Q74 / 105 · JEE Main 2008 · Physics
A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is $11 \mathrm{kms}^{-1}$, the escape velocity from the surface of the planet would be
- A$1.1 \mathrm{kms}^{-1}$
- B$11 \mathrm{kms}^{-1}$
- Correct answerC$110 \mathrm{kms}^{-1}$
- D$0.11 \mathrm{kms}^{-1}$
Solution
$$
v_{\mathrm{esc}}=\sqrt{\frac{2 \mathrm{GM}}{R}}=\sqrt{\frac{2 \mathrm{G} \times 10 \mathrm{M}}{\mathrm{R} / 10}}=10 \times 11=110 \mathrm{~km} / \mathrm{s}
$$
Hence, the correct answer is option: 'C'.