Step 1: Define a geostationary satellite.
A geostationary satellite appears stationary to an observer on Earth. It must orbit above the equator and complete exactly one orbit in the same time Earth rotates once on its axis.
Step 2: Recall Earth's rotation period.
The Earth completes one full rotation in 24 hours.
Step 3: Match orbital period to Earth's rotation period.
For the satellite to appear fixed: $\omega_{\text{satellite}} = \omega_{\text{Earth}}$, so $T_{\text{satellite}} = 24$ h.
Step 4: Understand the uniqueness of this orbit.
By Kepler's third law $T^2 \propto r^3$, $T = 24$ h determines a unique orbital radius of about 42,000 km. All geostationary satellites share this altitude.
Step 5: Contrast with lower orbits.
Low Earth Orbit has $T \approx 90$ min; geostationary orbit is far higher, hence $T = 24$ h.
Step 6: State the final answer.
\[ \boxed{T = 24 \text{ hours}} \]