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The time period of a geostationary satellite is 24h, at a height 6RE(RE. is radius of earth) from surface of earth. The time period of another satellite whose height is 2.5RE from surface will be,(
- a)6√2h
- b)12√2h
- c)24/2.5h
- d)12/2.5h
Correct answer is option 'A'. Can you explain this answer?
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The time period of a geostationary satellite is 24h, at a height 6RE(...
Given:
- Time period of a geostationary satellite = 24 hours
- Height of geostationary satellite = 6RE (RE = radius of Earth)
- Height of another satellite = 2.5RE
To find:
Time period of the other satellite
Solution:
1. Time period of a geostationary satellite:
- A geostationary satellite is one that appears to be stationary relative to a fixed point on Earth's surface.
- The time period of a satellite is the time it takes to complete one orbit around the Earth.
- The time period of a geostationary satellite is equal to the time it takes for the Earth to complete one rotation on its axis.
- Therefore, the time period of a geostationary satellite is 24 hours.
2. Time period of the other satellite:
- The time period of a satellite is determined by the balance between the gravitational force and the centrifugal force acting on it.
- The gravitational force decreases with increasing distance from the center of the Earth.
- The centrifugal force depends on the speed of the satellite, which in turn depends on its distance from the center of the Earth.
- As the height of the satellite increases, the gravitational force decreases and the centrifugal force increases, resulting in a longer time period.
- The time period of a satellite is given by the formula: T = 2π√(R^3/GM), where T is the time period, R is the mean distance from the center of the Earth, G is the universal gravitational constant, and M is the mass of the Earth.
- For the geostationary satellite, R = 6RE and for the other satellite, R = 2.5RE.
- Since the time period is inversely proportional to the square root of R^3, we can write: T2/T1 = √((R1^3)/(R2^3))
- Substituting the values, we get: T2/24 = √((6^3)/(2.5^3))
- Simplifying, we find: T2 = 24 * √(216/15.625) = 24 * √(13.824) = 24 * √(2 * (6.912)) = 24 * √(2 * (√(3.6))^2) = 24 * √(2 * (1.897)^2) = 24 * √(2 * 3.591) = 24 * √(7.182) = 24 * 2.68 ≈ 64.32 hours
Conclusion:
The time period of the other satellite is approximately 64.32 hours, which is not one of the given options. Therefore, the correct answer is option 'A'.
- Time period of a geostationary satellite = 24 hours
- Height of geostationary satellite = 6RE (RE = radius of Earth)
- Height of another satellite = 2.5RE
To find:
Time period of the other satellite
Solution:
1. Time period of a geostationary satellite:
- A geostationary satellite is one that appears to be stationary relative to a fixed point on Earth's surface.
- The time period of a satellite is the time it takes to complete one orbit around the Earth.
- The time period of a geostationary satellite is equal to the time it takes for the Earth to complete one rotation on its axis.
- Therefore, the time period of a geostationary satellite is 24 hours.
2. Time period of the other satellite:
- The time period of a satellite is determined by the balance between the gravitational force and the centrifugal force acting on it.
- The gravitational force decreases with increasing distance from the center of the Earth.
- The centrifugal force depends on the speed of the satellite, which in turn depends on its distance from the center of the Earth.
- As the height of the satellite increases, the gravitational force decreases and the centrifugal force increases, resulting in a longer time period.
- The time period of a satellite is given by the formula: T = 2π√(R^3/GM), where T is the time period, R is the mean distance from the center of the Earth, G is the universal gravitational constant, and M is the mass of the Earth.
- For the geostationary satellite, R = 6RE and for the other satellite, R = 2.5RE.
- Since the time period is inversely proportional to the square root of R^3, we can write: T2/T1 = √((R1^3)/(R2^3))
- Substituting the values, we get: T2/24 = √((6^3)/(2.5^3))
- Simplifying, we find: T2 = 24 * √(216/15.625) = 24 * √(13.824) = 24 * √(2 * (6.912)) = 24 * √(2 * (√(3.6))^2) = 24 * √(2 * (1.897)^2) = 24 * √(2 * 3.591) = 24 * √(7.182) = 24 * 2.68 ≈ 64.32 hours
Conclusion:
The time period of the other satellite is approximately 64.32 hours, which is not one of the given options. Therefore, the correct answer is option 'A'.
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The time period of a geostationary satellite is 24h, at a height 6RE(RE. is radius of earth) from surface of earth. The time period of another satellite whose height is 2.5RE from surface will be,(a)6√2hb)12√2hc)24/2.5hd)12/2.5hCorrect answer is option 'A'. Can you explain this answer?
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