At what depth below the surface of the earth, | vedclass

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Question

(a) Radius of earth $R = 6400\, km $

$h = \frac{R}{4}$ Acceleration due to gravity at a height h ${g_h} = g{\left( {\frac{R}{{R + h}}} \right)^2}$

Vedclass · Accepted Answer

$4.2 	imes {10^6}\,m$

Vedclass · Answer

(a) Radius of earth $R = 6400\, km $

$h = \frac{R}{4}$ Acceleration due to gravity at a height h ${g_h} = g{\left( {\frac{R}{{R + h}}} ight)^2}$

$ = g{\left( {\frac{R}{{R + \frac{R}{4}}}} ight)^2}$

$ = \frac{{16}}{{25}}g$ 

At depth $'d'$ value of acceleration due to gravity

${g_d} = \frac{1}{2}{g_h}$  (According to problem) 

$&rArr;$ ${g_d} = \frac{1}{2}\left( {\frac{{16}}{{25}}} ight)g$ $ \Rightarrow g\left( {1 - \frac{d}{R}} ight)$

$ = \frac{1}{2}\left( {\frac{{16}}{{25}}} ight)\;g$

By solving we get $d = 4.3 	imes {10^6}\,m$

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

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