## Essential University Physics: Volume 1 (4th Edition) Clone

We treat the earth as a solid sphere. We know that there are 86,400 seconds in a day. Thus, the earth is rotating and $7.272205\times10^{-5}$ radians per second. If the day were a second less, the earth would rotate at $7.272289\times10^{-5}$ radians per second. Thus, we find: $\Delta K = \frac{1}{2}(\frac{2}{5}M_ER_E^2)(( 7.272121\times10^{-5})^2-(7.272289\times10^{-5})^2)=2.97\times10^{24}$ Dividing this by $16\times10^{12}W$ and then converting to seconds, we find: $\Delta t \approx 12,000\ years$