If a planet of radius $R$ spins with an angular velocity $\omega$ about an axis through the North Pole, what is the ratio of the normal force experienced by a person at the equator to that experienced by a person at the North Pole? Assume a constant gravitational field $g$ and that both people are stationary relative to the planet and are at sea level.

$ \textbf{(A)}\ g/R\omega^2$

$\textbf{(B)}\ R\omega^2/g $

$\textbf{(C)}\ 1- R\omega^2/g$

$\textbf{(D)}\ 1+g/R\omega^2$

$\textbf{(E)}\ 1+R\omega^2/g $
1. **Determine the normal force at the North Pole:**
   At the North Pole, the person is not experiencing any centrifugal force due to the planet's rotation. Therefore, the only force acting on the person is gravity. The normal force \( F_{N_1} \) at the North Pole is given by:
   \[
   F_{N_1} = mg
   \]
   where \( m \) is the mass of the person and \( g \) is the acceleration due to gravity.

2. **Determine the normal force at the Equator:**
   At the Equator, the person experiences a centrifugal force due to the planet's rotation. The centrifugal force \( F_c \) is given by:
   \[
   F_c = m \omega^2 R
   \]
   where \( \omega \) is the angular velocity and \( R \) is the radius of the planet. The normal force \( F_{N_2} \) at the Equator is the difference between the gravitational force and the centrifugal force:
   \[
   F_{N_2} = mg - F_c = mg - m \omega^2 R
   \]

3. **Calculate the ratio of the normal forces:**
   The ratio of the normal force at the Equator to the normal force at the North Pole is:
   \[
   \frac{F_{N_2}}{F_{N_1}} = \frac{mg - m \omega^2 R}{mg}
   \]
   Simplifying the expression:
   \[
   \frac{F_{N_2}}{F_{N_1}} = \frac{mg - m \omega^2 R}{mg} = 1 - \frac{R \omega^2}{g}
   \]

4. **Conclusion:**
   The ratio of the normal force experienced by a person at the Equator to that experienced by a person at the North Pole is:
   \[
   \boxed{1 - \frac{R \omega^2}{g}}
   \]