Question

A body is launched from the earth's surface an angle $\alpha ={30}^o$ to the horizontal at a speed $v_0=\sqrt{\frac{1.5GM}{R}}$. Neglecting air resistance and earth's rotation, Radius of curvature of trajectory at its top point is $\frac{xR}{100}$. The value of $x$ is:

Answer 1

Conserving angular momentum about the center of the earth, we get

$m{v}_{0}Rsin\alpha =mvx$

$vx=v_{0}Rsin\alpha$

$vx=\frac{3}{2}\sqrt{\frac{GMR}{2}}$

Also, from energy conservation,

$\frac{3GMm}{4R}-\frac{GMm}{R}=\frac{1}{2}m{v}^2-\frac{GMm}{x}$

$\frac{1}{2}m{v}^2-\frac{GMm}{x}=-\frac{1}{4}\frac{GMm}{R}$

Radius of curvature will be

$R_C=\frac{m{v}^2}{GMm/x^2}=\frac{v^2{x}^2}{GM}=\frac{9}{8}R=1.13R$

Answer is $x=113$