Gravitation Ques 2
- Gravitational acceleration on the surface of a planet is $\frac{\sqrt{6}}{11} g$, where $g$ is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $1 / 3$ times that of the earth. If the escape speed on the surface of the earth is taken to be $11 \mathrm{kms}^{-1}$, the escape speed on the surface of the planet in $\mathrm{kms}^{-1}$ will be
(2010)
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Answer:
Correct Answer: 2.( 3 )
Solution:
- $g=\frac{G M}{R^2}=\frac{G\left(\frac{4}{3} \pi R^3\right) \rho}{R^2}$ or $g \propto \rho R \quad$ or $\quad R \propto \frac{g}{\rho}$ Now escape velocity, $v_e=\sqrt{2 g R}$ or
$ \begin{gathered} \text { or } \quad v_e \propto \sqrt{g R} \text { or } v_c \propto \sqrt{g \times \frac{g}{\rho}} \propto \sqrt{\frac{g^2}{\rho}} \\ \therefore \quad\left(v_c\right)_{\text {planet }}=\left(11 \mathrm{~km} \mathrm{~s}^{-1}\right) \sqrt{\frac{6}{121} \times \frac{3}{2}}=3 \mathrm{~km} \mathrm{~s}^{-1} \end{gathered} $
$\therefore$ The correct answer is 3 .
BETA
