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A stone weight $100 \mathrm{~N}$ on the surface of the earth. The ratio of its weight at a height of half the radius of the earth to a depth of half the radius of the earth will be approximately
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The correct answer is:
$0.9$
Weight of stone at the surface of the earth, $w=100 \mathrm{~N}$ At height $h$ from the surface of earth, gravitational acceleration,
$$
g_{h}=\frac{g}{\left(1+\frac{h}{R}\right)^{2}}
$$
At depth $d$ from the surface of the earth, gravitational acceleration,
$$
g_{d}=g\left(1-\frac{d}{R}\right)
$$
According to question,
$\frac{\text { Weight at } h=R / 2}{\text { Weight at } d=R / 2}=\frac{m_{h}(\text { at } h=R / 2)}{m g_{d}(\text { at } h=R / 2)}$
$=\frac{\frac{g}{\left(1+\frac{h}{R}\right)^{2}}(\text { at } h=R / 2)}{g\left(1-\frac{d}{R}\right) \text { at }(d=R / 2)}$
$=\frac{\frac{1}{\left(1+\frac{R / 2}{R}\right)^{2}}}{1-\frac{R / 2}{R}}=\frac{4 / 9}{1 / 2}=\frac{8}{9}=0.9$
$$
g_{h}=\frac{g}{\left(1+\frac{h}{R}\right)^{2}}
$$
At depth $d$ from the surface of the earth, gravitational acceleration,
$$
g_{d}=g\left(1-\frac{d}{R}\right)
$$
According to question,
$\frac{\text { Weight at } h=R / 2}{\text { Weight at } d=R / 2}=\frac{m_{h}(\text { at } h=R / 2)}{m g_{d}(\text { at } h=R / 2)}$
$=\frac{\frac{g}{\left(1+\frac{h}{R}\right)^{2}}(\text { at } h=R / 2)}{g\left(1-\frac{d}{R}\right) \text { at }(d=R / 2)}$
$=\frac{\frac{1}{\left(1+\frac{R / 2}{R}\right)^{2}}}{1-\frac{R / 2}{R}}=\frac{4 / 9}{1 / 2}=\frac{8}{9}=0.9$
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