Question

A projectile is fired vertically upwards from the surface of the earth with a velocity $K{V}_e$, where $V_e$ is the escape velocity and $K<1$. If $R$ is the radius of the earth, the maximum height to which it will rise measured from the centre of the earth will be (neglect air resistance) :

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

Answer 1

The correct option is B $\frac{R}{1-K^2}$

Let the height attained by the projectile above the earth surface be $h$ and radius of earth be $R$.

Thus distance of maximum height attained from the center of earth $r=R+h$ ....(1)

Potential energy of the projectile at maximum height $P.E=\frac{mgh}{1+\frac{h}{R}}$

According to the law of conservation of energy : $K.E=P.E$
$\frac{1}{2}m{V}^2=\frac{mgh}{1+\frac{h}{R}}$
$\frac{1}{2}m(K{V}_e{)}^2=\frac{mg(r-R)}{1+\frac{(r-R)}{R}}$ (using 1)

Escape velocity at earth's surface $V_e=\sqrt{2gR}$
$\frac{1}{2}m{K}^2\left(2gR\right)=\frac{mg(r-R)}{1+\frac{(r-R)}{R}}$
$K^{2}R[1+\frac{(r-R)}{R}]=r-R$
$K^{2}r=r-R$
$R=(1-K^2)r$
$⟹r=\frac{R}{1-K^2}$