Asked by Anonymous
(a) Find the speed of a satellite moving around the earth in a circular orbit that has a radius equal to six times the earth's radius of 6.38 106 m.
(b) Find the satellite's orbital period.
(b) Find the satellite's orbital period.
Answers
Answered by
drwls
Becasue of the inverse swquare law of gravity, the value of g at that orbital altitude will be 1/36 of the value at the earth's surface, which make is g/36. Thus
Centripetal acceleration =
V^2/(6*Re) = g/36
where Re is the radius of the Earth.
V^2 = Re*g/6 = 1.043*10^7
V = 3.32*10^3 m/s
For the orbit period T:
2 pi Re = V*T
T = 1.241*10^4 s = 207 minutes
Centripetal acceleration =
V^2/(6*Re) = g/36
where Re is the radius of the Earth.
V^2 = Re*g/6 = 1.043*10^7
V = 3.32*10^3 m/s
For the orbit period T:
2 pi Re = V*T
T = 1.241*10^4 s = 207 minutes
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