A particle of mass ‘m’ is projected with a velocity v=kVe (k < 1) from the surface of the earth. (Ve=escape velocity) The maximum height above the surface reached by the particle is :

Correct Answer/Option:
$\frac{R{k}^2}{1-k^2}$

Step-by-Step Derivation:

Let the Earth be a uniform sphere of mass M and radius R.
A particle of mass m is projected from the surface of the Earth with a velocity:
$v=k{v}_e$
where $v_e$ is the escape velocity from the Earth's surface, defined as:
$v_e=\sqrt{\frac{2GM}{R}}$

The initial total mechanical energy ($E_i$) of the particle at the Earth's surface (at distance R from the center) is the sum of its kinetic energy and gravitational potential energy:
$E_i=\frac{1}{2}m{v}^2-\frac{GMm}{R}$

Substitute the value of velocity $v=k{v}_e$ into the energy equation:
$E_i=\frac{1}{2}m{\left(k{v}_e\right)}^2-\frac{GMm}{R}$
$E_i=\frac{1}{2}m{k}^2\left(\frac{2GM}{R}\right)-\frac{GMm}{R}$
$E_i=\frac{k^{2}GMm}{R}-\frac{GMm}{R}$
$E_i=-\frac{GMm}{R}(1-k^2)$

Let the maximum height reached by the particle above the surface of the Earth be h. At this maximum height, the particle momentarily stops, meaning its final kinetic energy is zero.
The distance of the particle from the center of the Earth at this point is $R+h$.
Therefore, the final total mechanical energy ($E_f$) is solely potential energy:
$E_f=-\frac{GMm}{R+h}$

By the principle of conservation of mechanical energy, $E_i=E_f$:
$-\frac{GMm}{R}(1-k^2)=-\frac{GMm}{R+h}$

Canceling the common term $-GMm$ from both sides gives:
$\frac{1-k^2}{R}=\frac{1}{R+h}$

Taking the reciprocal of both sides:
$R+h=\frac{R}{1-k^2}$

Solving for h:
$h=\frac{R}{1-k^2}-R$
$h=R(\frac{1}{1-k^2}-1)$
$h=R\left(\frac{1-(1-k^2)}{1-k^2}\right)$
$h=\frac{R{k}^2}{1-k^2}$