Which of the following statements is/are correct about a satellite moving in a geostationary orbit?
Show Hint
- The orbit lies in the equatorial plane
- The orbit is circular about the center of the Earth
- The time period of motion is 90 minutes
- The satellite is visible from all parts of the Earth
The Correct Option is A
Solution and Explanation
Step 1: Definition of a geostationary orbit (GEO).
A geostationary satellite appears fixed to an observer on the Earth because its angular velocity about the Earth's axis equals that of the Earth's rotation.
This requires: (i) zero inclination (orbit in the equatorial plane), (ii) zero eccentricity (circular), and (iii) orbital period equal to one sidereal day.
Step 2: Check each statement against GEO requirements.
(A) GEO must have inclination \(i=0^\circ\), i.e., the orbit lies in the equatorial plane. \(\Rightarrow\) True.
(B) For a satellite to remain over the same ground longitude, its orbital speed must be constant; hence the orbit is circular and centered at the Earth's center (geocentric). \(\Rightarrow\) True.
(C) The orbital period of GEO is not 90 minutes. Using Kepler's third law, \[ T = 2\pi\sqrt{\frac{a^3}{\mu}},\quad a\approx 42{,}164\,\text{km},\;\mu=3.986\times10^{14}\,\text{m}^3\text{s}^{-2}. \] This gives \(T\approx 86{,}164\,\text{s}\approx 23\,\text{h}\,56\,\text{min}\,(= \text{sidereal day})\approx 1436\,\text{min}.\)
A 90-minute period corresponds to low Earth orbit (LEO), not GEO. \(\Rightarrow\) False.
(D) A GEO satellite is not visible from all points on Earth. From geometry, the maximum latitude from which GEO is visible is roughly \[ \phi_{\max} \approx \cos^{-1}\!\left(\frac{R_E}{a}\right)\approx \cos^{-1}\!\left(\frac{6378}{42164}\right)\approx 81^\circ, \] so it is not visible from the polar regions, nor from the opposite hemisphere beyond the Earth's horizon. \(\Rightarrow\) False.
\[ \boxed{\text{Correct statements: (A), (B).}} \]
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