At what distance from the centre of the earth | vedclass

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(D) The acceleration due to gravity at a height $h$ above the surface of the earth is given by $g' = g \left( \frac{R}{R+h} \right)^2$.We are given th

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$1.414\, R$

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(D) The acceleration due to gravity at a height $h$ above the surface of the earth is given by $g' = g \left( \frac{R}{R+h} \right)^2$.We are given that $g' = \frac{g}{2}$.Substituting this into the formula: $\frac{g}{2} = g \left( \frac{R}{R+h} \right)^2$.Taking the square root on both sides: $\frac{1}{\sqrt{2}} = \frac{R}{R+h}$.Rearranging the terms: $R+h = \sqrt{2} R$.Since $1.414 \approx \sqrt{2}$,we have $R+h = 1.414 R$.Here,$R+h$ represents the distance from the centre of the earth. Thus,the required distance is $1.414 R$.

At what distance from the centre of the earth,the value of acceleration due to gravity $g$ will be half that on the surface? ($R =$ radius of earth)

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