A satellite orbiting above the equator of the earth is geosynchronous if the period is T = 24 hours (in this case, the satellite stays over a fixed point on the equator). Use Kepler’s Third Law to show that in a circular geosynchronous orbit, the distance from the center ofthe earth is R ≈ 42,246 km. Then compute the altitude h of the orbit above the earth’s surface. The earth has mass M ≈ 5.974 × 10^(24) kg and radius R ≈ 6371 km.
A satellite orbiting above the equator of the earth is geosynchronous if the period is T = 24 hours (in this case, the satellite stays over a fixed point on the equator). Use Kepler’s Third Law to show that in a circular geosynchronous orbit, the distance from the center ofthe earth is R ≈ 42,246 km. Then compute the altitude h of the orbit above the earth’s surface. The earth has mass M ≈ 5.974 × 10^(24) kg and radius R ≈ 6371 km.
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter11: Gravity, Planetary Orbits, And The Hydrogen Atom
Section: Chapter Questions
Problem 4P
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A satellite orbiting above the equator of the earth is geosynchronous if the period is T = 24 hours (in this case, the satellite stays over a fixed point on the equator). Use Kepler’s Third Law to show that in a circular geosynchronous orbit, the distance from the center of
the earth is R ≈ 42,246 km. Then compute the altitude h of the orbit above the earth’s surface. The earth has mass M ≈ 5.974 × 10^(24) kg and radius R ≈ 6371 km.
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