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Question

(b)

Acceleration due to gravity above the surface of earth at a height $h$ is given $g^{\prime}=g\left(1-\frac{2 h}{R_	heta}ight)$ here, $g^{\pr

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(b)

Acceleration due to gravity above the surface of earth at a height $h$ is given $g^{\prime}=g\left(1-\frac{2 h}{R_	heta}ight)$ here, $g^{\prime}=0.98 \,g$

$\Rightarrow 0.98=1-\frac{2 h}{R_e}$

$\Rightarrow \frac{2 h}{R_e}=0.02$

$h=0.01 R_e$

$=0.01 	imes 6400 \,km$

$=64 \,km$

At what height above the surface of earth the value of $"g"$ decreases by $2 \%$ ..... $km$ [radius of the earth is $6400 \,km$ ]

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