Since we have agreed flying off a spinning ball proves the Earth can't rotate, so we can return to the main issue.

Stop lying, we have not agreed to that at all.

That is the baseless assertion put forward by flat Earthers or other geocentrists.

The main issue is that that is a load of garbage.

It has been explained why that is a load of garbage and you have shown know problems with this explanation.

Again:

People love appealing to the linear velocity, but spinning a tiny ball at that velocity and a large ball at that velocity produce vastly different forces, making the comparison meaningless.

If you want to see if the ball's gravity can hold it you need to spin a ball with the same density as Earth at the same angular velocity.

Here is why:

For gravity at the surface of roughly spherical object of mass

*M* and radius

*r*, we can find the acceleration

*a*=

*G M*/

*r*^{2}.

Where

*M*=

*ρ V*=(4/3) π

*ρ r*^{3}.

Thus we can rewrite the acceleration due to gravity as

*a*=(4/3) π

*G ρ r*^{3}/

*r*^{2}=(4/3) π

*G ρ r*.

Notice that this is linearly dependent upon r. That means as you increase the size of the roughly spherical object, assuming you keep the density the same, the acceleration due to gravity at the surface will increase proportionally with the radius. Double the radius, double the acceleration due to gravity.

Now, in order to prevent the water flying off, the acceleration due to gravity needs to be greater than the acceleration required to maintain a circular path.

For a path of radius

*r*, with an angular velocity of

*ω*, we have

*a*=

*ω*^{2} *r*.

So again, we see that this is linearly proportional to r. If you double the radius, you double the acceleration required.

This means that the angular velocity is the natural way to compare. We can even find the limiting condition where the water would just stick and above this angular velocity water wouldn't stick (assuming gravity is the only force holding it there rather than surface tension or cohesion). This is when the 2 accelerations are equal, i.e.:

(4/3) π

*G ρ r*=

*ω*^{2} *r*and thus:

(4/3) π

*G ρ*=

*ω*^{2}or:

*ω*=sqrt((4/3) π

*G ρ*)

For an object like Earth, with an average density of 5515.3 kg/m

^{3}, this works out to be an angular velocity of 0.0012 s

^{-1}.

This can also be expressed as a period of 5060 s or 84 min, or in rpm as 0.71 rpm.

This is much slower than typically done, but much much faster (roughly 20 x) than the rotational speed of Earth.

Using velocity does not work like this, and such doing is extremely dishonest or irrational, or both.

If we use velocity we end up with

*a*=

*v*^{2}/

*r*, with it now being inversely proportional to r. Doubling the radius would half the acceleration.

Again, if we find the limiting case we end up with:

(4/3) π

*G ρ r*=

*v*^{2}/

*r*and thus to find

*v* we end up with:

*v*=sqrt((4/3) π

*G ρ*)

*r*with the velocity now dependent upon

*r*.

Now if we do this for Earth, with a radius of roughly 6 371 000 m, we end up with a velocity of ~7900 m/s or roughly 28 500 km/hr, again, roughly 20x the speed Earth rotates at.

So Earth is clearly rotating no where near fast enough to have all the water fly off.

If instead we rotate a small ball, say a 30 cm wide basketball (which has a density much less than that of Earth), we have

*r*=0.15 m, you end up with the velocity needed to throw it off being a mere 0.19 mm/s, which is much slower than people rotate it in their demonstrations.

More importantly, the velocity required to compare it directly to Earth is such that the ratio of the 2 are the same.

As we have shown that this velocity is linearly proportional to radius, we just need to use that same proportion with the velocity.

So instead of the very large Earth spinning at 1600 km/hr, we have the tiny ball spinning at 0.01 mm/s.

This is much slower than the experiments show.

But don't worry, there can be a partial saving grace.

If you instead spin the ball such that the axis of rotation is horizontal, and just try to demonstrate that the water doesn't fly off the top as it is held down by Earth's gravity, you can go much faster. However now note that you also have Earth's rotation to deal with which does technically make it a bit more complicated.

Now we are comparing the acceleration due to Earth with the acceleration due to rotation.

So now, we can directly compare the acceleration due to the rotation of Earth with the acceleration due to the object.

Remember, using angular velocity, this is linearly proportional to radius, however it is the angular velocity squared.

So the keep the acceleration the same, the angular velocity squared is inversely proportional to r.

So multiplying the radius by 4 cuts the angular velocity in half.

This means the limiting angular velocity would be ~8/s, corresponding to a period of ~0.7 s, and 77 rpm. But that is still much slower than people spin the balls. To compare it to Earth however you would need an angular velocity of 0.47/s, so a period of 13 seconds or 4.5 rpm.

So these experiments with spinning quite rapidly do not show any problem with the real rotating Earth as they are spinning much to fast and anyone who is just appealing to the linear velocity either has no idea what they are talking about or is intentionally being dishonest by misrepresenting Earth.

It means that when the water flowing from the tap falls from 30 centimeters, it falls as much as 1 centimeter forward.

No, it doesn't.

I means that at the equator, where you actually experience that, instead of the water accelerating downwards at a rate of 9.8ish m/s, it instead falls at 9.8ish-0.036 m/s.

That 0.036 is insignificant compared to the 9.8. You aren't going to notice it, especially as if you haven't travelled you would have gotten used to it. Again, it varies by more than this across the surface of Earth.

At other locations, the acceleration due to rotation is smaller and isn't aligned with gravity so you have the effect of the location direction of gravity being skewed, resulting in Earth being an oblate spheroid instead of a perfect sphere.

This also results in variations in the value of

*g* around Earth, which has been measured, and can be done so by simply measuring the period of a pendulum in various locations.

The amount things will drift is tiny and is summed up by the Coriolis effect, and that is also observed, by typically requires very large scales to do so.

As a first order approximation, lets say you are pouring water from a tap or teapot into something 0.1 m below, and again, we will be at the equator to make it most significant.

Now the Earth is moving at a velocity of 464 m/s. Meanwhile, the additional height of the tap/pot means it is travelling at a staggering 464 m/s, or roughly 7 μm/s faster.

Now, the water/tea falls at a rate of 9.8 m/s

^{2}, so it takes a total of 0.14 seconds to fall.

That means that the water/tea moves a staggering 1 μm forwards in the time taken to fall into the cup.

YOU WILL NOT NOTICE THAT!

If you wish to observe the effect of this moving forwards, set up a Foucault's pendulum. Note the plane it starts to swing in and then note how this plane moves.