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The Hubble Space Telescope (HST) orbits 569 km above Earth's surface. If HST has a tangential speed of 7,750
m/s, how long is HST's orbital period? The radius of Earth is 6.38 x 106m. Make sure to use correct significant
figures.

Respuesta :

Answer:

Mean radius of earth = 6.37E6 meters

Orbital radius = .569E6 meters

Total radius = (6.37 + .57) * 10^6 m = 6.94E6 m

P = S / t = 2 * pi * 6.94E6 / 7.75E3 sec = 5630 sec = 93.8 min