Answer
We note that, since $$v=2 \pi r / T,$$ the centripetal acceleration may be written as $$a= 4 \pi^{2} r / T^{2} .$$ To express the result in terms of $g,$ we divide by 9.8 $\mathrm{m} / \mathrm{s}^{2}$ . The acceleration associated with Earth's $\operatorname{spin}(T=24 \mathrm{h}=86400 \mathrm{s})$ is $$a=g \times \frac{4 \pi^{2}\left(6.37 \times 10^{6} \mathrm{m}\right)}{(86400 \mathrm{s})^{2}\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)}=3.4 \times 10^{-3} \mathrm{g}$$
Work Step by Step
We note that, since $$v=2 \pi r / T,$$ the centripetal acceleration may be written as $$a= 4 \pi^{2} r / T^{2} .$$ To express the result in terms of $g,$ we divide by 9.8 $\mathrm{m} / \mathrm{s}^{2}$ . The acceleration associated with Earth's $\operatorname{spin}(T=24 \mathrm{h}=86400 \mathrm{s})$ is $$a=g \times \frac{4 \pi^{2}\left(6.37 \times 10^{6} \mathrm{m}\right)}{(86400 \mathrm{s})^{2}\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)}=3.4 \times 10^{-3} \mathrm{g}$$
