If a satellite orbiting the Earth is 9 times closer to the Earth than the Moon, what is the time period of rotation of the satellite? Given rotational time period of Moon =27 days and gravitational attraction between the satellite and the moon is neglected.

We begin with a satellite orbiting Earth that is 9 times closer than the Moon and seek its time period.

We know from Kepler’s third law that $$T^2 \propto r^3$$.

Since the satellite is 9 times closer: $$r_s = \frac{r_m}{9}$$.

Substituting into the proportionality gives $$\frac{T_s^2}{T_m^2} = \frac{r_s^3}{r_m^3} = \left(\frac{1}{9}\right)^3 = \frac{1}{729}$$.

Taking the square root yields $$T_s = T_m \times \frac{1}{\sqrt{729}} = 27 \times \frac{1}{27} = 1$$ day.

The correct answer is Option 2: 1 day.