What is the height from the surface of earth, | vedclass

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Question

(A) The acceleration due to gravity at a height $h$ above the surface of the earth is given by the formula $g_h = g \left( \frac{R_E}{R_E + h} \right)

Vedclass · Accepted Answer

$6400$

Vedclass · Answer

(A) The acceleration due to gravity at a height $h$ above the surface of the earth is given by the formula $g_h = g \left( \frac{R_E}{R_E + h} \right)^2$,where $g$ is the acceleration due to gravity at the surface and $R_E$ is the radius of the earth.Given that $g_h = \frac{g}{4}$,we substitute this into the equation:$\frac{g}{4} = g \left( \frac{R_E}{R_E + h} \right)^2$$\frac{1}{4} = \left( \frac{R_E}{R_E + h} \right)^2$Taking the square root of both sides:$\frac{1}{2} = \frac{R_E}{R_E + h}$$R_E + h = 2R_E$$h = R_E$Given $R_E = 6400 \ km$,therefore $h = 6400 \ km$.

What is the height from the surface of earth,where acceleration due to gravity will be $\frac{1}{4}$ of that of the earth (in $km$)? $(R_E = 6400 \ km)$

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