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A satellite is revolving in a circular orbit at a distance of 2620 km from the surface of the earth. The time
period of revolution of the satellite is (Radius of the earth = 6380 km, mass of the earth = 6 × 1024 kg,
G = 6.67 × 10-11 N-m2/kg2)
- a)2.35 hours
- b)23.5 hours
- c)3.25 hours
- d)32.5 hours
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A satellite is revolving in a circular orbit at a distance of 2620 km ...
To find the time period of revolution of the satellite, we can use the formula for the period of a satellite in circular orbit:
T = 2π * √(r^3 / GM)
Where:
T = time period of revolution
r = distance from the center of the Earth to the satellite
G = gravitational constant
M = mass of the Earth
Given:
r = 2620 km + 6380 km = 9000 km = 9,000,000 m
G = 6.67430 × 10^-11 m^3/kg/s^2
M = 6 × 10^24 kg
Substituting the values into the formula:
T = 2π * √(9,000,000^3 / (6.67430 × 10^-11 * 6 × 10^24))
T = 2π * √(729,000,000,000,000,000,000,000,000,000 / 40.0452 × 10^13)
T = 2π * √(18,203,666,626,556,560,000)
T ≈ 2π * 4,269,928
T ≈ 26,789,908 seconds
Therefore, the time period of revolution of the satellite is approximately 26,789,908 seconds.
T = 2π * √(r^3 / GM)
Where:
T = time period of revolution
r = distance from the center of the Earth to the satellite
G = gravitational constant
M = mass of the Earth
Given:
r = 2620 km + 6380 km = 9000 km = 9,000,000 m
G = 6.67430 × 10^-11 m^3/kg/s^2
M = 6 × 10^24 kg
Substituting the values into the formula:
T = 2π * √(9,000,000^3 / (6.67430 × 10^-11 * 6 × 10^24))
T = 2π * √(729,000,000,000,000,000,000,000,000,000 / 40.0452 × 10^13)
T = 2π * √(18,203,666,626,556,560,000)
T ≈ 2π * 4,269,928
T ≈ 26,789,908 seconds
Therefore, the time period of revolution of the satellite is approximately 26,789,908 seconds.
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A satellite is revolving in a circular orbit at a distance of 2620 km ...
2.35 hrs................
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A satellite is revolving in a circular orbit at a distance of 2620 km from the surface of the earth. The timeperiod of revolution of the satellite is (Radius of the earth = 6380 km, mass of the earth = 6 × 1024kg,G = 6.67 × 10-11N-m2/kg2)a)2.35 hoursb)23.5 hoursc)3.25 hoursd)32.5 hoursCorrect answer is option 'A'. Can you explain this answer?
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A satellite is revolving in a circular orbit at a distance of 2620 km from the surface of the earth. The timeperiod of revolution of the satellite is (Radius of the earth = 6380 km, mass of the earth = 6 × 1024kg,G = 6.67 × 10-11N-m2/kg2)a)2.35 hoursb)23.5 hoursc)3.25 hoursd)32.5 hoursCorrect answer is option 'A'. Can you explain this answer? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about A satellite is revolving in a circular orbit at a distance of 2620 km from the surface of the earth. The timeperiod of revolution of the satellite is (Radius of the earth = 6380 km, mass of the earth = 6 × 1024kg,G = 6.67 × 10-11N-m2/kg2)a)2.35 hoursb)23.5 hoursc)3.25 hoursd)32.5 hoursCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A satellite is revolving in a circular orbit at a distance of 2620 km from the surface of the earth. The timeperiod of revolution of the satellite is (Radius of the earth = 6380 km, mass of the earth = 6 × 1024kg,G = 6.67 × 10-11N-m2/kg2)a)2.35 hoursb)23.5 hoursc)3.25 hoursd)32.5 hoursCorrect answer is option 'A'. Can you explain this answer?.
A satellite is revolving in a circular orbit at a distance of 2620 km from the surface of the earth. The timeperiod of revolution of the satellite is (Radius of the earth = 6380 km, mass of the earth = 6 × 1024kg,G = 6.67 × 10-11N-m2/kg2)a)2.35 hoursb)23.5 hoursc)3.25 hoursd)32.5 hoursCorrect answer is option 'A'. Can you explain this answer? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about A satellite is revolving in a circular orbit at a distance of 2620 km from the surface of the earth. The timeperiod of revolution of the satellite is (Radius of the earth = 6380 km, mass of the earth = 6 × 1024kg,G = 6.67 × 10-11N-m2/kg2)a)2.35 hoursb)23.5 hoursc)3.25 hoursd)32.5 hoursCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A satellite is revolving in a circular orbit at a distance of 2620 km from the surface of the earth. The timeperiod of revolution of the satellite is (Radius of the earth = 6380 km, mass of the earth = 6 × 1024kg,G = 6.67 × 10-11N-m2/kg2)a)2.35 hoursb)23.5 hoursc)3.25 hoursd)32.5 hoursCorrect answer is option 'A'. Can you explain this answer?.
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A satellite is revolving in a circular orbit at a distance of 2620 km from the surface of the earth. The timeperiod of revolution of the satellite is (Radius of the earth = 6380 km, mass of the earth = 6 × 1024kg,G = 6.67 × 10-11N-m2/kg2)a)2.35 hoursb)23.5 hoursc)3.25 hoursd)32.5 hoursCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A satellite is revolving in a circular orbit at a distance of 2620 km from the surface of the earth. The timeperiod of revolution of the satellite is (Radius of the earth = 6380 km, mass of the earth = 6 × 1024kg,G = 6.67 × 10-11N-m2/kg2)a)2.35 hoursb)23.5 hoursc)3.25 hoursd)32.5 hoursCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A satellite is revolving in a circular orbit at a distance of 2620 km from the surface of the earth. The timeperiod of revolution of the satellite is (Radius of the earth = 6380 km, mass of the earth = 6 × 1024kg,G = 6.67 × 10-11N-m2/kg2)a)2.35 hoursb)23.5 hoursc)3.25 hoursd)32.5 hoursCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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