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A ball is launched from the top of Mt. Everest which is at elevation of $9000 \mathrm{~m}$. 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
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nearly equal to $\mathrm{g}$.
$\frac{m v^{2}}{r}=m g^{\prime}$ (where $\mathrm{g}$ ' is nearly equal to $\mathrm{g}$ )
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