MHT CET · Physics · Gravitation
If the density of a small planet is the same as that of earth, while the radius of the planet is \(0.2\) times that the earth, the gravitational acceleration on the surface of that planet is
- A \(0.2 \mathrm{~g}\)
- B \(0.4 \mathrm{~g}\)
- C \(2 \mathrm{~g}\)
- D \(4 \bar{g}\)
Answer & Solution
Correct Answer
(A) \(0.2 \mathrm{~g}\)
Step-by-step Solution
Detailed explanation
\(g=\frac{4}{3} \pi G R \rho\)
and
\(g^{\prime}=\frac{4}{3} \pi G R^{\prime} \rho\)
\(\begin{array}{l}\therefore\frac{g^{\prime}}{g}=\frac{R^{\prime}}{R}=0.2 \\g^{\prime}=0.2 g\end{array}\)
and
\(g^{\prime}=\frac{4}{3} \pi G R^{\prime} \rho\)
\(\begin{array}{l}\therefore\frac{g^{\prime}}{g}=\frac{R^{\prime}}{R}=0.2 \\g^{\prime}=0.2 g\end{array}\)
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