1. 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 between the Earth and the satellite.

2. Therefore,

$T \propto r^{3/2}$

3. Initially, the time period is

$T_1 = 5\text{ hours}$

and the new separation is

$r_2 = 4r_1$

4. Using the proportionality,

$\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{3/2}$

Substitute $r_2/r_1 = 4$:

$\frac{T_2}{5} = 4^{3/2}$

5. Now,

$4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$

So,

$\frac{T_2}{5} = 8$

$T_2 = 5 \times 8 = 40\text{ hours}$

6. Hence, the new time period is

$\boxed{40\text{ hours}}$

So the correct option is C.