Geosynchronous Satellites
A geosynchronous satellite is a satellite that orbits the Earth at the same rotational speed as the Earth’s rotation. This means that the satellite appears to be stationary relative to an observer on the ground. These satellites are widely used for communication, weather monitoring, and navigation systems.
The Motion of Satellites around Earth
Let us consider a satellite assumed to be revolving around the earth in a circular orbit of radius ‘r’, having the centre of the earth as its centre; let us see the kinematics of the satellite moving around the earth.
Here, the test mass is a satellite, and the source mass is earth. The velocity with which a satellite orbits around the earth is given by the orbital velocity,
\(\begin{array}{l}{{v}_{o}}=\sqrt{\frac{GM}{r}}\end{array} \)
Time Period of a Satellite
The time taken by the satellite to complete one revolution around the earth is known as the time period of a satellite. The time period of a satellite is the ratio of the total distance travelled by the satellite around the earth to the orbital velocity.
\(\begin{array}{l}T=\frac{Total~distance~travelled~by~the~satellite}{Orbital~velocity}\end{array} \)
\(\begin{array}{l}T=\frac{2\pi r}{{{v}_{O}}}\end{array} \)
\(\begin{array}{l}T=\frac{2\pi r}{\sqrt{\frac{GM}{r}}}\end{array} \)
\(\begin{array}{l}\Rightarrow T=\frac{2\pi }{\sqrt{GM}}{{\left( r \right)}^{{}^{3}/{}_{2}}}\end{array} \)
Squaring both sides
\(\begin{array}{l}{{T}^{2}}=\frac{4{{\pi }^{2}}}{GM}{{\left( r \right)}^{3}}\end{array} \)
T2 ∝ r3
which is Kepler’s 3rd law.
The constant of proportionality in the above equation depends only on the source mass but not on the test mass.
Q: What is the time period of a geosynchronous satellite?
A: The time period of a geosynchronous satellite is 24 hours (or exactly one Earth day).
Q: Why does a geosynchronous satellite have a time period of 24 hours?
A: Because the satellite is synchronized with the Earth’s rotation, ensuring it remains above the same point on the equator. This makes it highly useful for applications like broadcasting and GPS tracking.
Height of a Geosynchronous Satellite
\(\begin{array}{l}{{T}^{2}}=\frac{4{{\pi }^{2}}}{GM}{{\left( r \right)}^{3}}\end{array} \)
If satellite is launched at height h from surface of earth, then put r = R + h in the above equation and we get
\(\begin{array}{l}{{T}^{2}}=\frac{4{{\pi }^{2}}}{GM}{{\left( r +h\right)}^{3}}\end{array} \)
where:
- T = Time period of the satellite (24 hours for a geosynchronous satellite)
- r = Orbital radius from Earth’s center
- G = Gravitational constant (6.674 × 10⁻¹¹ Nm²/kg²)
- M = Mass of Earth (5.972 × 10²⁴ kg)
- R = 6.4 x 106 m
By solving for r, the height of the satellite above Earth’s surface is found to be approximately 35,786 km.
Q: What is the altitude of a geosynchronous satellite?
A: The altitude of a geosynchronous satellite is approximately 35,786 km above the Earth’s surface.
Q: How is the height of a geosynchronous satellite calculated?
A: The height of a geosynchronous satellite is derived using Kepler’s Third Law, which relates the orbital period (T) of a satellite to its orbital radius (r):
\(\begin{array}{l}{{T}^{2}}=\frac{4{{\pi }^{2}}}{GM}{{\left( r +h\right)}^{3}}\end{array} \)
where:
- T = Time period of the satellite (24 hours for a geosynchronous satellite)
- r = Orbital radius from Earth’s center
- G = Gravitational constant (6.674 × 10⁻¹¹ Nm²/kg²)
- M = Mass of Earth (5.972 × 10²⁴ kg)
By solving for r, the height of the satellite above Earth’s surface is found to be approximately 35,786 km.
FAQs on Geosynchronous Satellites
Q1: How is a geosynchronous satellite different from a geostationary satellite?
A: A geostationary satellite is a special type of geosynchronous satellite that orbits exactly over the equator in a circular orbit, whereas a geosynchronous satellite may have an inclined orbit and appear to move in the sky.
Q2: Why are geosynchronous satellites placed at 35,786 km altitude?
A: At this altitude, the satellite’s orbital period matches the Earth’s rotation period, making it useful for continuous communication and weather monitoring.
Q3: Can geosynchronous satellites be used for GPS?
A: No, GPS satellites are placed in Medium Earth Orbit (MEO) at about 20,200 km altitude to provide global coverage.
Multiple-Choice Questions (MCQs) with Answers
Q1: What is the approximate height of a geosynchronous satellite from the Earth’s surface?
- (A) 500 km
- (B) 10,000 km
- (C) 35,786 km ✅
- (D) 100,000 km
Explanation: The correct height of a geosynchronous satellite is 35,786 km above Earth’s surface as per Kepler’s Third Law.
Q2: What is the orbital period of a geosynchronous satellite?
- (A) 12 hours
- (B) 24 hours ✅
- (C) 48 hours
- (D) 7 days
Explanation: A geosynchronous satellite must complete one full orbit in 24 hours to stay synchronized with Earth’s rotation.
Q3: Which of the following is true for a geosynchronous satellite?
- (A) It moves in a highly elliptical orbit
- (B) It moves at a constant height of 35,786 km ✅
- (C) Its time period is less than 12 hours
- (D) It does not follow Kepler’s laws
Explanation: A geosynchronous satellite remains at 35,786 km and follows Kepler’s laws of planetary motion.
Q4: Why are geosynchronous satellites mainly used for communication purposes?
- (A) They move very fast
- (B) They provide continuous coverage to a fixed location ✅
- (C) They are closer to the Earth
- (D) They do not require power to function
Explanation: Since geosynchronous satellites remain above a fixed point, they ensure uninterrupted communication services.
Q5: What is the key requirement for a satellite to be geostationary?
- (A) It must orbit at a height of 10,000 km
- (B) It must orbit in a circular path over the equator ✅
- (C) It must have a period of 12 hours
- (D) It must be placed in Low Earth Orbit (LEO)
Explanation: A geostationary satellite is a special case of a geosynchronous satellite that orbits directly above the equator in a circular orbit.
Test Your Knowledge
Conclusion
A geosynchronous satellite orbits Earth at 35,786 km with a time period of 24 hours, making it ideal for communication and weather observation. Understanding these concepts is crucial for JEE, NEET, and CBSE Board Class 11 exams.
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Written by: Neeraj Anand
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