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

We can find the angular speed of the earth as it rotates once each day: $\omega = \frac{2\pi~rad}{(24)(3600~s)}$ $\omega = 7.27\times 10^{-5}~rad/s$ For the surfer, the radius of rotation is $6.4\times 10^6~m$. Therefore: $v = \omega ~r$ $v = (7.27\times 10^{-5}~rad/s)(6.4\times 10^6~m)$ $v = 465.3~m/s$ For the climber, the radius of rotation is $6.403\times 10^6~m$. Therefore: $v = \omega ~r$ $v = (7.27\times 10^{-5}~rad/s)(6.403\times 10^6~m)$ $v = 465.5~m/s$ The climber's speed is faster than the surfer's speed by 20 cm/s.