Gravitation

JEE Main level, single correct, medium

110 of 223

Question 110 of 223

Earth has mass 8 times and radius 2 times that of a planet. If the escape velocity from the earth is 11.2 km/s11.2 \mathrm{~km} / \mathrm{s}, the escape velocity in km/s\mathrm{km} / \mathrm{s} from the planet will be :

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Answer: (C) 5.65.6

Vescape =2GMR(Vescape )Planet (Vescape )Earth =(MPME)×(RERP)=12( Vescape )Planet =12( Vescape )Earth =5.6 km/s\begin{aligned} & \mathrm{V}_{\text {escape }}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \\ & \frac{\left(\mathrm{V}_{\text {escape }}\right)_{\text {Planet }}}{\left(\mathrm{V}_{\text {escape }}\right)_{\text {Earth }}}=\sqrt{\left(\frac{\mathrm{M}_{\mathrm{P}}}{\mathrm{M}_{\mathrm{E}}}\right) \times\left(\frac{\mathrm{R}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{P}}}\right)}=\frac{1}{2} \\ & \left(\mathrm{~V}_{\text {escape }}\right)_{\text {Planet }}=\frac{1}{2}\left(\mathrm{~V}_{\text {escape }}\right)_{\text {Earth }}=5.6 \mathrm{~km} / \mathrm{s}\end{aligned}

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