When one moves from a point 16 text{ km} belo | vedclass

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(B) The acceleration due to gravity at a depth $d$ is given by $g_d = g(1 - \frac{d}{R})$.The acceleration due to gravity at a height $h$ is given by

Vedclass · Accepted Answer

$0.25$

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(B) The acceleration due to gravity at a depth $d$ is given by $g_d = g(1 - \frac{d}{R})$.The acceleration due to gravity at a height $h$ is given by $g_h = g(1 - \frac{2h}{R})$.We are moving from depth $d = 16 	ext{ km}$ to height $h = 16 	ext{ km}$.The change in $g$ is $\Delta g = g_h - g_d = g(1 - \frac{2h}{R}) - g(1 - \frac{d}{R})$.Substituting the values: $\Delta g = g(1 - \frac{2 	imes 16}{6400}) - g(1 - \frac{16}{6400}) = g(1 - \frac{32}{6400} - 1 + \frac{16}{6400}) = g(-\frac{16}{6400}) = -\frac{g}{400}$.The percentage change $\alpha$ is given by $|\frac{\Delta g}{g}| 	imes 100 = |-\frac{1}{400}| 	imes 100 = 0.25 \%$.Thus, the value of $\alpha$ is $0.25$.

When one moves from a point $16 \text{ km}$ below the earth's surface to a point $16 \text{ km}$ above the earth's surface, the change in $g$ is approximately $\alpha \%$. The value of $\alpha$ is . . . . . . . (Take radius of the earth $R = 6400 \text{ km}$.)

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