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.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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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