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(C) Given,escape velocity on the surface of the earth is $v_e = \sqrt{\frac{2GM_e}{R_e}} = 11.2 \ km/s$. At an altitude $h = \frac{R_e}{3}$,the distan

Vedclass · Accepted Answer

$9.7$

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(C) Given,escape velocity on the surface of the earth is $v_e = \sqrt{\frac{2GM_e}{R_e}} = 11.2 \ km/s$. At an altitude $h = \frac{R_e}{3}$,the distance from the center of the earth is $r = R_e + h = R_e + \frac{R_e}{3} = \frac{4R_e}{3}$. To escape from this point,the kinetic energy of the rocket must be equal to the magnitude of the gravitational potential energy at that distance. $\frac{1}{2}mv_{e1}^2 = \frac{GM_em}{r} = \frac{GM_em}{4R_e/3} = \frac{3GM_em}{4R_e}$. Since $v_e^2 = \frac{2GM_e}{R_e}$,we have $\frac{GM_e}{R_e} = \frac{v_e^2}{2}$. Substituting this into the energy equation: $\frac{1}{2}v_{e1}^2 = \frac{3}{4} \left(\frac{v_e^2}{2}ight) = \frac{3}{8}v_e^2$. $v_{e1}^2 = \frac{3}{4}v_e^2 \Rightarrow v_{e1} = \frac{\sqrt{3}}{2}v_e$. Given $v_e = 11.2 \ km/s$,$v_{e1} = \frac{1.732}{2} 	imes 11.2 = 0.866 	imes 11.2 \approx 9.7 \ km/s$.

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