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Question
The orbit of a satellite is exactly 35780 km above the earth's surface and its tangential velocity is 3.08 km/s.
How much time the satellite will take to complete one revolution around the earth?
(Radius of earth = 6400 km.)
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Solution
Given:
Height of the satellite above the earth’s surface = 35780 km.
Tangential velocity of the satellite = 3.08 km/sec
Suppose the satellite takes T seconds to complete one revolution around the Earth. The distance travelled during this one revolution is equal to the circumference of the circular orbit. If r is the radius of the orbit, the satellite will travel a distance of 2πr during one revolution. Thus, the time required for one complete revolution can be obtained as follows:
v = `"distance"/"time"`
= `"circumference"/"time"`
= `(2pir)/"T"` ...[∵ x = (r + h)]
T = `(2pir)/v`
= `2pi("R" + h)`
= `(2 xx 3.14 xx (6400 + 35780))/3.08`
= 86003.38 sec
= 23.89 hrs.
= 23 hrs 54 minutes.
Thus, a satellite takes 23 hours and 54 minutes to complete one revolution around the earth.
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