The variation of acceleration due to gravity | vedclass

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Question

with depth

$g_{1}=g\left(1-\frac{(d)}{R}\right)$

As depth $d$ goes on increasing $g_{1}$ goes on decreasing, it remains maximum at the surface o

Vedclass · Accepted Answer

Vedclass · Answer

with depth

$g_{1}=g\left(1-\frac{(d)}{R}ight)$

As depth $d$ goes on increasing $g_{1}$ goes on decreasing, it remains maximum at the surface of Earth. The above equation is in the form of straight line.

With height

$g_{2}=g\left(1-\frac{2 h}{R}ight)$

$=g-\frac{2 g h}{R}$

$g_{2} \propto \frac{1}{R}(	ext { Hyperbola })$

Acceleration due to gravity goes on decreasing as the $h$ above Earth surface increases.

Hence the correct answer is option $B$.

The variation of acceleration due to gravity $g$ with distance $d$ from centre of the earth is best represented by ($R =$ Earth's radius)

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