All objects on the surface of the Earth rotate with it about the North-South axis. Let $R$ be the radius of the Earth and $\lambda$ be the latitude of an object. The object undergoes uniform circular motion in a plane parallel to the equatorial plane.

The radius of its circular path, $r$, is the perpendicular distance from the object to the axis of rotation. From the geometry of a cross-section of the Earth, this radius is given by:

$$r(\lambda) = R \cos(\lambda)$$

[Q1] Where are the centers of their circular paths located? The center of any object's circular path lies on the Earth's axis of rotation.

• For the object in Beijing at latitude $\lambda_B = 40^\circ$ N, its circular path is the 40° N latitude circle. The center of this circle is on the Earth's axis of rotation.
• For the object at the Equator at latitude $\lambda_E = 0^\circ$, its circular path is the Equator itself. The center of this circle is the center of the Earth.

[Q2] What is the ratio of the radii of their circular paths (Beijing to Equator)? The radius for Beijing is $r_B = R \cos(40^\circ)$. The radius for the Equator is $r_E = R \cos(0^\circ) = R$. The ratio is:

$$\frac{r_B}{r_E} = \frac{R \cos(40^\circ)}{R} = \cos(40^\circ)$$

[Q3] What is the ratio of their angular velocities? The Earth rotates as a rigid body, so all points on its surface have the same angular velocity, $\omega$.

$$\omega_B = \omega_E = \omega_{Earth}$$

Therefore, the ratio is:

$$\frac{\omega_B}{\omega_E} = 1$$

[Q4] What is the ratio of their linear velocities? The linear velocity is given by $v = \omega r$. The ratio of the linear velocities is:

$$\frac{v_B}{v_E} = \frac{\omega_B r_B}{\omega_E r_E}$$

Since $\omega_B = \omega_E$, the ratio simplifies to the ratio of the radii:

$$\frac{v_B}{v_E} = \frac{r_B}{r_E} = \cos(40^\circ)$$

[Q5] What is the ratio of their centripetal accelerations? The centripetal acceleration is given by $a_c = \omega^2 r$. The ratio of the centripetal accelerations is:

$$\frac{a_B}{a_E} = \frac{\omega_B^2 r_B}{\omega_E^2 r_E}$$

Since $\omega_B = \omega_E$, the ratio simplifies to the ratio of the radii:

$$\frac{a_B}{a_E} = \frac{r_B}{r_E} = \cos(40^\circ)$$