The weight of a body of mass m decreases by 1 | vedclass

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(B) The acceleration due to gravity at a height $h$ above the Earth's surface is given by $g_h = g(1 - \frac{2h}{R})$.The fractional change in weight

Vedclass · Accepted Answer

$0.5\%$ decrease

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(B) The acceleration due to gravity at a height $h$ above the Earth's surface is given by $g_h = g(1 - \frac{2h}{R})$.The fractional change in weight (or gravity) is $\frac{\Delta g}{g} = \frac{2h}{R}$.Given $\frac{\Delta g}{g} 	imes 100\% = 1\%$,therefore $\frac{2h}{R} = 0.01$,which implies $\frac{h}{R} = 0.005$.For a depth $d = h$ inside a mine,the acceleration due to gravity is $g_d = g(1 - \frac{d}{R})$.The fractional change in weight at depth $d$ is $\frac{\Delta g}{g} = \frac{d}{R}$.Since $d = h$,the fractional change is $\frac{h}{R} = 0.005$.Converting to percentage: $0.005 	imes 100\% = 0.5\%$.Thus,the weight decreases by $0.5\%$.

The weight of a body of mass $m$ decreases by $1\%$ when it is raised to a height $h$ above the Earth's surface. If the body is taken to a depth $h$ in a mine,the change in its weight is:

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