A geostationary satellite orbits the Earth with a period equal to:
AAbout 23 h 56 min (sidereal day)
B12 hours (typical GPS period)
CExactly 24 hours (solar day)
DAbout 27.3 days (lunar period)
Answer & Solution
Correct answer: A. About 23 h 56 min (sidereal day)
1. To remain over the same point on Earth, the satellite's period must equal the Earth's rotation period RELATIVE TO THE STARS.
2. That is the sidereal day, 23 h 56 min 4 s, not the 24-hour solar day.
3. From Kepler's 3rd law, this fixes the orbital radius to ≈ 4.2 × 10⁷ m, i.e. ≈ 36 000 km above the surface.
4. 12 h is the GPS-orbit period; 27.3 days is the Moon's sidereal period.
_Source: NCERT Class 11 Physics Ch 7 "Gravitation", §7.9 (geostationary satellites)_
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