets, the ratio of acceleration due to gravity at the surface of A to its value at the surface of B is
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Verified Answer
The correct answer is:
$1: 4$
Given:
$\rho_1: \rho_2=1: 8$ and $m_1=m_2$
We know that, $\rho=\frac{\mathrm{m}}{\mathrm{V}}$
$$
\begin{aligned}
& \Rightarrow \rho_1 V_1=\rho_2 V_2 \\
& \Rightarrow \rho_1 \frac{4}{3} \pi R_1^3=\rho_2 \frac{4}{3} \pi R_2^3 \\
& \Rightarrow\left(\frac{R_1}{R_2}\right)=\frac{\rho_2}{\rho_1}=\frac{8}{1} \\
& \Rightarrow\left(\frac{R_1}{R_2}\right)=\frac{2}{1}
\end{aligned}
$$
As we know that the acceleration due to gravity, $g=\frac{G M}{R^2}$
$$
\begin{aligned}
& g=\frac{G}{R^2} \times \rho \times \frac{4}{3} \pi R^3=\frac{4}{3} G \pi \rho R \\
& \Rightarrow \frac{g_1}{g_2}=\left(\frac{\rho_1}{\rho_2}\right)\left(\frac{R_1}{R_2}\right) \\
& \therefore \frac{g_1}{g_2}=\frac{1}{8} \times 2=\frac{1}{4} \quad \text { New soln }
\end{aligned}
$$
$\rho_1: \rho_2=1: 8$ and $m_1=m_2$
We know that, $\rho=\frac{\mathrm{m}}{\mathrm{V}}$
$$
\begin{aligned}
& \Rightarrow \rho_1 V_1=\rho_2 V_2 \\
& \Rightarrow \rho_1 \frac{4}{3} \pi R_1^3=\rho_2 \frac{4}{3} \pi R_2^3 \\
& \Rightarrow\left(\frac{R_1}{R_2}\right)=\frac{\rho_2}{\rho_1}=\frac{8}{1} \\
& \Rightarrow\left(\frac{R_1}{R_2}\right)=\frac{2}{1}
\end{aligned}
$$
As we know that the acceleration due to gravity, $g=\frac{G M}{R^2}$
$$
\begin{aligned}
& g=\frac{G}{R^2} \times \rho \times \frac{4}{3} \pi R^3=\frac{4}{3} G \pi \rho R \\
& \Rightarrow \frac{g_1}{g_2}=\left(\frac{\rho_1}{\rho_2}\right)\left(\frac{R_1}{R_2}\right) \\
& \therefore \frac{g_1}{g_2}=\frac{1}{8} \times 2=\frac{1}{4} \quad \text { New soln }
\end{aligned}
$$
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