At what height over the earth's pole,the free | vedclass

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(A) The acceleration due to gravity at a height $h$ above the surface of the earth is given by $g_h = g \left(1 - \frac{2h}{R}\right)$ for $h \ll R$.G

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(A) The acceleration due to gravity at a height $h$ above the surface of the earth is given by $g_h = g \left(1 - \frac{2h}{R}ight)$ for $h \ll R$.Given that the acceleration decreases by $1\%$,we have $\frac{\Delta g}{g} = 0.01$.Using the relation $\frac{\Delta g}{g} = \frac{2h}{R}$,we get $0.01 = \frac{2h}{R}$.Substituting $R = 6400 \, km$,we have $h = \frac{0.01 	imes 6400}{2} = \frac{64}{2} = 32 \, km$.Thus,the height is $32 \, km$.

At what height over the earth's pole,the free fall acceleration decreases by one percent? (Assume the radius of earth to be $6400 \, km$)

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