The Coriolis Force on the Rotating Earth
~6 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with vector cross products and the general expression for fictitious forces in a rotating frame, as derived in the previous lecture.
Why this matters
The Coriolis force explains why large-scale weather systems and ocean currents deflect systematically across the globe.
While the centrifugal force modifies the effective gravity we feel, the Coriolis force is velocity-dependent and acts perpendicular to the motion. This distinction is crucial for understanding atmospheric dynamics, where it balances pressure gradients to create geostrophic winds, and for long-range ballistics, where it causes projectiles to deviate from their expected paths.
The Intuitive Picture
Imagine standing at the North Pole and throwing a ball towards the equator. As the ball travels south, it moves into a region of the Earth that is rotating faster (in terms of linear speed) than the pole, because the radius of rotation increases. However, the ball retains its initial angular momentum. From the perspective of an observer on the rotating Earth, the ball appears to lag behind the surface, deflecting to the right in the Northern Hemisphere. This is not a real force in the inertial sense, but a kinematic effect of describing motion in a rotating coordinate system. In the Southern Hemisphere, the deflection is to the left. This systematic deflection is the hallmark of the Coriolis effect.
Decomposing the Earth's Rotation
To apply the general rotating frame equation to Earth, we must resolve the angular velocity vector \(\vec{\omega}\) into components relative to a local coordinate system. Let us define a local frame at latitude \(\lambda\): \(\hat{\mathbf{e}}_r\) points radially outward from the Earth's center, \(\hat{\mathbf{e}}_\theta\) points north along the meridian, and \(\hat{\mathbf{e}}_\phi\) points east along the parallel. The Earth's rotation axis is tilted relative to the local vertical by the angle \(\lambda\).
The angular velocity vector \(\vec{\omega}\) can be decomposed into a vertical component \(\omega \sin\lambda\) and a horizontal component \(\omega \cos\lambda\) pointing north. Thus, \(\vec{\omega} = \omega \sin\lambda\, \hat{\mathbf{e}}_r + \omega \cos\lambda\, \hat{\mathbf{e}}_\theta\). This decomposition is essential because the Coriolis force depends on the cross product \(\vec{\omega} \times \vec{v}_R\). The vertical component of \(\vec{\omega}\) is responsible for the horizontal deflections (east-west and north-south), while the horizontal component contributes to vertical effects, though these are typically negligible for atmospheric motion.
Calculating the Coriolis Acceleration
The Coriolis acceleration is given by \(\vec{a}_C = -2\vec{\omega} \times \vec{v}_R\). Let the velocity in the rotating frame be \(\vec{v}_R = u\, \hat{\mathbf{e}}_\phi + v\, \hat{\mathbf{e}}_\theta + w\, \hat{\mathbf{e}}_r\), where \(u\) is the eastward velocity, \(v\) is the northward velocity, and \(w\) is the vertical velocity. We compute the cross product term by term. The dominant term for horizontal motion arises from the vertical component of \(\vec{\omega}\), which is \(\omega \sin\lambda\, \hat{\mathbf{e}}_r\).
Writing \(\vec{\omega} = (0,\ \omega\cos\lambda,\ \omega\sin\lambda)\) and \(\vec{v}_R = (u, v, w)\) in (east, north, up) components and expanding \(-2\vec{\omega}\times\vec{v}_R\) gives the components below. Applying this to the velocity components, the Coriolis acceleration components in the local frame are derived as follows. The eastward component of acceleration is \(a_{C,\phi} = 2\omega \sin\lambda\, v - 2\omega\cos\lambda\, w\); for horizontal motion (\(w = 0\)) this is \(2\omega \sin\lambda\, v\). The northward component is \(a_{C,\theta} = -2\omega \sin\lambda\, u\). The vertical component is \(a_{C,r} = 2\omega \cos\lambda\, u\). Note the sign conventions: a positive \(v\) (northward motion) results in a positive \(a_{C,\phi}\) (eastward deflection) in the Northern Hemisphere (\(\lambda > 0\)).
The Coriolis Parameter
In meteorology and oceanography, the coefficient \(f = 2\omega \sin\lambda\) is known as the Coriolis parameter. It quantifies the strength of the Coriolis effect at a given latitude. At the equator (\(\lambda = 0\)), \(f = 0\), meaning there is no horizontal Coriolis deflection. At the poles (\(\lambda = \pm 90^\circ\)), \(|f| = 2\omega\), which is the maximum possible value. The sign of \(f\) determines the direction of deflection: positive in the Northern Hemisphere and negative in the Southern Hemisphere. This parameter is central to the geostrophic balance, where the Coriolis force balances the pressure gradient force, leading to winds that flow parallel to isobars rather than directly from high to low pressure.
The derivation
Rotating Frames: Centrifugal and Coriolis Forces
If \(S\) is an inertial frame and \(R\) is a frame rotating relative to \(S\) with angular velocity \(\vec{\omega}(t)\) about a common origin, then a particle of mass \(m\) obeys, in the rotating frame, \(m\,\vec{a}_R = \vec{F} - 2m\,\vec{\omega}\times\vec{v}_R - m\,\vec{\omega}\times(\vec{\omega}\times\vec{r}) - m\,\dot{\vec{\omega}}\times\vec{r}\), where \(\vec{F}\) is the real (inertial) force and the three added terms are the Coriolis, centrifugal, and Euler fictitious forces.
A common misconception is that the Coriolis force causes water to swirl in opposite directions in bathtubs or toilets. In reality, the Coriolis parameter \(f\) is extremely small for small-scale, short-duration flows. The frictional forces and the initial conditions of the water dominate the rotation direction. The Coriolis effect is only significant for large-scale systems (like hurricanes) that persist for hours or days and span hundreds of kilometers. C
Worked example
Problem. A projectile is fired horizontally eastward with a speed of \(v = 100\ \mathrm{m/s}\) at a latitude of \(\lambda = 45^\circ\). Calculate the initial northward Coriolis acceleration. Use \(\omega = 7.29 \times 10^{-5}\ \mathrm{rad/s}\).
- Identify the relevant component of the Coriolis acceleration. For eastward motion (\(u = v\), \(v_{north} = 0\)), the northward component is \(a_{C,\theta} = -2\omega \sin\lambda\, u\).
- Calculate the Coriolis parameter \(f = 2\omega \sin\lambda\). With \(\lambda = 45^\circ\), \(\sin 45^\circ = 1/\sqrt{2} \approx 0.707\). Thus, \(f = 2(7.29 \times 10^{-5})(0.707) \approx 1.03 \times 10^{-4}\ \mathrm{s^{-1}}\).
- Compute the acceleration. \(a_{C,\theta} = -f \cdot u = -(1.03 \times 10^{-4}\ \mathrm{s^{-1}})(100\ \mathrm{m/s})\).
- Perform the multiplication: \(a_{C,\theta} \approx -0.0103\ \mathrm{m/s^2}\). The negative sign means the deflection is southward — to the right of the eastward motion, as expected in the Northern Hemisphere.
Units check. \(\omega\) is in \(\mathrm{rad/s}\) (dimensionless rad), \(\sin\lambda\) is dimensionless, and \(u\) is in \(\mathrm{m/s}\). The product is \(\mathrm{m/s^2}\), which is correct for acceleration.
Three quick tests
- Why is the Coriolis force zero at the equator?
Answer
The vertical component of Earth's angular velocity is zero at the equator, so the cross product with horizontal velocity yields no horizontal deflection. - If a wind blows from the North to the South in the Northern Hemisphere, which way does it deflect?
Answer
To the west. In the Northern Hemisphere the Coriolis force deflects motion to the right, and the right of a southward velocity is west. - How does the magnitude of the Coriolis parameter \(f\) change as you move from the equator to the pole?
Answer
It increases linearly with \(\sin\lambda\), reaching its maximum value \(2\omega\) at the poles.
Problem set for PU-101 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.