## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Let $R'$ be the radius of the earth after it shrinks. Note that the mass $M$ of the earth does not change. We can find the value of $\frac{R'}{R}$, where $R$ is the original radius of the earth. $\frac{G~M}{(R')^2} = 3\times \frac{G~M}{R^2}$ $(\frac{R'}{R})^2 = \frac{1}{3}$ $\frac{R'}{R} = \sqrt{\frac{1}{3}}$ $\frac{R'}{R} = 0.577$ If the radius shrinks to 0.577 of the original radius, the free-fall acceleration would be three times its present value.