Question
The diameters of two planets are in the ratio $4: 1$ and the ratio of their density is $1: 2$. What will be the ratio of gravitational accelerations on the planets?
✓
Answer
We know that,
$\begin{array}{l}
g_1=\frac{GM_1}{R_1{ }^2}.....(1) \\
g_2=\frac{GM_2}{R_2{ }^2}......(2) \\
\frac{g_1}{g_2}=\frac{M_1}{M_2} \times\left(\frac{R_2}{R_1}\right)^2 \\
=\frac{\frac{4}{3} \pi R_1{ }^3 \cdot d_1}{\frac{4}{3} \pi R_2{ }^3 \cdot d_2} \times \frac{\left(R_2\right)^2}{\left(R_1\right)^2} \\
\because \text { Mass }=\text { Volume } \times \text { density } \\
\frac{g_1}{g_2}=\frac{R_1}{R_2} \times \frac{d_1}{d_2} \\
\text { For diameters }=\frac{2 R_1}{2 R_2} \times \frac{d_1}{d_2}
\end{array}$
(Here, $d_1$ and $d_2$ are densitites.)
$\begin{array}{cc}
& \frac{g_1}{g_2}=\frac{4}{1} \times \frac{1}{2}=\frac{2}{1} \\
\because & \text { Ratio of diameters }=4: 1 \\
& \text { and Ratio of densities }=1: 2 \\
\therefore \quad & g_1: g_2=2: 1 \quad Ans.
\end{array}$
$\begin{array}{l}
g_1=\frac{GM_1}{R_1{ }^2}.....(1) \\
g_2=\frac{GM_2}{R_2{ }^2}......(2) \\
\frac{g_1}{g_2}=\frac{M_1}{M_2} \times\left(\frac{R_2}{R_1}\right)^2 \\
=\frac{\frac{4}{3} \pi R_1{ }^3 \cdot d_1}{\frac{4}{3} \pi R_2{ }^3 \cdot d_2} \times \frac{\left(R_2\right)^2}{\left(R_1\right)^2} \\
\because \text { Mass }=\text { Volume } \times \text { density } \\
\frac{g_1}{g_2}=\frac{R_1}{R_2} \times \frac{d_1}{d_2} \\
\text { For diameters }=\frac{2 R_1}{2 R_2} \times \frac{d_1}{d_2}
\end{array}$
(Here, $d_1$ and $d_2$ are densitites.)
$\begin{array}{cc}
& \frac{g_1}{g_2}=\frac{4}{1} \times \frac{1}{2}=\frac{2}{1} \\
\because & \text { Ratio of diameters }=4: 1 \\
& \text { and Ratio of densities }=1: 2 \\
\therefore \quad & g_1: g_2=2: 1 \quad Ans.
\end{array}$
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