QuestionIf Earth spins faster, find the period $T$ for centripetal acceleration $a = 9.8 \, \mathrm{m/s}^{2}$ using $a = \frac{4\pi^{2}r}{T^{2}}$. Use $r = 6.38 \times 10^{6}$ m. Output $T$ in minutes.

Studdy Solution
Calculate the period in minutes.
$$_{\mathrm{min}} = \frac{5056.5\, \mathrm{s}}{60} =84.275\, \mathrm{min}$$
So, if the Earth were spinning such that the centripetal acceleration is equal to $9. \, \mathrm{m/s}^{2}$, the period of one full rotation of the Earth would be approximately84.275 minutes.