Satellites feel the gravitational pull of the earth. They remain in orbit becaus

Question

Satellites feel the gravitational pull of the earth. They remain in orbit because of their velocity, which acts to counteract gravity. (The satellite wants to fly out in a straight line, but gravity forces it to curve towards the earth.) Consider a communications satellite that needs to be 27000 km above the Earth's surface.

(a) Assuming the satellite travels in a perfect circle, what is the radius of the satellite's travel. (The radius of the earth is 6375 km.)


(b) At the satellite's altitude, the acceleration of gravity is 0.357 m/s2. What is the magnitude of the tangential velocity that the satellite must have to remain in orbit?


(c) How much time will the satellite take to orbit the earth?

Explanation / Answer

The radius of the satellite's travel is the radius of the earthplus the height the satellite needs to be above the earth. a) 6375km + 36000km = 42375 km In this problem the acceleration of gravity is the same as thecentripetal acceleration so it can be solve by the equation centripetal acceleration= v^2/ r 0.222 m/s^2 = (v^2)/(42375000 m) v^2 = 0.222(42375000) v^2 = 9407250 v = 3067.124 m/s b) the tangential velocity is 3067.124 m/s The time to orbit the earth once is the period, T. T = (2pr)/v = (266249977.4)/(3067.124) = 86807.699 seconds = 1446.795 minutes = 24.113 hours C) the satellite will take 24.113 hours or 1446.795 minutes.Or youcould say 86807.699 seconds

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