Question

A body is projected horizontally near the surface of the earth with $\sqrt{1.5}$ times the orbital velocity. The maximum height up to which it will rise above the surface of the earth is 'n' times the radius of earth, then find the value of n.

Answer 1

Apply angular momentum and energy conservation.
By angular momentum conservation,
$m{v}_{0}R=m{v}^{\prime }{R}^{\prime }$
$\frac{v_{0}R}{R^{\prime }}=v^{\prime }$ ...(i)
$v_0=\sqrt{1.5}\sqrt{\frac{GM}{R}}$
$R^{\prime }\to$ Maximum height
By energy conservation,
$-\frac{GMm}{R}+\frac{1}{2}m{V}_0^2$
$=-\frac{GMm}{R^{\prime }}+\frac{1}{2}m{v}^{\prime 2}$ ...(ii)
From equation (i) and (ii),
$\frac{-GMm}{R}+\frac{1}{2}m\times 1.5\times \frac{GM}{R}$
$=\frac{-GMm}{R^{\prime }}+\frac{1}{2}m\times 1.5\times \frac{GM}{R}\times {\left(\frac{R}{R^{\prime }}\right)}^2$
$\Rightarrow \frac{-GMm}{R}+\frac{3}{4}\frac{GMm}{R}=\frac{-GMm}{R^{\prime }}+\frac{3}{4}\frac{GMmR}{(R^{\prime }{)}^2}$
$\Rightarrow \frac{-1}{4}\frac{GMm}{R}=\frac{-GMm}{R^{\prime }}+\frac{3}{4}\frac{GMm}{R}\left(\frac{R^2}{(R^{\prime }{)}^2}\right)$
$R^{\prime 2}-4R{R}_1+3R^2=0$
$R^{\prime }=3R$
$h=R^{\prime }-R=3R-R$
$h=2R$

Why this question? Tip:Angular momentum conservation can always be applied in binary mass systems, since the gravitational force is central in nature.