A satellite is said to be in geosynchronous (or geostationary) orbit if, as it o

Question

A satellite is said to be in geosynchronous (or geostationary) orbit if, as it orbits, it remains directly above a fixed point on the surface of the Earth. Its orbital period must be 23.93 hrs (= Earth's sidereal day) and its orbital radius (from centre of Earth) 42,240 km. Suppose that satellite A is successfully launched into a geostationary orbit. However, due to a miscalculation. satellite B ends up in a circular orbit 1000 km below its intended geostationary orbit. [In other words, rA = 42,240 km and rB = 41,240 km. ]
(a) What is the orbital velocity of satellite A? (km/hr)
What is the orbital velocity of satellite B (km/hr)?
What is the orbital period (hours) of satellite B?
(b) At some point in time, satellite B passes directly beneath satellite A. How much time will pass before the two satellites are once again configured this way (with B directly beneath A, but not necessarily at the exact same location)?

Explanation / Answer

The orbital radius of A is rA = 42,240 km The period of the A is TA = 23.93 hr The orbital radius of B is rB = 41,240 km The period of the B is TB 1) The orbital speed of A is vA = [GM/(R+rA)] G = 6.67*10-11 Nm2/kg2 is universal gravitational constant M = 5.97*1024 kg is mass of the Earth R = 6400 km is the radius of the Earth vA = [(6.67*10-11 Nm2/kg2)(5.97*1024 kg)/(6400 km+42,240 km)]      = 2.86 km/s The orbital speed of A is vB = [GM/(R+rB)] vB = [(6.67*10-11 Nm2/kg2)(5.97*1024 kg)/(6400 km+41,240 km)]      = 91.4 km/s According to Kepler's third law, TA2/TB2 = rA3/rB3 (23.93 hr)2 / TB2 = (42,240 km)3/(41240 km)3TB = 23.08 hr 2)    The time passed for the two sattelites to be in the same position again is t = 23.93 hr/(23.93 hr - 23.08 hr) = 28.15 hr

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