At what height above the surface of the Earth | vedclass

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(B) The acceleration due to gravity at a height $h$ above the surface of the Earth is given by the formula $g' = g(1 - \frac{2h}{R_e})$,where $R_e$ is

Vedclass · Accepted Answer

$64$

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(B) The acceleration due to gravity at a height $h$ above the surface of the Earth is given by the formula $g' = g(1 - \frac{2h}{R_e})$,where $R_e$ is the radius of the Earth.Given that the value of $g$ decreases by $2 \%$,the new value $g'$ is $98 \%$ of $g$,so $g' = 0.98g$.Substituting this into the formula: $0.98g = g(1 - \frac{2h}{R_e})$.Dividing both sides by $g$: $0.98 = 1 - \frac{2h}{R_e}$.Rearranging the terms: $\frac{2h}{R_e} = 1 - 0.98 = 0.02$.Solving for $h$: $h = 0.01 	imes R_e$.Given $R_e = 6400 \, km$,we have $h = 0.01 	imes 6400 \, km = 64 \, km$.

At what height above the surface of the Earth does the value of $g$ decrease by $2 \%$? [Radius of the Earth is $6400 \, km$]

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