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  • If earth has a mass nine times and radius twice to that of a planet P. Then  <img alt="\frac{v_e}{3} \sqrt{x} \mathrm{~ms}^{-1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfr

If earth has a mass nine times and radius twice to that of a planet P. Then  \frac{v_e}{3} \sqrt{x} \mathrm{~ms}^{-1}  will be the

minimum velocity required by a rocket to pull out of gravitational force of P, where  v_e  is escape velocity on earth. The value of x is

Option: 1

1


Option: 2

3


Option: 3

18


Option: 4

2


Answers (1)

best_answer

\begin{aligned} & \mathrm{M}_{\mathrm{E}}=9 \mathrm{M}_{\mathrm{P}} \\ & \mathrm{R}_{\mathrm{E}}=2 \mathrm{R}_{\mathrm{P}} \\ & \mathrm{V}_{\mathrm{c}}^1=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{P}}}{R_{\mathrm{P}}}}=\sqrt{\frac{2 \mathrm{G} \frac{\mathrm{M}_{\mathrm{E}}}{9}}{\frac{\mathrm{R}_{\mathrm{E}}}{2}}} \\ & =\sqrt{\frac{2 \mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}} \times \sqrt{\frac{2}{9}} \\ & V_c^1=\frac{V_c}{3} \sqrt{2} \\ & \end{aligned}

 

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