Question

A projectile is fired vertically from the Earth with a velocity $k{v}_e$ where $v_e$ is the escape velocity and $k$ is a constant less than unity. The maximum height to which projectile rises, as measured from the centre of Earth, is

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

Answer 1

Answer: (C)

$\frac{R}{1-k^2}$

Applying conservation of energy principle, we get
$\frac{1}{2}m{k}^2{v}_e^2-\frac{GMm}{R}=-\frac{GMm}{r}$
$\Rightarrow \frac{1}{2}m{k}^2\frac{2GM}{R}-\frac{GMm}{R}=-\frac{GMm}{r}$
$\Rightarrow \frac{k^2}{R}-\frac{1}{R}=-\frac{1}{r}\Rightarrow \frac{1}{r}=\frac{1}{R}-\frac{k^2}{R}$
$\Rightarrow \frac{1}{r}=\frac{1}{R}(1-k^2)\Rightarrow r=\frac{R}{1-k^2}$