At what height above the earth's surface does | vedclass

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(A) Given: $g_{h} = 1\% 	ext{ of } g_{s} = 0.01 g_{s}$.The formula for acceleration due to gravity at height $h$ is $g_{h} = g_{s} \left( \frac{R}{R+

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$9$

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(A) Given: $g_{h} = 1\% 	ext{ of } g_{s} = 0.01 g_{s}$.The formula for acceleration due to gravity at height $h$ is $g_{h} = g_{s} \left( \frac{R}{R+h} ight)^{2}$.Substituting the given value: $0.01 g_{s} = g_{s} \left( \frac{R}{R+h} ight)^{2}$.Dividing both sides by $g_{s}$: $0.01 = \left( \frac{R}{R+h} ight)^{2}$.Taking the square root on both sides: $0.1 = \frac{R}{R+h}$.Rearranging the equation: $0.1(R + h) = R$.$0.1R + 0.1h = R$.$0.1h = 0.9R$.$h = \frac{0.9}{0.1} R = 9R$.Thus,at a height of $9R$,the acceleration due to gravity falls to $1\%$ of its value at the surface.

At what height above the earth's surface does the acceleration due to gravity fall to $1\%$ of its value at the earth's surface (in $,R$)?

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