for the tennis ball, the gravitational acceleration inward is 3.32 x 10-9 m/s2 (or 0.00000000332 meters per second squared), and the centrifugal acceleration will be roughly 376 m/s2. The net acceleration is the difference between those two quantities, or approximately 376 m/s2 outward. So naturally the water flies off the tennis ball.
Next, repeat the same calculation for a drop of water at the Earth’s surface. The gravitational acceleration is 9.82 m/s2 inward, while the centrifugal acceleration is 0.03 m/s2 outward. Thus, the net acceleration is 9.79 m/s2 inward. The net force on the drop of water is inward, and it does not fly off the Earth, as summarized in Fig. II.14. The “tennis ball” analogy is an extremely poor one, because the gravitational force from the Earth is a billion times larger than that for the tennis ball, while the outward centrifugal force from the tennis ball is 10,000 times larger than that of the Earth. The gravitational force from the Earth is much greater than for the tennis ball because of the huge difference in the masses of Earth and tennis ball. The centrifugal acceleration depends on the square of the angular speed ω times the radius of the spinning object. For a tennis ball ω will correspond to about 1-20 complete revolutions/second, while for Earth ω corresponds to one complete revolution/day, 100, 000 or more times smaller than for the tennis ball . The huge difference in the square of ω more than compensates for the much larger radius of the Earth. The tennis ball analogy is relevant only for people who reject the concept of gravity!
Of course, Flat-Earthers could discover this by talking to any physicist; the calculation is trivial. The fact that they continue to push this “analogy” shows that they have no interest in answering their rhetorical question.
Here is a classic demonstration that the Earth is rotating, and that it makes one full (360o) rotation every day. Take a heavy plumb bob (you can purchase one at a hardware store), and attach it to a long wire. Carefully suspend the plumb bob so that it is hanging straight down. Then give it a push so that it starts swinging straight ahead. Watch this “Foucault Pendulum” for a long time. The direction of motion of the pendulum will change. In the Northern Hemisphere, the pendulum will rotate in the clockwise direction. In the Southern Hemisphere, the pendulum will rotate in a counter-clockwise direction. At the Equator the pendulum does not rotate.
After a certain amount of time, the direction of the pendulum will return to its initial orientation (but not at the Equator). A Foucault pendulum set in motion at the Equator will remain in the same orientation. At the North and South Poles, the pendulum will rotate by 360º in one day (but will rotate in opposite directions). At any other latitude, the pendulum will rotate with angular speed ω = 360º sin φ/day, where φ is the magnitude of the latitude. Thus, when φ = 30º (North or South latitude), it will take 2 days for the Foucault pendulum to make one full revolution, while for φ = 45º it will take 1.41 days to rotate.
The rotation of the Foucault pendulum is a direct demonstration of the rotation of the Earth. Flat-Earthers claim that the “rotation” is not reproducible, that the rotation is random, and that the pendulum will eventually come to rest and have to be re-started. The last of these statements is correct; the combined effects of friction and air resistance will cause the pendulum to slow down. The pendulum needs to be “re-started” at regular intervals. It is also true that great care must be taken to insure that the pendulum begins its trajectory with no initial sideways motion. Unless the initial conditions are very carefully set, the circular motion from the Earth’s rotation will mix with another normal mode to produce elliptical motion.
However, the pendulum does not have to be observed for an exceptionally long time to estimate its period. And the fact that the period increases as one goes from the North Pole to the Equator (where it does not rotate), and that the direct