By writing this article I want to give some more accurate calculation of the Coriolis acceleration on the Globe. I would like to prove, this way, that you cannot detect Coriolis on Earth because the planet is not spinning. I want to check an affirmation I made time ago. It’s about an airplane taking off from the north pole and from there flying toward the equator in an exclusively south direction. While keeping its speed constant, it should detect the Earth rotating beneath. The rotation speed of the Earth should gradually increase to reach at the equator the incredible speed of almost 1700km/h. This is the speed of the Earth due to its rotation around its axis.
So, I will try to make the exact calculation of the Coriolis acceleration. The expected result should show that the earth moves at 1700km/h under the airplane. Why is this an important calculation? Because there’s a number of guys who underestimate Coriolis. They minimize this acceleration and pretend an airplane can easily correct its speed in order to reach the landing airport. On the other hand, it should be clear that, with such a quick spinning under it, an airplane will not be in the possibility to land. The only possibility left is to return to the exact starting point, and then try a landing there.
Peripheral speed on the pole
On the pole, the peripheral speed of the Earth is zero and the airplane is purely rotating around the earth’s axis. When it takes off it has no peripheral speed and when moving toward south the earth won’t pull it anymore. So it will be flying independently from the Earth.
We know the Coriolis acceleration will affect a body proceeding in a rectilinear way. In the same time, this same body will move with a uniform speed V and in the radial direction over a rotating body. Radial means from the center toward the external circumference or vice versa. We can calculate this acceleration with the formula ac=2xVxω. On the globe, an airplane that moves radially moves in south-north direction or vice versa.
Two different components
The Coriolis acceleration is formed by two different components. Only one of this is in relation with the peripheral speed of the Earth and is the one that we want. Let’s see these two components. This part is a little more mathematical so, if you want, you can directly skip to the final considerations.
A Platform with a guide
Let’s consider a point P moving with constant speed V along a platform with a guide that constrains the point to move in a rectilinear way. We can express the speed as meaning that the speed is the derivative in respect of time of the distance from the center.
The rotation speed of the platform is i.e. the derivative of the angle. If we consider an interval of time dt, the point P moves for a distance along the guide. So, its rotation radius changes and with it the peripheral speed changes from =ω(r+rdt).
The variation of speed, i.e. the acceleration, according to the general formula will be:
The second component
This is the first component of the Coriolis acceleration, and it is the component we are interested in. And just to be complete, we have to consider the other component too. In the time interval dt the radial speed V changes direction and the angle with the horizontal changes of . This is a very small angle, thus the variation of the vector is given by .=ωVdt
The acceleration term that arises is
If we sum together the two terms we have exactly the Coriolis acceleration, but we are interested mainly in the first component, i.e. in half of the Coriolis acceleration.
Let’s try to apply the calculation to our airplane that flies with a speed of 500km/h toward the equator.
We have that the globe spins with one turn per day i.e
The airplane moves at 500km/h but we have to consider its projection on the globe radius and calculate the speed of the point P*. It is rather an easy calculation. From the north pole to the equator on a globe Earth there are 10000km.
With a speed of 500km/h, we have a 20 hours journey. The radius of the globe is 6378km. The speed of the point p* will be:
With this speed I can calculate half the acceleration of Coriolis:
A speed greater than the sound speed in the air
Using the time of 20 hours i.e. 72000seconds we can calculate the final speed of the Earth at the equator. This is the speed of the Earth moving under the Airplane.
Is this a small speed? It is much greater than the sound speed in the air. It is clearly impossible for the airplane to land.
An objection often made is that the atmosphere that spins with the earth pushes the airplane. Below I copy an old study I did on this argument.
A very small acceleration
When you try the calculation, you will notice that, for an airplane or helicopter that is flying at an average speed of 500 km/h, the Coriolis acceleration is less than 0,0065 m/s2. It is a very small acceleration. If you consider a lateral surface, offered to the wind by the helicopter (10 m2 for a total mass of 5000 kg), you will reach a needed lateral force of the wind of 33 N, that really does not seem so great. So, you could infer, it would be possible for the atmosphere to produce a sort of lateral and very constant wind forcing the helicopter to move while avoiding the Coriolis effect. But is it really like this?
A force of 33 Newton on a 10 square meter surface generates a pressure of 3.3 Pascal [N/m2]. You can use the formula
where ρ is the air density (1,25 kg/m3), useful to calculate the speed of the wind generating this amount of pressure and which the helicopter should constantly feel on its side.
We find a side wind speed of 2.3 m/s that should be constantly applied to the helicopter in the east direction (the rotation direction of the Earth). But look at the following picture.
The speed of winds
We imagined that helicopter starting its journey from the north pole. As you can deduce from the picture, for a long distance, polar winds blow westward and not eastward, which would be needed to win the Coriolis acceleration. How much will be the speed of these winds?
You have to consider that winds have a logarithmic profile with the altitude, so they pass from low speed at soil level to high speed at higher levels according to the formula:
Where Vz and Vs are the speeds respectively at z height and at soil and hz and hs are the respective heights. n is a coefficient that describes the nature of terrain at soil level.
This formula generates a wind speed profile of this kind:
So, if we have a wind of 3 m/s at 10 meters height, we can easily have a 7-12 m/s wind at 2000 meters height. This wind can nullify or even win this hypothetical push of the atmosphere on the helicopter. The result will be that, once more, a helicopter should feel the earth rotating at high speed under itself, due to Coriolis.