2.On the surface of the Earth we find that the acceleration of gravity is \"g\"
ID: 1398484 • Letter: 2
Question
2.On the surface of the Earth we find that the acceleration of gravity is "g" or 9.8 m/s2. Geosynchronous orbit (i.e., one revolution per 24 hour period) occurs above the equator at a distance about 22,250 miles directly above the Earth's surface. Which answer below most closely approximates what a 100 kg astronaut would weigh (i.e., what is the gravitational force of the Earth on the astronaut) at this altitude? Remember that Newton's universal law of gravitation references the center of the Earth.
Explanation / Answer
His altitude is 22250 mi = 35,800 km
The total distance from the center is
r = 35800 km + 6371 km = 42171 km = 4.217E7 m
Thus, the weight is, as M = 5.972E24 kg (mass of earth)
F = G M m / r^2
F = 22.4 N [ANSWER]
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