# 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?

Dec 24, 2014

We have the following situation:

Where:
Earth radius: ${R}_{T} = 6 , 371 k m$
Eight of satellite: $h = 100 k m$
Velocity of satellite: $| \vec{v} | = v = 7 , 850 \frac{m}{s}$

The distance that the satellite must cover in one revolution is equal to the circumference: $2 \pi r = 2 \pi \left({R}_{T} + h\right)$
The time needed to do that is: $t = \frac{2 \pi \left({R}_{T} + h\right)}{v}$

Remember thar: $\left({R}_{T} + h\right) = 6 , 371 + 100 = 6 , 471 k m = 6 , 471 \cdot {10}^{3} m$
and:
$t = 2 \pi \frac{{R}_{T} + h}{v} = \frac{2 \pi \cdot 6 , 471 \cdot {10}^{3}}{7 , 850} = 5 , 179 s \approx 1 h 26 \min 19 s$