Question
What is the height at which the value of g is the same as at a depth of $\frac{\text{R}}{2}?$
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Answer
At depth $\frac{\text{R}}{2},\text{g}'=\text{g}\Big(1-\frac{\text{R}}{2\text{R}}\Big)=\frac{\text{g}}{2}$At height $\text{x},\text{g}'=\text{g}\Big(1-\frac{2\text{x}}{\text{R}}\Big)$
$\therefore\text{g}\Big(1-\frac{2\text{x}}{\text{R}}\Big)=\frac{\text{g}}{2}$
$\frac{1}{2}=\frac{2\text{x}}{\text{R}}$
$\therefore\text{x}=\frac{\text{R}}{4}$
$\therefore\text{g}\Big(1-\frac{2\text{x}}{\text{R}}\Big)=\frac{\text{g}}{2}$
$\frac{1}{2}=\frac{2\text{x}}{\text{R}}$
$\therefore\text{x}=\frac{\text{R}}{4}$
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