An earth satellite S has an orbit radius which is 4 times that of a communication satellite C. The period of revolution of S is

To find the period of revolution of the satellite $S$, we can apply Kepler's Third Law of Planetary Motion.

Kepler's Third Law states that the square of the orbital period ($T$) of a satellite is directly proportional to the cube of its orbital radius ($r$):
$T^2\propto r^3$

Let the orbital radius of the communication satellite $C$ be $r_C$ and its period of revolution be $T_C$.
A standard geostationary communication satellite has an orbital period of 1 day:
$T_C=1\text{day}$

Let the orbital radius of the satellite $S$ be $r_S$ and its period of revolution be $T_S$.
We are given that the orbital radius of $S$ is 4 times that of $C$:
$r_S=4r_C$

Using the ratio form of Kepler's Third Law:
${\left(\frac{T_S}{T_C}\right)}^2={\left(\frac{r_S}{r_C}\right)}^3$

Substitute the given relationship $\frac{r_S}{r_C}=4$ into the equation:
${\left(\frac{T_S}{T_C}\right)}^2=4^3$
${\left(\frac{T_S}{T_C}\right)}^2=64$

Taking the square root on both sides:
$\frac{T_S}{T_C}=\sqrt{64}$
$\frac{T_S}{T_C}=8$

Now, solving for $T_S$:
$T_S=8\times T_C$
Since $T_C=1\text{day}$:
$T_S=8\times 1\text{day}=8\text{days}$

Therefore, the period of revolution of the satellite S is 8 days.