Physics - Gravitation Question with Solution | TestHub

By conservation of mechanical energy:

$K_i+U_i=K_f+U_f$

$\frac{1}{2}m{v}^2-\frac{GMm}{R}=0-\frac{GMm}{R+h}$

Given $v=K{V}_e$ and $V_e=\sqrt{\frac{2GM}{R}}$, we have $v^2=K^2\frac{2GM}{R}$.

Substituting $v^2$ in the energy equation:

$\frac{1}{2}m\left(K^2\frac{2GM}{R}\right)-\frac{GMm}{R}=-\frac{GMm}{R+h}$

$\frac{GMm{K}^2}{R}-\frac{GMm}{R}=-\frac{GMm}{R+h}$

Dividing by $GMm$:

$\frac{K^2-1}{R}=-\frac{1}{R+h}$

$\frac{1-K^2}{R}=\frac{1}{R+h}$

$R+h=\frac{R}{1-K^2}$

$h=\frac{R}{1-K^2}-R=\frac{R{K}^2}{1-K^2}$

Answer: $\frac{R{K}^2}{1-K^2}$