At what altitude will the acceleration due to | vedclass

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(B) The acceleration due to gravity at the surface of the earth is given by $g_0 = \frac{GM}{R^2}$,where $M$ is the mass of the earth and $R$ is its r

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$R$

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(B) The acceleration due to gravity at the surface of the earth is given by $g_0 = \frac{GM}{R^2}$,where $M$ is the mass of the earth and $R$ is its radius.At an altitude $h$,the acceleration due to gravity $g_h$ is given by $g_h = \frac{GM}{(R+h)^2}$.We are given that $g_h = 25\% 	ext{ of } g_0$,which means $g_h = \frac{1}{4} g_0$.Substituting the expressions for $g_h$ and $g_0$,we get $\frac{GM}{(R+h)^2} = \frac{1}{4} \frac{GM}{R^2}$.Canceling $GM$ from both sides,we have $\frac{1}{(R+h)^2} = \frac{1}{4R^2}$.Taking the square root of both sides,we get $\frac{1}{R+h} = \frac{1}{2R}$.Cross-multiplying gives $2R = R + h$,which simplifies to $h = R$.

At what altitude will the acceleration due to gravity be $25\%$ of that at the earth's surface (given radius of earth is $R$)?

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