First slide

Acceleration due to gravity and its variation

Question

At what depth below the surface of the earth, acceleration due to gravity is will be half its value 1600 km above the surface of the earth?

Moderate

A

$4 .2 \times 10^{6} m$
B

$3 .19 \times 10^{6} m$
C

$1 .59 \times 10^{6} m$
D

None of these

Solution

Radius of earth $R=6400km⇒h= \frac{R}{4}$
Acceleration due to gravity at a height h
$g_{h} = g {(\frac{R}{R + h})}^{2} = g {(\frac{R}{R + \frac{R}{4}})}^{2} = \frac{16}{25} g$
At depth d, value of acceleration due to gravity
$g_{d} = \frac{1}{2} g_{h}$ (According to problem)
$\Rightarrow g_{d} = \frac{1}{2} (\frac{16}{25}) g \Rightarrow g (1 - \frac{d}{R}) = \frac{1}{2} (\frac{16}{25}) g$
By solving we get $d = 4 .3 \times 10^{6} m$

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