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(B) Let $M$ be the mass of the Earth and $R$ be its radius. The mass of the Sun is $M_s = 3 	imes 10^5 M$ and the distance of the Sun from the Earth

Vedclass · Accepted Answer

$v_s = 42 	ext{ km s}^{-1}$

Vedclass · Answer

(B) Let $M$ be the mass of the Earth and $R$ be its radius. The mass of the Sun is $M_s = 3 	imes 10^5 M$ and the distance of the Sun from the Earth is $d = 2.5 	imes 10^4 R$.The total energy of the rocket at the surface of the Earth is $E = \frac{1}{2}mv_s^2 - \frac{GMm}{R} - \frac{GM_sm}{d}$.For the rocket to escape the system,the final energy must be at least $0$.$\frac{1}{2}mv_s^2 = \frac{GMm}{R} + \frac{G(3 	imes 10^5 M)m}{2.5 	imes 10^4 R}$.Since $v_e^2 = \frac{2GM}{R}$,we have $\frac{GM}{R} = \frac{v_e^2}{2}$.Substituting this into the energy equation: $\frac{1}{2}v_s^2 = \frac{v_e^2}{2} + \frac{3 	imes 10^5}{2.5 	imes 10^4} 	imes \frac{v_e^2}{2}$.$v_s^2 = v_e^2 + 12 v_e^2 = 13 v_e^2$.$v_s = v_e \sqrt{13} = 11.2 	imes 3.605 \approx 40.38 	ext{ km s}^{-1}$.The closest value is $42 	ext{ km s}^{-1}$.

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