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7.Gravitation
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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 $g$
(KVPY-2015)
Solution
(d)
Let orbital radius of ball is $r$ then orbital velocity of ball is
$v=\sqrt{\frac{G M}{r}}$
$\text { Here, }r=R+h$
$\Rightarrow r=6400 km +9 km$
$\text { or } r \approx 6400 km$
$\Rightarrow r=R \text { (radius of earth) }$
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 $g$.
Standard 11
Physics
