221 – The Earth and the atmosphere are motionless

We have see in the first article, “Coriolis proves the Earth is motionless“, that the Coriolis acceleration is not manifestating on the Earth as it should do. Many people however makes the obiection that the atmosphere is pulled in rotation together with the Earth and acts on the helicopter with a lateral force that nullifies the Coriolis acceleration.

I want to answer in this article to this obiection. When you try the calculation (I will write it in another article), you will notice that, for an airplane or helicopter that is flying at an average speed of 500 km/h, the Coriolis acceleration is about 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 much. 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?

Consider now to have two different helicopters that start together their travel from the North Pole to the equator. These helicopters have the same mass (5000kg) and can develop the same speed of 500 km/h. This means that at a certain time they will have run the same distance. The only difference is the geometry. One is more compact and offers a side surface to the push of the atmosphere of 10 m2, while the other offers a side surface of 15 m2.

two helicopters


The Coriolis acceleration that the atmosphere should create to maintain the two helicopters with the same peripheral speed of the Earth is given by the formula:
AC = 2wVh
where Ac is the Coriolis acceleration w is the angular speed of the Earth and Vh is the speed of the two helicopters. Since Vh is the same it is clear that it is needed the same acceleration for the two helicopters. The force necessary to develop such an acceleration is given by Fc=m∙Ac.
Since the mass of the two helicopters is the same the needed force is the same.

The pressure due to the force impressed by the atmosphere on the side of the two helicopters however is different. Since one helicopter is bigger of the other A1>A2 we will have that P1<P2 according to the formula P=F/A. But the pressure generated by the atmosphere is linked by the relative speed of the atmosphere with the helicopter with the formula below with which we can calculate the necessary speed of the atmosphere.

where r is the air density (1,25 kg/m3).

Since P1 and P2 are different we find that two different speeds of the atmosphere V1<V2 are needed. This is not possible because in the same point (where we have both the helicopters) we can have only one single speed of the atmosphere. This is a clear evidence that the atmosphere doesn’t act over the helicopter to pull it with the Earth in rotation.

Let’s consider now the Foucault pendulum.

Its rotating movement during the oscillations is considered to be a proof of the rotation of the
Earth. Why, in this case, the atmosphere doesn’t act on it, blocking the rotation of the pendulum in connection with the Earth? If you consider the objection to be good in the case of the helicopter, it should be valid for the pendulum as well. This is a clear demonstration that the atmosphere around the earth doesn’t exert any influence in order to nullify the Coriolis acceleration.

But what else could be added on the topic? Regarding the Foucault pendulum, I mean. This has always been used as an evidence of the Earth’s rotation. It is because, in the course of its oscillation, it doesn’t follow the earth’s meridian. This would imply that you too, when sitting inside the helicopter, in the case it stopped for a brief time in the air, you yourself, I mean, should be able to behold the Earth moving beneath your feet. Anyway, this never happens.

The Foucault pendulum

A lot of experiments have shown that Foucault pendulum works as expected only if launched in a very carefully chosen direction, with a specific initial force. A random launch will not produce the expected rotating movement. The conclusion is that the Foucault pendulum cannot be considered a proof of the Earth’s rotation.

To stay on the topic, we could even discuss Guglielmini’s experience. He launched many lead balls from a tower 100 m tall in Boulogne. Story tells that, in his experiment, he found that the ball had fallen down with a displacement of 17 cm far from the tower basis, toward est. The explanation is that the tower top, being 100 m high has a peripheral speed of rotation greater than the basis, according to the formula V=ω ∙ r.

This experiment underlines, once more, that a body in the air is not pulled by the Earth or the atmosphere but it moves with the peripheral speed of the point from where it has started its motion, in this case the tower’s top. So, if Guglielmini’s experience has to be considered valid, also our consideration about the airplane that finds it impossible to follow the Earth should be considered valid. But, is Guglielmini’s experiment true? Let’s see.

Guglielmini’s experiment from the Asinelli tower in Boulogne

The bowl has a peripheral speed that is the peripheral speed of the top of the tower. The bottom of the tower moves at a lower speed because it is nearer to the center of the globe.
There is then a triangular speed profile like that in the picture below. Science states that the bowl falls down with the peripheral speed of the top of the tower andthus it moves eastward during its falling.

As a consequence of this experiment, if a helicopter moves on its vertical and stays stopped, let’s say, for one hour on the same place, it should feel the Earth moving under its belly. This should happen because, when the helicopter rises up, it maintains the peripheral speed of the Earth, but at an altitude of, let’s say, 2000 meters, it should have a greater speed to keep stationary on the same point.

Let’s make the calculation by supposing that the helicopter is on an airport situated at the equator. The equator has a peripheral speed of 1670 km/h (1669,3 to be a little more precise) at a radius of 6378 km. The helicopter flies staying on the vertical at 2000 meters altitude. Let’s calculate the speed that the helicopter should have at that altitude, to remain on the
vertical. 𝑉𝑝 = 𝜔 ∙ 𝑟 where Vp is the peripheral speed and ω is the rotation speed expressed
in radiants per second.

ωearth=7,27E–5 rad/sec = 6.94E–4 rpm
Vp=7,27E–5 ∙ (6378000+2000)=464 m/s ∙ 3.6= 1670.28 km/h
Δ=1670.28–1669.3≈1 km/h

The difference, as you can immediately notice, is very small: just 1 km/h, but it means that, if the helicopter stays in the same position, at that altitude, for one hour, the Earth will move of 1 km under the helicopter itself. Ask to pilots: this does not happen, in any case. Another proof that the Guglielmini’s experience is false and that the Earth doesn’t move.


http://www.aracneeditrice.it/index.php/pubblicazione.html?item=9788825530308

Rispondi

Inserisci i tuoi dati qui sotto o clicca su un'icona per effettuare l'accesso:

Logo di WordPress.com

Stai commentando usando il tuo account WordPress.com. Chiudi sessione /  Modifica )

Google photo

Stai commentando usando il tuo account Google. Chiudi sessione /  Modifica )

Foto Twitter

Stai commentando usando il tuo account Twitter. Chiudi sessione /  Modifica )

Foto di Facebook

Stai commentando usando il tuo account Facebook. Chiudi sessione /  Modifica )

Connessione a %s...