A satellite orbits the Earth at a height of 100km above the surface, at a velocity of 7,850m/s. How many hours does it take to complete 1 revolution around the Earth?

1 Answer
Dec 24, 2014

We have the following situation:
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Where:
Earth radius: R_T=6,371 kmRT=6,371km
Eight of satellite: h=100 kmh=100km
Velocity of satellite: |vecv|=v=7,850 m/sv=v=7,850ms

The distance that the satellite must cover in one revolution is equal to the circumference: 2pir=2pi(R_T+h)2πr=2π(RT+h)
The time needed to do that is: t=(2pi(R_T+h))/vt=2π(RT+h)v

Remember thar: (R_T+h)=6,371+100=6,471 km=6,471*10^3 m(RT+h)=6,371+100=6,471km=6,471103m
and:
t=2pi(R_T+h)/v=(2pi*6,471*10^3)/(7,850)=5,179 s~~1h 26min 19s