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

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Question

(d)

Let orbital radius of ball is $r$ then orbital velocity of ball is

$v=\sqrt{\frac{G M}{r}}$

$	ext { Here, }r=R+h$

$\Rightarrow r=6400

Vedclass · Accepted Answer

nearly equal to $g$

Vedclass · Answer

(d)

Let orbital radius of ball is $r$ then orbital velocity of ball is

$v=\sqrt{\frac{G M}{r}}$

$	ext { Here, }r=R+h$

$\Rightarrow r=6400 km +9 km$

$	ext { or } r \approx 6400 km$

$\Rightarrow r=R 	ext { (radius of earth) }$

Now, acceleration of ball in orbit is

$a=\frac{v^2}{r}=\frac{G M}{r^2} \approx \frac{G M}{R^2}$ or $a \approx g$

So, acceleration of ball is 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

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