The Earth is assumed to be a sphere of radius $R$. A platform is arranged at a height $R$ from the surface of the Earth. The escape velocity of a body from this platform is $f v$, where $v$ is its escape velocity from the surface of the Earth. The value of $f$ is:

Question

Aicharya · Accepted Answer

We know that Escape velocity
$V_e=\sqrt{2 g R}$
Escape velocity of the body from platform
$\begin{aligned}
& \text {P. } \mathrm{E}+\mathrm{K} \cdot \mathrm{E}=0 \\
& -\frac{G M m}{2 \mathrm{R}}+\frac{1}{2} m V^2=0 \\
& \Rightarrow \quad V=\sqrt{\frac{G M}{R^2} \cdot R}=\sqrt{g R}
\end{aligned}$
This confirms that:
$\begin{aligned}
& fv =\frac{V}{\sqrt{2}} \\
\therefore f & =\frac{1}{\sqrt{2}} .
\end{aligned}$ .

The Earth is assumed to be a sphere of radius $R$. A platform is arranged at a height $R$ from the surface…

Solution