Asked by allie
the moon has a period of 27.3 days and a mean distance of 3.90x10^8 m from the center of the earth. find the period of a satellite in orbit 6.7x10^6 m from the center of earth.
thank you!
thank you!
Answers
Answered by
ASingh
Two body system witah a same center (earth), by using Keplers law
rm³/Pm² = rs³/Ps²
r = distance, P = period, m = moon, s = satellite
(3.90*10^5)³/(27.3)² = (6.70*10^3)³/Ps²
Ps² = (6.70*10^3)³(27.3)²/(3.90*10^5)³
Ps² = 0.003779
Ps = 0.0615 days = 1.48 hours
(b) r = 6700 km
Earth's radius = 6378 km
altitude = h = r - Re
h = 6700 - 6378 = 332 km
rm³/Pm² = rs³/Ps²
r = distance, P = period, m = moon, s = satellite
(3.90*10^5)³/(27.3)² = (6.70*10^3)³/Ps²
Ps² = (6.70*10^3)³(27.3)²/(3.90*10^5)³
Ps² = 0.003779
Ps = 0.0615 days = 1.48 hours
(b) r = 6700 km
Earth's radius = 6378 km
altitude = h = r - Re
h = 6700 - 6378 = 332 km
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