An earth satellite moves in a circular orbit at a speed of 5000 m/s.

Asked by maham237 @ 23/12/2020 in Physics viewed by 281 People

part a what is its orbital period? express your answer using two significant figures

Answered by maham237 @ 23/12/2020

r = GM/v^2
r is the distance of the satellite from the center of Earth in meters
G is newtons constant, 6.67428 x 10^-11
M is the mass of Earth, 5.9742 x 10^24
v is the orbital speed, which we know is 5,000 m/s
r = 6.67428 x 10^-11 x 5.9742 x 10^24/5,000^2
r = 8,137,445 meters from the center of the Earth
and : 8,137,445 - 6,378,100 = 1,749,345 meters from the surface of the Earth
Now that we know the distance we can calculate the orbital period.
Time = Distance/speed
distance is the circumference of the orbit, 2Ď€ x 8,137,445 = 51,129,074 meters
Time = (51,129,074)/5,000
Time = 10,225 seconds = 2.84 hours
so, orbital period is 2.84 hrs

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An earth satellite moves in a circular orbit at a speed of 5000 m/s.

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