QJEEAdv10Ph4c5c: JEE Advanced 2010 Physics - Class 11 JEEAdv10Ph4c5c Solution: Gravitational acceleration on the surface of a planet is ,…

On the planet,

${g_p} = {{G{M_p}} \over {R_p^2}} = {G \over {R_p^2}}\left( {{4 \over 3}\pi R_p^3{\rho _p}} \right) = {4 \over 3}G\pi R_p^{}{\rho _p}$

On the earth,

${g_e} = {{G{M_e}} \over {R_e^2}} = {G \over {R_e^2}}\left( {{4 \over 3}\pi R_e^3{\rho _e}} \right) = {4 \over 3}G\pi R_e^{}{\rho _e}$

$\therefore$ ${{{g_p}} \over {{g_e}}} = {{R_p^{}{\rho _p}} \over {R_e^{}{\rho _e}}}$ or ${{{R_p}} \over {{R_e}}} = {{g_p^{}{\rho _p}} \over {g_e^{}{\rho _e}}}$ ..... (i)

On the planet, ${v_p} = \sqrt {2{g_p}{R_p}}$

On the earth, ${v_e} = \sqrt {2{g_e}{R_e}}$

$\therefore$ ${{{v_p}} \over {{v_e}}} = \sqrt {{{{g_p}{R_p}} \over {{g_e}{R_e}}}} = {{{g_p}} \over {{g_e}}}\sqrt {{{{\rho _e}} \over {{\rho _p}}}}$ (Using (i))

Here, ${\rho _p} = {2 \over 3}{\rho _e}$ , ${g_p} = {{\sqrt 6 } \over {11}}{g_e}$

$\therefore$ ${{{v_p}} \over {{v_e}}} = {{\sqrt 6 } \over {11}}\sqrt {{3 \over 2}}$

or, ${v_p} = 11 \times {{\sqrt 6 } \over {11}} \times \sqrt {{3 \over 2}}$ ( $\because$ v_e = 11 km s ^{$-$ 1} (Given))

= 3 km s ^{$-$ 1}