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

If $v_e$ is the escape velocity then

$\frac{1}{2}{v}_e^2=\frac{GMm}{R}$

Now the projectile is fired with speed $k{v}_e$.

Then loss in kinetic energy is gain in potential energy

$⟹\frac{1}{2}m(k{v}_e{)}^2=GMm(\frac{1}{R}-\frac{1}{h})$

$⟹k^2\left(\frac{GMm}{R}\right)=GMm(\frac{1}{R}-\frac{1}{h})$

$⟹h=\frac{R}{1-k^2}$.