[Physics] Why would the apple fall under the tree

centrifugal forcecoriolis-effectnewtonian-mechanicsreference frames

Here is an extremely naive question: Why would the apple fall under the tree?

I am puzzled by this, because the conventional answer that the gravity between the apple and the earth pulling apple down is not satisfactory to me. My thought process goes as follows:

  1. We know that in an appropriate reference frame, we can view the apple falling down from the tree as free fall. Therefore we know $F=mg, a=g$ and the apple should fall to the ground in $\sqrt{\frac{2h}{g}}$ time period since $h=\frac{1}{2}gt^{2}$.

  2. However, the picture is not so clear when we consider earth's rotation. For convenience I ignore the earth movement around the sun. We know that the centripedal force and the gravity are given by
    $$
    F_{1}=m\omega^{2}R, F_{2}=c\frac{mM}{R^{2}}
    $$
    where $c$ is some constant. Therefore, the reason the apple falls to the tree must be the gravity is much stronger than the centrialfugal force required when the apple rotates with the tree. If the centrialfugal force is equal to the gravity, than the apple should be staying in the same spot at the tree. If the centrialfugal force needed is greater than gravity, then the apple would not stay in the free and would flying away from the earth.

  3. Now image a pear falling from the middle of the tree. From our everyday experience, the pear would fall in the same spot as the apple. However, since the pear is closer to the earth, the centripedal force it experiences is less, and the gravity it experiences is greater, too (Here we denote $R'$ for the pear's distance from the earth center, $m_{1}$ for its mass):
    $$
    P_{1}=m_{1}\omega^{2}R', P_{2}=c\frac{Mm_{1}}{R'^{2}}
    $$
    Therefore it is not difficult to see that the acceleration the apple and the pear experiences must be different because of the height. The pear must fall faster. However, since the apple and the pear moves from the same tree, they must have the same angular velocity. In particular when $R$ goes very large, the gravity would be too small and the object would fly away from earth.

  4. But I feel this explanation is unclear. Now instead of using a constant $g$ denoting the acceleration the apple experiences, we have:
    $$
    g(R)=c\frac{M}{R^{2}}-\omega^{2}R
    $$
    Therefore the apple should somehow deviate from the tree. Intuitively, since at the bottom of the tree the apple would not move at all, and at very high the apple would flying away, at a middle height the apple should have a moderate but measurable deviation.
  5. My question is, how can we calculate the deviation from the height of tree exactly? The above calculation assumed the angular velocity of the earth is constant; in reality if the tree is tall enough, the angular velocity might be changing subtlely as well. But cast this aside for the moment. If we assume the earth is a sphere and we know $h, c, \omega, M$, etc, can we compute it? Should I expect that an apple falling from the empire state building would move to a different spot than an apple falling down from my hand?
  6. The problem is difficult to me because assume we know the position, velocity and the exterior force exerted on the apple at some moment:
    $$
    F_1-F_2=F(t_0), V=V(t_0), r=r(t_0)
    $$
    (at least we know when $t=0$), we would not know where the apple is at the next moment unless we do some calculation. At moment $t_{0}$ we know what the angular velocity is; but when the apples falls, its velocity changes. And its angular velocity $\omega=\frac{V}{R}$ would also changes. Therefore we would have to solve a differential equation (a non-linear second order ODE) to compute the answer. And the amount of deviation is simply unclear to me.

Due to the extreme naive nature of the question, all answers are welcome. If I made some stupid mistake in the derivation, please do not hesitate to point it out. Maybe this strange phenomenon I thought would happen never happens in real life because I made some mistake.

Best Answer

Your logic is right on, it's just your arithmetic that needs work.

First, of course, you need to assume that everything happens in a vacuum. Air resistance will dominate any other effects for the sort of distances you have indicated. Also, let's assume (just to make calculations easier, that this takes place on the equator, at sea level. Earth's equatorial radius is 6378 km, and the tangential velocity is ~ 464 m/sec (but let's call it 464 m/sec exactly). Now, let us take an apple at a height of 3 meters, and a pear at a height of 10 meters.

At an altitude of 6378.003 km, the tangential velocity is 464.000218 m/sec, and at 6378.01 km the velocity is 464.000728 m/sec. For an altitude of 3 meters, the time to fall to earth is .782 seconds, and from 10 meters it is 1.429 seconds. Ignoring the fact that the base of the tree does not move in a straight line, but rather in a circle, it should be obvious that the apple will fall .00017 meters from vertical, and the pear will fall .00104 meters from vertical. Good luck with the measurements.

For larger heights and longer drop times, various other aspects of the earth's rotation start to show up, such as Coriolis effects, but we don't need to go into that here.

Notice that at higher latitudes the effect is decreased, since the effective radius of motion is proportional to the cosine of the latitude. At the North and South poles, the deviation is zero. I hope it's pretty obvious to you that trying to measure the effect of the precession of the earth's axis would be, ahem, a challenge.

And finally, although you mentioned it in passing, dropping a pineapple from about 22,400 miles is possible, but the deviation is not measurable.

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