The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value then its new time period will become

Kepler's Third Law establishes $T^2 \propto r^3$ for orbital motion. Substituting the given initial period $T_1 = 24$ hours and radius ratio $\frac{r_2}{r_1} = \frac{1}{4}$ into $\frac{T_2^2}{T_1^2} = \left(\frac{r_2}{r_1}\right)^3$ yields:

$T_2^2 = (24)^2 \cdot \left(\frac{1}{4}\right)^3 = 576 \cdot \frac{1}{64} = 9$

Taking the square root gives the new period $T_2 = 3$ hours.