1. Use Kepler’s third law for bodies orbiting the same central mass

For a satellite and the Moon both revolving around the Earth,

$T^2 \propto r^3$

So,

$\left(\frac{T_s}{T_m}\right)^2 = \left(\frac{r_s}{r_m}\right)^3$

where:

• $T_s$ = time period of satellite
• $T_m = 27$ days = time period of Moon
• $r_s$ = orbital radius of satellite
• $r_m$ = orbital radius of Moon

2. Interpret “9 times closer to Earth than the Moon”

This means the satellite is at a distance equal to one-ninth that of the Moon from Earth:

$r_s = \frac{r_m}{9}$

Hence,

$\frac{r_s}{r_m} = \frac{1}{9}$

3. Substitute into Kepler’s law

$\left(\frac{T_s}{27}\right)^2 = \left(\frac{1}{9}\right)^3$

$\left(\frac{T_s}{27}\right)^2 = \frac{1}{729}$

Taking square root,

$\frac{T_s}{27} = \frac{1}{27}$

Thus,

$T_s = 27 \times \frac{1}{27} = 1 \text{ day}$

4. Check options

• A: $3$ days $\to$ incorrect
• B: $27$ days $\to$ incorrect
• C: $81$ days $\to$ incorrect
• D: $1$ day $\to$ correct

Therefore, the correct answer is D: 1 day.