#### Answer

The climber's speed is faster than the surfer's speed by 20 cm/s

#### Work Step by Step

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.