Homework Help: Coriolis force

1. Dec 28, 2017

Kosta1234

1. The problem statement, all variables and given/known data
A plane is flying in latitude 25S to the west.
due to the Coriolis force will he fly to the southern pole or to the equator?
and the same question but that he is flying to the west in 25N?

2. Relevant equations

3. The attempt at a solution
I'm not sure, but think that Coriolis force is acting on a body when it's in the southern part of Earth to the left (so he will go to the southern pole) and in the northern part to the right (so he will go to the northern pole)
So, in both cases he will go to the poles?

Thank you.

2. Dec 28, 2017

Orodruin

Staff Emeritus
Yes.

However, "go to" is a bit misleading. The Coriolis force will be acting in the direction of the pole, that does not mean that the plane will end up at the pole.

3. Dec 28, 2017

FactChecker

The equation for the Coriolis acceleration is ac = -2Ω×v, where v is the velocity vector and Ω is the angular velocity vector. The acceleration ac would be perpendicular to Ω, so I don't think it would go toward either pole. CORRECTION: THIS POST IS WRONG. It is explained by @Orodruin and @ehild below.

Last edited: Dec 28, 2017
4. Dec 28, 2017

Orodruin

Staff Emeritus
In general when Coriolis forces are talked about in reference to motion on the Earth's surface, one is actually talking about the projection of the Coriolis force on the tangent space of the Earth. With the motion described here, this component is indeed directed towards the poles. Of course, if you look at the full Coriolis force, it will generally also have a component orthogonal to the Earth's surface.

5. Dec 28, 2017

FactChecker

I'm having trouble seeing how the projection would have a North/South component. CORRECTION: This is explained by @Orodruin and @ehild below.

Last edited: Dec 28, 2017
6. Dec 28, 2017

Orodruin

Staff Emeritus
The angular velocity is $\vec \omega = \omega \vec e_z = \omega [\cos(\theta) \vec e_r - \sin(\theta) \vec e_\theta]$. A westward velocity vector is given by $\vec v = -v \vec e_\varphi$. It follows that
$$\vec a_c = 2\vec v \times \vec \omega = - 2v\omega \vec e_\varphi \times [\cos(\theta) \vec e_r - \sin(\theta) \vec e_\theta] = - 2v\omega [\cos(\theta) \vec e_\theta + \sin(\theta) \vec e_r].$$
The projection onto the tangent plane of the sphere (spanned by $\vec e_\theta$ and $\vec e_\varphi$ at every point) is therefore $-2v\omega \cos(\theta) \vec e_\theta$ and $\vec e_\theta$ is a unit vector in the south direction. For $\theta > \pi/2$, $\cos(\theta) > 0$ and this vector is therefore pointing towards the north and for $\theta > \pi/2$, $\cos(\theta) < 0$ and this vector is therefore pointing to the south.

At the equator $\theta = \pi/2$ and therefore $\cos(\theta) = 0$, leading to no Coriolis force in the tangent plane at all (as you would expect).

Edit: Removed some typos (1 -> 0).
Edit 2: Some clarifications.

Edit 3: The perhaps easiest way to see that the force must have a component in the tangent plane (everywhere but at the equator) is to note that the angular velocity is not in the tangent plane. Hence, its component orthogonal to the tangent plane will give a vector in the tangent plane when crossed with any velocity vector in the tangent plane.

Last edited: Dec 28, 2017
7. Dec 28, 2017

ehild

In addition to @Orodruin's explanation: The velocity vector points out of the plane of the picture. The Coriolis force points towards the rotation axis, and it has a radial component (towards the center of Earth), and one in the plane tangent of the Earth surface.

8. Dec 28, 2017

Orodruin

Staff Emeritus
Which, luckily, is exactly what we see here:

Edit: I wanted to draw a picture, but I was too lazy and went for the maths instead.

9. Dec 28, 2017

scottdave

Perhaps you could think of it as acting like a Foucault Pendulum, which moves in a counterclockwise direction (relative the the Earth) in the Southern Hemisphere. A plane moving West would tend to veer toward the South Pole.

In the Northern Hemisphere, it will move clockwise. A plane moving West would tend to veer toward the North Pole.

10. Dec 28, 2017

FactChecker

I see. Thanks. I will/have corrected my posts above.

11. Dec 29, 2017

Kosta1234

Thank you everyone!