Maximum height reached by a rocket fired with a speed equal to $50 \%$ of the escape speed from the surface of the earth is ( $R$ - Radius of the earth)

Question

Aicharya · Accepted Answer

$\mathrm{v}_{\mathrm{i}}=\frac{1}{2} \mathrm{v}_{\mathrm{e}}=\frac{1}{2} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}, \mathrm{U}_{\mathrm{f}}=0$ Applying conservation of mechanical energy, $K_i+V_i=K_f+U_f$ $\begin{aligned} & \Rightarrow \frac{1}{2} \mathrm{mv}_{\mathrm{i}}^2+\frac{-\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}}=0+\frac{-\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}+\mathrm{h}} \ & \Rightarrow \frac{1}{2} \mathrm{~m}\left(\frac{1}{2} \frac{2 \mathrm{GM}_{\mathrm{m}}}{\mathrm{R}}ight)-\frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}}=-\frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}+\mathrm{h}} \ & \Rightarrow \frac{3}{4 \mathrm{R}}=\frac{1}{\mathrm{R}+\mathrm{h}} \ & 	herefore \mathrm{h}=\frac{\mathrm{R}}{3}\end{aligned}$

Maximum height reached by a rocket fired with a speed equal to $50 \%$ of the escape speed from the surface…

Solution