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(D) Let $h$ be the distance above and below the surface of the earth where the weight of the body is the same. This implies the acceleration due to gr

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$\frac{\sqrt{5} R - R}{2}$

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(D) Let $h$ be the distance above and below the surface of the earth where the weight of the body is the same. This implies the acceleration due to gravity at height $h$ $(g_h)$ must be equal to the acceleration due to gravity at depth $h$ $(g_d)$.The formula for acceleration due to gravity at height $h$ is $g_h = g \left( 1 + \frac{h}{R} ight)^{-2} \approx g \left( 1 - \frac{2h}{R} ight)$ (for $h \ll R$). However,for a general solution,we use $g_h = \frac{g R^2}{(R+h)^2}$.The formula for acceleration due to gravity at depth $h$ is $g_d = g \left( 1 - \frac{h}{R} ight)$.Equating the two: $\frac{g R^2}{(R+h)^2} = g \left( 1 - \frac{h}{R} ight)$.$\frac{1}{(1 + h/R)^2} = 1 - \frac{h}{R}$.Let $x = \frac{h}{R}$. Then $\frac{1}{(1+x)^2} = 1 - x$.$1 = (1-x)(1+x)^2 = (1-x)(1 + 2x + x^2) = 1 + 2x + x^2 - x - 2x^2 - x^3$.$1 = 1 + x - x^2 - x^3$.$x^3 + x^2 - x = 0$.Since $x 
eq 0$,we have $x^2 + x - 1 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$.Since $x$ must be positive,$x = \frac{\sqrt{5} - 1}{2}$.Therefore,$h = \frac{\sqrt{5} - 1}{2} R = \frac{\sqrt{5} R - R}{2}$.

At what distance above and below the surface of the earth will a body have the same weight? (Take the radius of the earth as $R$.)

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