To solve this problem, we need to determine the maximum height reached by a particle projected with a velocity $v = k V_e$ from the surface of the Earth.$\\$The escape velocity $V_e$ is given by:$\\ V_e = \sqrt{\frac{2GM}{R}} \\$where $G$ is the gravitational constant, $M$ is the mass of the Earth, and $R$ is the radius of the Earth.$\\$The initial kinetic energy of the particle is:$\\ K.E._{initial} = \frac{1}{2} m (k V_e)^2 = \frac{1}{2} m k^2 V_e^2 \\$The initial potential energy of the particle at the Earth's surface is:$\\ P.E._{initial} = -\frac{GMm}{R} \\$At the maximum height, the particle's velocity is zero, so its kinetic energy is zero.$\\$Let $h$ be the maximum height reached above the Earth's surface.$\\$The potential energy at this height is:$\\ P.E._{final} = -\frac{GMm}{R + h} \\$By conservation of energy, the initial total energy equals the final total energy:$\\ \frac{1}{2} m k^2 V_e^2 - \frac{GMm}{R} = -\frac{GMm}{R + h} \\$Substitute $V_e^2 = \frac{2GM}{R}$ into the equation:$\\ \frac{1}{2} m k^2 \left(\frac{2GM}{R}\right) - \frac{GMm}{R} = -\frac{GMm}{R + h} \\$Simplify:$\\ m k^2 \frac{GM}{R} - \frac{GMm}{R} = -\frac{GMm}{R + h} \\$Factor out $GMm: \\ GMm \left(\frac{k^2}{R} - \frac{1}{R}\right) = -\frac{GMm}{R + h} \\$Cancel $GMm$ from both sides:$\\ \frac{k^2 - 1}{R} = -\frac{1}{R + h} \\$Rearrange to solve for $h: \\ R + h = -\frac{R}{k^2 - 1} \\ h = -\frac{R}{k^2 - 1} - R \\ h = R \left(\frac{-1}{k^2 - 1} - 1\right) \\ h = R \left(\frac{-1 - (k^2 - 1)}{k^2 - 1}\right) \\ h = R \left(\frac{-k^2}{k^2 - 1}\right) \\ h = \frac{R k^2}{1 - k^2} \\$Therefore, the maximum height above the surface reached by the particle is $\frac{R k^2}{1 - k^2}. \\$This corresponds to Option 4.