The acceleration due to gravity at an altitud | vedclass

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(C) The acceleration due to gravity at height $h$ is given by $g' = g \left( \frac{R}{R + h} \right)^2$. Given that $g' = \frac{g}{2}$,we have $\frac{

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$1600$

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(C) The acceleration due to gravity at height $h$ is given by $g' = g \left( \frac{R}{R + h} \right)^2$. Given that $g' = \frac{g}{2}$,we have $\frac{1}{2} = \left( \frac{R}{R + h} \right)^2$. Taking the square root on both sides,we get $\frac{1}{\sqrt{2}} = \frac{R}{R + h}$. This implies $R + h = R\sqrt{2}$,so $h = R(\sqrt{2} - 1)$. Given $R = 4000 \ mile$ and $\sqrt{2} \approx 1.414$,we have $h = 4000(1.414 - 1) = 4000(0.414) = 1656 \ mile$. Rounding to the nearest provided option,$h \approx 1600 \ mile$.

The acceleration due to gravity at an altitude $h$ above the Earth's surface is half of its value on the surface of the Earth. If the radius of the Earth $R = 4000 \ mile$,the altitude $h$ is approximately ......... $mile$.

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