Consider two identical metallic spheres of radius R each having charge Q and mass m. Their centers have an initial separation of 4R. Both the spheres are given an initial speed of u towards each other. The minimum value of u, so that they can just touch each other is :
(Take $$k= \frac{1}{4\pi \epsilon_{0}}$$ and assume $$kQ^{2}$$ > $$Gm^{2}$$ where G is the Gravitational constant)

Two identical metallic spheres each of radius $$R$$, charge $$Q$$, and mass $$m$$ are initially separated by a center-to-center distance of $$4R$$ and are given an initial speed $$u$$ towards each other; we seek the minimum value of $$u$$ for them to just touch.

We denote the initial separation as $$d_i = 4R$$ and the final separation upon just touching as $$d_f = 2R$$.

The interaction between the spheres involves electrostatic repulsion with potential energy $$U_e = \frac{kQ^2}{d}$$ and gravitational attraction with potential energy $$U_g = -\frac{Gm^2}{d}$$, so that the total potential energy at separation $$d$$ is $$ U(d) = \frac{kQ^2}{d} - \frac{Gm^2}{d}. $$

Since the spheres have equal mass and equal and opposite velocities, they momentarily come to rest at the point of touching, making the final kinetic energy zero. By conservation of energy, $$ \text{Initial KE} + \text{Initial PE} = \text{Final KE} + \text{Final PE}. $$

The initial kinetic energy is $$2\times \frac12 mu^2 = mu^2$$ and the initial potential energy at $$d_i=4R$$ is $$\frac{kQ^2}{4R} - \frac{Gm^2}{4R}$$. The final potential energy at $$d_f=2R$$ is $$\frac{kQ^2}{2R} - \frac{Gm^2}{2R}$$, and the final kinetic energy is zero. Hence, $$ mu^2 + \frac{kQ^2}{4R} - \frac{Gm^2}{4R} = \frac{kQ^2}{2R} - \frac{Gm^2}{2R}. $$

Rearranging to isolate $$mu^2$$ gives $$ mu^2 = \frac{kQ^2}{2R} - \frac{Gm^2}{2R} - \frac{kQ^2}{4R} + \frac{Gm^2}{4R} = kQ^2\Bigl(\frac{1}{2R}-\frac{1}{4R}\Bigr) - Gm^2\Bigl(\frac{1}{2R}-\frac{1}{4R}\Bigr) = \frac{1}{4R}\bigl(kQ^2 - Gm^2\bigr). $$ Dividing by $$m$$ yields $$ u^2 = \frac{kQ^2 - Gm^2}{4mR} = \frac{kQ^2}{4mR}\Bigl(1 - \frac{Gm^2}{kQ^2}\Bigr), $$ and thus $$ u = \sqrt{\frac{kQ^2}{4mR}\Bigl(1 - \frac{Gm^2}{kQ^2}\Bigr)}. $$

The correct answer is Option (4): $$\sqrt{\frac{kQ^{2}}{4mR}\left(1-\frac{Gm^{2}}{kQ^{2}}\right)}$$.