Is it possible to derivate Coriolis force using scalars only?

Question

I am a self learnt dude and i have seen and studied the derivation of coriolis force of the earth using vector algebra;

But I tried to derive it using only scalars and also started doing it; but got stuck.. Here is what i tried :

If an object is moved in a direction perpenndicular to the angular velocity of the sphere; it's linear velocity stays constant; But the angular velocity also stays constant. this causes the object to deflect a bit towards the direction of the rotation of the sphere. This is the concept i will use to derive the formula

Radius of motion of the object on earth on a latitude $\varphi$: $$ R_E \cos \varphi$$

Linear velocity of the object: $$\omega_E R_E \cos \varphi$$

When the object's latitude changes (increases in this case)($d\varphi$ refers to change in latitude): $$\omega _ E R_E \cos(\varphi + d\varphi)$$

Relative motion of the object(motion due to the fictitious coriolis force): $$\omega_E R_E \cos \varphi - \omega_E R_E \cos (\varphi + d\varphi)$$

and when using identity of cosine, it becomes:$$\omega_E R_E \cos \varphi - \omega_E R_E \cos \varphi \cos d\varphi - \omega _ E R_E \sin \varphi \sin d\varphi $$

and after factoring: $$(\omega_E R_E \cos \varphi)(1-\cos d\varphi - \tan \varphi \sin d\varphi)$$

Now let the speed of the body moving perpendicularly on earth be $v$

and the time taken for all of this events : $dt$

now we know that : $vdt=R_E d \varphi$

therefore, $d\varphi = \frac{vdt}{R_E}$

(lets call $\frac{v}{R_E}$ as $k$)

subsitute that into the main equation and find acceleration(change in velocity/time) : $$\frac {(\omega_E R_E \cos \varphi)(1-\cos d\varphi - \tan \varphi \sin d\varphi)}{dt}$$

which becomes :

$$ (\omega_E R_E \cos \varphi) \left( \frac{1- \cos k dt - \tan \varphi \sin k dt}{dt}\right)$$

I am unable to simplfy beyond this, Can anybody help?

Answer 1

I start with giving a link to a Oct. 2024 answer by me with derivation of the magnitude of the rotation effect.
That derivation uses scalars only.

Note:
The above linked to answer is for the context of the rotation-of-Earth-effect that is taken into account in Oceanography and Meteorology. That is, the context of mass that is free to move parallel to the local Earth surface, but the weight is carried by the Earth surface, hence the mass is constrained in height relative to the Earth surface.

As overarching category for the fluid dynamics of oceanic currents and the fluid dynamics of air mass along the surface of the Earth I will use the expression 'Geophysical Fluid Dynamics' (GFD)

We have for the case of GFD: the weight of the fluid is resting on the Earth surface. (The fluid tends to flow in layers, the weight of each layer carried by the layers below it.)

The reason to state the above explicitly: the derivation of the magnitude of the rotation-of-Earth-effect that I linked to is specifically for the case of GFD.

The thing to be aware of: orbital motion of satelites around the Earth is a different case; the weight of the satellite is not carried by the Earth surface.

You write:

If an object is moved in a direction perpendicular to the angular velocity of the sphere; it's linear velocity stays constant; But the angular velocity also stays constant.

The last statement 'the angular velocity also stays constant' is incorrect.

The fluid is circumnavigating the Earth's axis.
That means: the dynamics is rotational dynamics, not linear dynamics.

Let's say we start with fluid that is circumnavigating the Earth's axis, co-rotating with the Earth. Then we impart to the fluid a velocity in north-south direction. Then the angular momentum of the fluid will remain constant.

Derivation of the rotation-of-Earth-effect

The derivation in the linked to answer gives the magnitude of the effect.

The direction of the effect is straightforward: the derivation demonstrates that the effect acts perpendicular to the instantaneous velocity relative to the rotating system.

General remarks about the rotation-of-Earth-effect that is taken into account in GFD:

As we know, the fluid is not in any way constrained in the direction parallel to the Earth's surface.

If the Earth would be perfectly spherical, and rotating, then all the fluid would flow to the equatorial region.

The reason that the fluid layer is on average the same thickness over the entire Earth: the Earth itself is not a perfect sphere, there is an equatorial bulge.

The Earth started out as a protoplanetary disk. Over time that disk contracted to a oblate spheroid. The Earth did not contract all the way to perfectly spherical shape. The magnitude of the equatorial bulge: the equatorial radius is about 21 kilometers larger than the polar radius.

It may look as if that difference is negligable, but it isn't.

Forces on a rotating celestial body

The image above is an animated GIF with three frames. In the diagram the equatorial bulge is very much exaggerated for the purpose of demonstration.

Blue arrow: gravity
Red arrow: buoyancy force
Green arrow: resultant force of gravity and buoyancy force.

At all latitudes other than the Equator and the poles the force of gravity and the buoyant force are not exactly opposite to each other.

The resultant force (in the diagram represented by the green arrow) provides the centripetal force that a fluid requires in order to remain co-rotating with the Earth at the same latitude.