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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.

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4 Answers 4

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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0.00104 m is a little over 1 mm. You should be able to measure that. –  LDC3 Aug 3 at 2:25
    
Yes, but there are a few problems. First, aerodynamic effects. Second, how do you tell where it ought to hit? That is, how do you measure vertical? 1 mm at 10 meters is 100 uradian, or about .006 degrees. Like I say, good luck. –  WhatRoughBeast Aug 3 at 10:44
    
A plumb line will give you the direction of gravity. Yes, an apple is a poor choice due to aerodynamics. –  LDC3 Aug 3 at 13:28
    
Yup. A plumb bob will do, assuming the tree doesn't sway. Which, in any wind it will, of course. And the release mechanism has to be perfectly aligned with the plumb line, and can impart no lateral or rotational motion to the "apple". –  WhatRoughBeast Aug 3 at 15:27

Ok, let's see if I get this right.

The circumference of the equator is 24902 miles. The Earth rotates 360° in 86,164.098 seconds. This means the surface of the Earth moves 1525.96 ft/sec at the equator.

Adding 1000 ft to the height give a circumference of 24903 miles. This means the lateral speed is 1526.03 ft/sec; a difference of about 0.87 inches/sec. It would take about 7.9 sec to fall, so the lateral distance would be 6.91 inches.

So yes, if you dropped an object from 1000 ft, it would be slightly off from directly down. The Empire State Building has a height of 1250 ft, so the distance from a plumb line would be a little over 7 inches.

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You're focusing on the centrifugal force. You forgot about the coriolis effect. Neither has much of an effect for an object dropped from the top of a tree, even a very, very tall tree.

The first thing you need to realize is that "down" (the direction in which a plumb line points) is generally not toward the center of the Earth. That is only true at the equator and the poles. Elsewhere, "down" is in the direction of the perceived gravitational force. That perceived gravitational force is the vector sum of the more or less inward force due to gravity plus the more or less outward force due to the centrifugal force. Gravity is not quite directed toward the center of the Earth, thanks to two effects, the centrifugal force and the Earth's not quite spherical shape.

An optimal tree will grow vertically, where "vertical" means the direction a plumb bob hangs. This optimal tree will have an immeasurably small curvature to it because gravitational acceleration decreases slightly and centrifugal acceleration increases slightly with height. Suppose this optimal tree is very leafy from bottom to top. A falling apple will never have a chance to build up speed and thus won't be subject to the coriolis effect. The apple will fall straight down.

Suppose instead that the optimal tree is instead mostly leaf-free, and it drops drag-free bullets instead of apples. The falling bullets will be subject to the coriolis effect. A drag-free object will be deflected eastward by $\sqrt{\frac{8h^3}g}\frac{\Omega \cos \lambda} 3$ where $h$ is the height of the fall, $\Omega$ is the Earth's rotational angular velocity, and $\lambda$ is the latitude. If the tree is 100 meters tall and is at the equator, a drag-free bullet will the ground with a 2.2 cm eastward deflection from vertical. Drag will reduce this deflection.

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Your reasoning is essentially correct, but it would have to be a really tall tree to see any significant deviation.

Consider that geosynchronous orbit is at an altitude of 36,000 km. So an apple falling from the top of a tree 36,000 km tall would stay in orbit and would not gain or lose altitude. An apple falling from a higher altitude than that would fly off into space, and an apple falling from a lower altitude would either go into an elliptical orbit, or crash into the earth at some point far from the tree.

The height of an actual tree, or even the Empire State Building, is just not high enough to see any significant deviation of the paths of falling objects from "straight down".

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