Solution:
Since \(g = \frac{GM}{R^2}\), for a small percentage change: \(\frac{\Delta g}{g} \times 100 = -2 \frac{\Delta R}{R} \times 100\). Since \(R\) decreases by \(1%\), \(g\) increases by \(2%\).
If the radius of the earth were to shrink by one percent, its mass remaining the same, the value of \(g\) on the earth’s surface would
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