Question
What is the acceleration due to gravity at the bottom of a sea 30km deep taking radius of the earth as 6.3 × 10km?
✓
Answer
$\text{g}'=\text{g}\Big(1-\frac{\text{d}}{\text{R}}\Big)$ $=9.8\Big(1-\frac{30\times1000}{6.3\times(1000)^2}\Big) $ $=9.8\Big(1-\frac{1}{210}\Big)$ $=9.8\Big(\frac{209}{210}\Big)=9.75\text{ms}^{-2}$
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