A ball is launched from the top of Mt. Everes | vedclass

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(D) The orbital radius of the ball is $r = R + h$,where $R$ is the radius of the Earth $(6400 \, km)$ and $h$ is the height $(9 \, km)$.Since $h \ll R

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nearly equal to $g$

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(D) The orbital radius of the ball is $r = R + h$,where $R$ is the radius of the Earth $(6400 \, km)$ and $h$ is the height $(9 \, km)$.Since $h \ll R$,we have $r \approx R$.The acceleration of an object in a circular orbit is given by the centripetal acceleration $a = v^2 / r$.Substituting the orbital velocity $v = \sqrt{GM/r}$,we get $a = (GM/r) / r = GM/r^2$.Since $r \approx R$,the acceleration $a \approx GM/R^2$.By definition,the acceleration due to gravity near the Earth's surface is $g = GM/R^2$.Therefore,the acceleration of the ball in orbit is nearly equal to $g$.

$A$ ball is launched from the top of Mt. Everest,which is at an elevation of $9000 \, m$. The ball moves in a circular orbit around the 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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