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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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The eacape velocity from earth is 11km/s. The escape velocity from a planet having twice the radius and same mean density as that of earth is

\[(a)\;22\;km/s \quad (b)\;11\;km/s \quad (c)\;5.5 \;Km/s \quad (d)\;none\;of\;these\]
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1 Answer

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Escape Velocity $v_e=\sqrt {\large\frac{2GM_e}{R_e}}$
$\qquad= \sqrt {\large\frac{2G \bigg(\Large\frac{4}{3} \large \pi \rho R_e^3d\bigg)}{R_e}}$
$\qquad=\sqrt {2G \bigg(\large\frac{4}{3} \normalsize \pi \rho R_e^2\bigg)d}$ => $v_e \alpha R$
Since density is same.
When radius is doubled escape velocity gets doubled
$2v_e=2 \times 11 =22 \;km/s$
Hence a is the correct answer.
answered Aug 12, 2013 by meena.p
edited Jun 26, 2014 by lmohan717

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