Will an object dropped from a high building displace due to the Earth's rotation?

Question

I read that in the 16th and 17th century, the question of whether the Earth rotates around its axis or all celestial bodies rotate around it was extensively debated. One of the anti-rotation arguments was that objects dropped from high places should move from the true vertical due to the ground having traveled meanwhile. Since the Earth does rotate, I'm trying to quantify the effect and whether it could be measured at the time. Would appreciate my argument being checked for basic sanity and correctness.

Suppose I climb a 100 meter tower and, standing near its edge, let go of a brick. The Earth rotates with the angular velocity $\omega = \frac{2\pi}{24\cdot 3600\ \mathrm{s}}$, and the speed of the base of the tower vs. the brick at the top before I let go of it are $R\omega$ vs. $(R+100)\omega$. This means that horizontally the brick is moving with the speed of $100\omega ~= 0.00727 \,\mathrm{m/sec}$ relative to the ground, while vertically it's dropping with the uniform acceleration $g=9.8\ \mathrm{m/s^2}$ and will hit the ground in $4.52$ seconds, having travelled $0.03\ \mathrm{m}$ horizontally. If I do the same from the Empire State Building ($381\ \mathrm{m}$), it comes out as about $22\,\mathrm{cm}$ (if the height grows by a factor of $X$, the distance travelled horizontally grows by $X^{3/2}$, the speed increase contributing $X$ and the time to drop $\sqrt{X}$).

So my questions:

1. Is this analysis basically sound? I realize I'm using the uniform vertical acceleration which is an approximation, and I'm using the circular motion of the Earth's surface when estimating velocities, but not when using them to find time and distance traveled. I guess the more exact calculation would be to treat the brick as a satellite on an elliptical orbit around the center of the Earth with the given initial position and velocity, find its orbit equation or simulate numerically, and find its intersection with the Earth's surface. It seems a bit daunting a task at the moment. Would that give substantially different results?

2. Is air resistance an important factor to take into consideration, for estimating the horizontal distance traveled? (if it is, I guess this still answers the same question for the Moon).

3. Assuming my estimates aren't too far-off, is that something that can be tested in a real experiment either now or in the 17th century? I think that'd depend on how exactly a true vertical we can guarantee the tower's wall to be?

Answer 1

This type of experiment has been attempted. Small objects (presumably small metal balls) were dropped down a vertical mineshaft. The hope was that the air mass in the shaft would be sufficiently stationary to not affect the falling motion too much. The spread was considerable, the balls hitting the bottom of the shaft in an area severa tens of centimeter across.

There was a bias in the distribution of the landing spots, consistent with what you would expect given the Earth's rotation, but that could easily have been a fluke.

Historical article by Alexandre Moatti titled 'Coriolis, the birth of a force'

According to the information in the article the attempt was by Ferdinand Reich, in 1833, and the mineshaft was 158 m. deep.

Google books has an integral scan of the report that Ferdinand Reich wrote. Use the search term "Fallversuche über die Umdrehung der Erde"

I also see mention of another vertical mineshaft setup, with a fall of 90 meters, also in the early 1800s

Feasibility:
What we see that even in the best of circumstances the noise is several times larger than the signal. (Spread of tens of centimeters while the expected effect is centimeters.)

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About air resistance: The effect of air resistance is very much not negligable. The air resistance slows down the fall, so that it takes longer, so there is more time to accumulate sideways displacement.

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On how to obtain a good approximation of the sideways displacement.

Let me refer to the objects that are released as 'pellets'.

For simplification take the case where the point of release is at the equator. The pellet is released vertically, but since the Earth is rotating the pellet does have a velocity with respect to the center of the Earth.

As you write: if we neglect air resistance effect then for the duration of the fall the motion of the pellet is orbital motion. It's not much of an orbit, it intersects the Earth surface within seconds, but still: for the duration of the fall the motion is orbital motion.

To emphasize that it's orbital motion:
Think for example about the orbit of Halley's comet. All the way from the outer distance to its point of closest approach to the Sun Halley's comet is being accelerated by the gravitational pull of the Sun. This acceleration is continuously increasing the angular velocity of Halley's comet.

So:
Initially the the falling pellet is moving parallel to the local plumb line, because initially the angular velocity of the pellet is the same as the angular velocity of the Earth as a whole. As the Earth's gravity pulls the pellet closer the angular velocity of the pellet increases.

(Again: initially the falling pellet is moving parallel to the local plumb line. The buildup of sideways displacement is not linear. This shows that a calculation of the sideways displacement that has the displacement increasing linearly is wrong.)

Obtaining a good first approximation for the angular velocity as a function of time:

As we know, in all forms of orbital motion there is conservation of angular momentum.

We have an expression for the radial velocity as a function of time (the pellet falling), so we need to work towards an expression that contains that radial velocity.

Conservation of angular momentum:

$$ \frac{d(\omega r^2)}{dt} = 0 $$

Differentiating:

$$ r^2 \frac{d\omega}{dt} + \omega \frac{d(r^2)}{dt} = 0 $$

With the chain rule we obtain a term $\frac{d(r)}{dt}$ which is what we are looking for.

$$ r^2 \frac{d\omega}{dt} + 2 r \omega \frac{d(r)}{dt} = 0 $$

Dividing by 'r', and rearranging:

$$ r \frac{d\omega}{dt} = - 2 \omega \frac{d(r)}{dt} $$

On the left we have a term $\frac{d\omega}{dt}$; that is angular acceleration.
So: $r\frac{d\omega}{dt}$ is an expression for the sideways acceleration.

On the right we have an expression with magnitude $2\omega\frac{d(r)}{dt}$, and in this case $\omega$ is the angular velocity of the pellet, which at the start is equal to the angular velocity of the Earth itself. The angular velocity of the pellet changes during the fall, but compared to the total angular velocity the change of angular velocity is small.

So for a good approximation we can treat the falling pellet as subject to a sideways acceleration as described by the follwing expression:

• $a_p$ acceleration component perpendicular to the radial direction
• $v_r$ velocity component in radial direction

$$ a_p = 2\omega v_r $$