Asked by Anonymous
1. At what distance above the Earth (mass of 5.97 x 1024 kg and radius of 6.38 x 106 m) would a satellite have a period of 125 min?
Answers
Answered by
Steve
don't you have a handy formula relating mass and radius to orbital period?
Just plug in your numbers. Watch the units.
Just plug in your numbers. Watch the units.
Answered by
Anonymous
F = G m M/r^2
Ac = F/m = G M/r^2 = v^2/r
so
G M = r v^2
125 min = 7500 s
v * 7500 = 2 pi r
v = pi r/3750
v^2 = pi^2 r^2/1.4*10^7
so
G M = 9.87 r^3 /1.4*10^7
but G = 6.67*10^-11 and M is 5.97*10^24
so
r^3 =(1.4*10^7/9.87)(6.67*10^-11*5.97*10^24)
height above earth = r - 6.38*10^6
Ac = F/m = G M/r^2 = v^2/r
so
G M = r v^2
125 min = 7500 s
v * 7500 = 2 pi r
v = pi r/3750
v^2 = pi^2 r^2/1.4*10^7
so
G M = 9.87 r^3 /1.4*10^7
but G = 6.67*10^-11 and M is 5.97*10^24
so
r^3 =(1.4*10^7/9.87)(6.67*10^-11*5.97*10^24)
height above earth = r - 6.38*10^6
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