For a satellite revolving around Earth, Kepler’s third law gives $T^2 \propto r^3$ where $T$ is the time period and $r$ is the distance (orbital radius) from the center of Earth.

Initially, $T_1 = 24\text{ hours}$ Let the initial separation be $r_1$.

The new separation is decreased to one fourth of the previous value: $r_2 = \frac{r_1}{4}$

Using $\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{r_2}{r_1}\right)^3$ we get $\left(\frac{T_2}{24}\right)^2 = \left(\frac{1}{4}\right)^3 = \frac{1}{64}$

Taking square root, $\frac{T_2}{24} = \frac{1}{8}$ so $T_2 = 24 \times \frac{1}{8} = 3\text{ hours}$

Therefore, the new time period is $\boxed{3\text{ hours}}$

Checking options:

• A: $12$ hours ❌
• B: $3$ hours ✅
• C: $6$ hours ❌
• D: $4$ hours ❌