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A ball is launched from the top of Mt. Everest which is at elevation of 9000 m. The ball moves in circular orbit around earth. Acceleration due to gravity near the earth’s surface is g. The magnitude of the ball’s acceleration while in orbit is –
(A) close to g/2
(B) zero
(C) much greater than g
(D) nearly equal to g
A ball is launched from the top of Mt. Everest which is at elevation of 9000 m. The ball moves in circular orbit around earth. Acceleration due to gravity near the earth’s surface is g. The magnitude of the ball’s acceleration while in orbit is –
(A) close to g/2
(B) zero
(C) much greater than g
(D) nearly equal to g
Correct option (D) nearly equal to g
Explanation:
At earth surface acceleration due to gravity
g =GM/R2
At height = 9000 m, Radius of orbit of ball is 6400 + 9 km
⇒ radius r > R
Radius is almost equal to radius of earth.
(v) orbital velocity of ball = √GM/r
Acceleration = v2/t ⇒GM/r2
as r is very near to R
∴ Acceleration = GM/R2 = g