Question

A projectile is projected with velocity $k{v}_e$ in vertically upward direction from the ground into the space.($v_e$ is escape velocity and k<1). If air resistance is considered to be negligible then the maximum height from the centre of earth to which it can go, will be : (R = radius of earth)

• A. $\frac{R}{k^2+1}$
• B. $\frac{R}{k^2-1}$
• C. $\frac{R}{1-k^2}$
• D. $\frac{R}{k+1}$

Answer 1

The correct option is C $\frac{R}{1-k^2}$

Decrease in Kinetic energy = Increase in Potential energy
$\frac{1}{2}m(k{v}_e{)}^2=\frac{mgh}{1+\frac{h}{R}}\Rightarrow \frac{1}{2}m{k}^{2}2gR=\frac{mgh}{1+\frac{h}{R}}\Rightarrow h=\frac{R{k}^2}{1-k^2}$
Height of Projectile from the earth's surface = h
Height from the centre $r=R+h=R+\frac{R{k}^2}{1-k^2}$
By solving $r=\frac{R}{1-k^2}$