A body is projected vertically, upwards from earth's surface with velocity $2 V_e$ where $V_e$ is the escape…
A body is projected vertically, upwards from earth's surface with velocity $2 V_e$ where $V_e$ is the escape velocity from earth's surface. The velocity when body escapes the gravitational pull is
$\sqrt{3} V_e$
$V_e$
$\sqrt{2} V_e$
$\frac{V_e}{\sqrt{3}}$
Solution
To escape the gravitational pull $F=\frac{G M m}{r^2}$, the body needs to be at
$\begin{aligned} & r \rightarrow \infty \\ & \Rightarrow-\left(\frac{G M m}{R}\right)+\frac{1}{2} m\left(V_e\right)^2=0 \\ & \Rightarrow V_e=\sqrt{\frac{2 G M}{R}}\end{aligned}$
Therefore, if the particle is projected with $2 V_e$ from the surface of earth, then at $r \rightarrow \propto$ the velocity of the projectile is $V$. Considering energy conservation:
$\begin{aligned} & \Rightarrow-\left(\frac{G M m}{R}\right)+\frac{1}{2} m\left(2 V_e\right)^2=0+\frac{1}{2} m V^2 \\ & \Rightarrow-\frac{G M}{\mathrm{R}}+\frac{1}{2}\left(2 V_e\right)^2=\frac{1}{2} V^2 \\ & \Rightarrow-\frac{G M}{\mathrm{R}}+\frac{4 G M}{R}=\frac{1}{2} V^2 \\ & \therefore V=\sqrt{\frac{6 G M}{R}}=\sqrt{3}\left(\sqrt{\frac{2 G M}{R}}\right)=\sqrt{3} V_e\end{aligned}$