Physics - Gravitation Question with Solution | TestHub

To escape the gravitational pull $F=\frac{GMm}{r^2}$, the body needs to be at $\begin{array}{rl}& r\to \infty \\ & \Rightarrow -\left(\frac{GMm}{R}\right)+\frac{1}{2}m{\left(V_e\right)}^2=0\\ & \Rightarrow V_e=\sqrt{\frac{2GM}{R}}\end{array}$ Therefore, if the particle is projected with $2V_e$ from the surface of earth, then at $r\to \propto$ the velocity of the projectile is $V$. Considering energy conservation: $\begin{array}{rl}& \Rightarrow -\left(\frac{GMm}{R}\right)+\frac{1}{2}m{\left(2V_e\right)}^2=0+\frac{1}{2}m{V}^2\\ & \Rightarrow -\frac{GM}{\mathrm{R}}+\frac{1}{2}{\left(2V_e\right)}^2=\frac{1}{2}{V}^2\\ & \Rightarrow -\frac{GM}{\mathrm{R}}+\frac{4GM}{R}=\frac{1}{2}{V}^2\\ & \therefore V=\sqrt{\frac{6GM}{R}}=\sqrt{3}\left(\sqrt{\frac{2GM}{R}}\right)=\sqrt{3}{V}_e\end{array}$