Explanation (Energy Conservation)

Initial separation = 4R

Final separation when just touching = 2R

Total initial kinetic energy:

\[ K_i = 2 \times \frac{1}{2}mu^2 = mu^2 \]

Electrostatic potential energy:

\[ U_e = \frac{kQ^2}{r} \]

Gravitational potential energy:

\[ U_g = -\frac{Gm^2}{r} \]

Total potential energy:

\[ U = \frac{kQ^2}{r} - \frac{Gm^2}{r} \]

Initial potential energy at r = 4R:

\[ U_i = \frac{kQ^2}{4R} - \frac{Gm^2}{4R} \]

Final potential energy at r = 2R:

\[ U_f = \frac{kQ^2}{2R} - \frac{Gm^2}{2R} \]

Using energy conservation (just touching → final KE = 0):

\[ mu^2 + U_i = U_f \]

\[ mu^2 = U_f - U_i \]

\[ mu^2 = \left(\frac{kQ^2}{2R} - \frac{Gm^2}{2R}\right) - \left(\frac{kQ^2}{4R} - \frac{Gm^2}{4R}\right) \]

\[ mu^2 = \frac{kQ^2}{4R} - \frac{Gm^2}{4R} \]

\[ u^2 = \frac{1}{m}\left(\frac{kQ^2 - Gm^2}{4R}\right) \]

\[ u = \sqrt{\frac{kQ^2}{4mR}\left(1 - \frac{Gm^2}{kQ^2}\right)} \]

Final Answer matches Option D

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