Given, escape velocity on the surface of the earth.
\(v_e=\sqrt{2 g R_e}=11.2 \mathrm{~km} / \mathrm{s}\)...(i)
where, \(R\) is the radius of earth.
Height of the rocket from the surface of earth, \(h=\frac{R_e}{3}\). If the rocket has to escape from the gravitational pull of the earth from \(h=\frac{R_\epsilon}{3}\). Then, the potential energy at the altitude,
\(\begin{aligned}
\frac{R_e}{3} & =\text { kinetic energy } \\
\frac{G M_e m}{R_e+h} & =\frac{1}{2} m v_{e_1}^2 \Rightarrow \frac{G M_e}{R_e+\frac{R_e}{3}}=\frac{1}{2} v_{e_1}^2
\end{aligned}\)
\(\left[v_{e_1}=\right.\) escape velocity of rocket from \(\left.\frac{R_e}{3}\right]\) \(\left(\begin{array}{rl}
M_e & =\text { mass of the earth and } \\
m & =\text { mass of the rocket }
\end{array}\right)\)
\(\begin{array}{rlrl}
\frac{g R_e^2}{4 R_e} & =\frac{1}{2} v_{e_1}^2 & \\
\Rightarrow \quad \frac{3}{4} g R_e & =\frac{v_{e_1}^2}{2} \quad \left[\because g=\frac{G M_e}{R_e^2}\right] \\
\frac{3}{4} \cdot 2 g R_e & =v_{e_1}^2 \\
\Rightarrow \quad \frac{\sqrt{3}}{2} \cdot \sqrt{2 g R_e} & =v_{e_1} \\
\frac{\sqrt{3}}{2} 11.2 & =v_{e_1} \quad \text {[From Eq. (i)] } \\
\Rightarrow \quad 9.7 & =v_{e_1} \\
\Rightarrow \quad v_{e_1} & =9.7 \mathrm{~km} / \mathrm{s}
\end{array}\)