QJEEAdv17Phab61: JEE Advanced 2017 Physics - Class 11 JEEAdv17Phab61 Solution: A rocket is launched normal to the surface of the Earth, away…

Given : Mass of the sun, M_s = 3 $\times$ 10⁵ $\times$ mass of the earth = 3 $\times$ 10⁵ M_e

Distance between the sun and the earth,

d = 2.5 $\times$ 10⁴ $\times$ radius of the earth = 2.5 $\times$ 10⁴ R_e,

Escape speed, v_e = 11.2 km s ^{$-$ 1}

$\therefore$ ${v_e} = \sqrt {{{2G{M_e}} \over {{R_e}}}} = 11.2$ km s ^{$-$ 1}

Minimum velocity required for the rocket to escape the given earth $-$ sun system = v_s

Using energy conservation for the given situation,

${1 \over 2}mv_s^2 = {{G{M_e}m} \over {{R_e}}} + {{G{M_s}m} \over {d + {R_e}}} = {{G{M_e}m} \over {{R_e}}} + {{G \times 3 \times {{10}^5}{M_e}m} \over {(2.5 \times {{10}^4}{R_e} + {R_e})}}$

$\therefore$ $v_s^2 = {{2G{M_e}} \over {{R_e}}} + {{24G{M_e}} \over {{R_e}}}$ ( $\because$ 2.5 $\times$ 10⁴ >> 1)

$\Rightarrow v_s^2 = v_e^2 + 12v_e^2 = 13v_e^2$

or ${v_s} = \sqrt {13}$ v_e = 40.38 km s ^{$-$ 1} $\approx$ 42 km s ^{$-$ 1}