Question

The initial velocity $v_1$ required to project a body vertically upward from the surface of the earth to just reach a height of $10R$, where $R$ is the radius of the earth, may be described in terms of escape velocity $v_e$ such that $v_i=\sqrt{\frac{x}{y}}\times v_e$.The value of $x$ will be_______.

Answer 1

Here $R=$radius of the earth
From energy conservation
$-\frac{G{m}_{e}m}{R}+\frac{1}{2}m{v}_i^2=-\frac{G{m}_{e}m}{11R}+0$

$\frac{1}{2}m{v}_i^2=\frac{10}{11}\frac{G{m}_{e}m}{R}$

$v_i=\sqrt{\frac{20}{11}\frac{G{m}_e}{R}}$

$v_i=\sqrt{\frac{20}{11}}{v}_e$
$[\because \text{escape velocity}{v}_e=\sqrt{\frac{G{m}_e}{R}}]$
Then the value of $x=20$