1. For a stationary orbit, the satellite’s orbital period equals the planet’s rotation period. The relation between orbital radius $r$, period $T$, and planet mass $M$ is

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

So,

$r^3 \propto M T^2$

2. Compare the given planet with Earth’s geostationary orbit.

For Earth:

• geostationary radius $r_E = 4.2 \times 10^4\,\text{km}$
• period $T_E = 24\,\text{h}$
• mass $M_E$

For the given planet:

• mass $M_P = \dfrac{M_E}{4}$
• stationary orbit period $T_P = 6\,\text{h}$

Using

$\frac{r_P^3}{r_E^3} = \frac{M_P T_P^2}{M_E T_E^2}$

Substitute values:

$\frac{r_P^3}{r_E^3} = \frac{\left(\frac{M_E}{4}\right)(6)^2}{M_E(24)^2}$

$= \frac{1}{4}\cdot \frac{36}{576}$

$= \frac{1}{4}\cdot \frac{1}{16} = \frac{1}{64}$

Thus,

$\frac{r_P}{r_E} = \left(\frac{1}{64}\right)^{1/3} = \frac{1}{4}$

3. Therefore,

$r_P = \frac{r_E}{4} = \frac{4.2 \times 10^4}{4}$

$r_P = 1.05 \times 10^4\,\text{km}$

4. Check options:

• A: $1.68 \times 10^5\,\text{km}$ ✗
• B: $1.4 \times 10^4\,\text{km}$ ✗
• C: $8.4 \times 10^4\,\text{km}$ ✗
• D: $1.05 \times 10^4\,\text{km}$ ✓

Hence the correct answer is Option D.