Asked by Anonymous
whenever two Apollo astronauts were on the surface of the moon, a third astronaut orbited the moon, assume the orbit to be circular and 100 km above the surface of the moon, where the acceleration due to gravity is 1.52 m/s^2. the radius of the moon is 1.70* 10^6 m.(a) determine the astronaut's orbital speed, and (b) the period of the orbit
Answers
Answered by
tchrwill
Orbital speed derives from
Vc = sqrt(µ/r) where µ = the Moon's gravitational constant, 4.9033x10^12m^3/sec.^2 and r = the Moon's radius, 1738km, and the orbital altitude = 1838km = 1,838,000m or
Vc = sqrt(4.9033x10^12/1,828,000m) = 1638m/sec.
The orbital period derives from
T = 2Pisqrt(r^3/µ) = 6.28sqrt[(1,838,000^3/4.9033x10^12]/60 = 118min.
Vc = sqrt(µ/r) where µ = the Moon's gravitational constant, 4.9033x10^12m^3/sec.^2 and r = the Moon's radius, 1738km, and the orbital altitude = 1838km = 1,838,000m or
Vc = sqrt(4.9033x10^12/1,828,000m) = 1638m/sec.
The orbital period derives from
T = 2Pisqrt(r^3/µ) = 6.28sqrt[(1,838,000^3/4.9033x10^12]/60 = 118min.
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