Ball free fall 100m at equator, where is the landing spot?

[Trikenstein]

TL;DR Summary

An object located at the equator is free falling. Does it fall straight with or without radial deviation?

We have a weekly fun random chat to break from the routine of WFH. A science quiz poped up. There is no consensus on the good answer. We need 3rd party opinion.

You are at the equator, dropping a steel ball into a dry well, the depth is 100 meters, the mass of the ball is 1kg. The starting point is the top center of the well. Where will the ball land at the bottom? Only the position of the contact point is asked. What happened after contact is not needed.

The initial conditions are ideal:

Air resistance/turbulence is ignored
The ball is a perfect sphere, released from the top center, zero spin, zero initial speed.
The well is a perfect cylinder, and perfectly vertical.
Environmental factors ignored (rain, wind, etc...)

There is no trick, no trap, no joke in the question. You are allowed to get additional data to support your answer. As long as it is compatible with known physical laws. For example, if you need the value of Earth gravity, please look it up.

There is no need to give an answer accurate to the nanometer. Only the reasoning is important. For example, micro-gravity variation could be ignored.

So far we have gathered several answers, which boil down to this:

• ANSWER1: landing spot is dead center at the bottom. The reasoning is because there is no external force influencing the sideway motion of the ball, therefore it drops straight down, with zero deviation from the center.
• ANSWER2: off-center, 32 millimeters toward the East. The submitter justifies by a half page of calculation. Please allow me to hold back temporarily the publishing of the details of this answer. Actually I am curious to know what other people would come up with if they have no prior knowledge of this answer.

Can you please help us to arbitrate the debate?

 

[andrewkirk]

Answer2 has the right idea, and the right direction, but I'm not going to try to verify the amount.

The ball will land slightly to the East because of the Coriolis effect. Its linear velocity relative to the centre of the Earth at the point from which it's dropped, caused by the rotation of the Earth, is greater than that velocity at the bottom of the well, because of the decrease in radius. So that excess velocity translates into a very slight Eastward movement as it falls.

 

[olivermsun]

I'm also going with ANSWER2, which seems correct to within rounding error. There is a geometric argument that allows the calculations to be simplified greatly(?), but I am curious to see the method that was submitted with ANSWER2.

 

[DaveC426913]

Yup. Answer 2 is closest. Rotation of Earth means ball does not land dead centre. Top of well and ball are moving slightly faster than bottom of well. Ball lands a little east.

 

[Baluncore]

I know the well is vertical because a plumb bob hangs everywhere with the string parallel to the walls. That is my definition of vertical. What is yours?
If instead I hang a link chain without a bob, will the chain axis follow the same straight line as the stretched string?

 

[Trikenstein]

I know the well is vertical because a plumb bob hangs everywhere with the string parallel to the walls. That is my definition of vertical. What is yours?
If instead I hang a link chain without a bob, will the chain axis follow the same straight line as the stretched string?

In my opinion, the plumb bob will not stay vertical (will not be parallel to the wall). It will swing. Same for the link chain.

See Foucault pendulum: https://en.m.wikipedia.org/wiki/Foucault_pendulum

 

[Baluncore]

In my opinion, the plumb bob will not stay vertical (will not be parallel to the wall). It will swing. Same for the link chain.

When you say "swing", do you mean it will not settle in a stable position, or that it will not meet your definition of vertical? How do you define vertical?

 

[Trikenstein]

When you say "swing", do you mean it will not settle in a stable position, or that it will not meet your definition of vertical? How do you define vertical?

I begin to see the subtlety of the question. In the context of the question, a possible vertical could be the line linking the top center to the bottom center of the well. If prolonged, that line would meet the center of the Earth. As such this "vertical" rotates at the same angular speed as that of the Earth. However, an observer standing inside the well, will see that "vertical" as immobile.

 

[andrewkirk]

Whether and how the chain or plumb bob swings depends on how it got there, ie on initial conditions. If we first secure it straight and tight in a vertical line* and then release it, it will stay motionless in that vertical line.
But if we lower either one gradually from the top, it will swing because it will get displaced to the East by the Coriolis force as it descends, while gravity counteracts that displacement to make it swing back towards the midpoint.
With the plumb bob it would end up in simple pendular motion.
I don't know about the chain. Possibly it will end up having an interesting wave motion, as gravity will be pushing lower parts towards the centre while coriolis is pushing higher parts to the East.

* as defined by the projection of a laser mounted on a universal joint at the bottom of the shaft with a weight below that forces the laser to point in the direction of the maximum gravitational gradient

 

[Baluncore]

23.19 mm East of centre.

Land surveying and geodesy are based on vertical being the direction of a static plumb line, (not a swinging pendulum). Extrapolation of that plumb-line passes through the zenith and the nadir. Settling a plumb line involves stopping it swinging, which can be very quick if you know what you are doing.
The cutting of a vertical well must have the axis and the walls parallel to the plumb line, or it would not be vertical.

 

[olivermsun]

How did you get 23 mm?

 

[Baluncore]

How did you get 23 mm?

It is quite simple. I left a factor of two out of a square root.
My corrected value has advanced by ≈√2, to become 32.85 mm East.
I want to see some numbers from others before I post my 8 line solution.
I now have 32.85 mm, do I hear any advance on 32.85 ?

[Edit: Quote of deleted post removed.]

 

[OmCheeto]

...
I now have 32.85 mm, do I hear any advance on 32.85 ?

I get 32.89

Guessing we used slightly different Earth radius values. I used 6.4e6 meters.

Curious if anyone else has expanded the problem to extend the well indefinitely and watched how far the ball deviates from a non-rotating Earth?

I get 3.53e-4 * time^3 = offset in meters for the first 2 minutes.
I assumed that since the Earth had only rotated 1/2 of 1 degree in those 2 minutes, the x direction of the gravitational force on the ball wouldn't be significant enough to add to the equation.

ps. Fun problem!

 

[DaveC426913]

Curious if anyone else has expanded the problem to extend the well indefinitely and watched how far the ball deviates from a non-rotating Earth?

As you get deeper, you will have to consider the diminishing value of g, as more and more of the Earth is above the ball and thus cancels out.

 

[OmCheeto]

As you get deeper, you will have to consider the diminishing value of g, as more and more of the Earth is above the ball and thus cancels out.

My calculations stated that the ball had only fallen 1% of the Earths radius after 2 minutes, so I ignored that also. I'm guessing that's why I stopped there, as, damn, I got yard work to do. Fall is coming, and that's a whole lotta maths.

 

[jbriggs444]

I want to see some numbers from others before I post my 8 line solution.

I am going to try to brute force it from the rotating frame.

In the inertial frame, we would be looking at the difference in horizontal velocities. But this would be confounded by an acceleration which changes direction during the fall. It is plausible that the competing analysis has ignored this factor. In the rotating frame the gravitational force has a constant direction. However, we do have the Coriolis force to contend with.

We can immediately factor out mass. The gravitational acceleration is $g$ vertically downward. The Coriolis acceleration is $2\vec{v} \times \vec{\omega}$ in an eastward direction. Where $\omega$ is, of course, the sidereal rotation rate of the Earth.

$\vec{v}$ is increasing due to gravitation. It will be given by $gt$. So we are looking at an eastward acceleration with magnitude $2 g \omega t$.

The effect of this acceleration on eastward velocity is the integral of this with respect to $t$. That will be $g \omega t^2$.

The effect of this eastward velocity on eastward displacement is the integral of the above with respect to $t$. That will be $\frac{1}{3}g \omega t^3$.

Now we need to know $t$ for the fall. This will depend on the fall height $h$ and the acceleration of gravity $g$. The formula is $t=\sqrt{\frac{2h}{g}}$.

Now we need to plug in the values...

Height = 100 m
g = 9.80665 m/s^2
omega = 7.29211505392385e-005 rad/s
Time of fall = 4.51600755751788 sec
Final vertical velocity = 44.2869055139327 m/s
Final horizontal acceleration = 0.00645890420779703 m/s^2
Coriolis deflection = 21.95416446254 millimeters

[I stuck some extra outputs into my code to allow for some simple plausibility checks]

 

[jbriggs444]

Now let me attempt the competing analysis from the inertial frame.

As a first approximation, we will assume that the Earth's surface is flat, the gravitational field is uniform and that the ball is dropped from a support that is moving eastward at a velocity corresponding to the rotation of the Earth on a platform at height $h$.

The horizontal velocity of the launch platform is given by $\omega h$.
The time of fall is $\sqrt{\frac{2h}{g}}$.
The resulting horizontal deflection is the product of the two.

Height = 100 m
g = 9.80665 m/s^2
omega = 7.29211505392385e-005 rad/s
Time of fall = 4.51600755751788 sec
Horizontal velocity of launch platform = 0.00729211505392385 m/s
Horizontal deflection = 32.93124669381 millimeters

Bingo. Somebody forgot to do the second approximation.

As a second approximation, we must realize that during the fall, the direction of the acceleration of gravity is changing. We can decompose the resulting acceleration into a vertical component and a horizontal component.

The vertical component will be essentially unchanged by the rotation.
The horizontal component will go as the sine of the rotation angle. We can use the small angle approximation and just say that it is directly proportional to the rotation angle. $a_x = g \sin(\omega t) = g \omega t$

We integrate once to get the effect of this acceleration on velocity. $v_x = \frac{1}{2} g \omega t^2$

We integrate again to get the effect of this velocity on deflection. $d_x = \frac{1}{6} g \omega t^3$

By no coincidence, this is 1/3 of the deflection predicted by the first approximation.

Correction for changing direction of gravity = 10.97708223127 millimeters
Corrected deflection = 21.95416446254 millimeters

[Apologies for excess significant digits. I plugged the numbers into a program and simply accepted the default output format]

I feel good enough about this analysis to say that the original post is erroneous. It's not 32 mm. It's 22 mm.

 

[berkeman]

This should be a pretty easy experiment to conduct using a typical university building and a laser level, etc., no? Road trip!

 

[Vanadium 50]

using a typical university building

The Harvard tower where the Pound-Rebka experiement was done would be appropriate.

10.97708223127

My eyes! All that extraneous precision hurts my eyes!

This seems fairly straightforward. Work in an inertial frame, and see where your coordinate system ends up.

The bottom is moving slower than the top because of the Earth's rotation. But only a little slower: 0.1/6000 slower. The Earth spins at 1000 mph = 450 m/s. That's 3/4 of a centimeter per second difference and in a 4.5 second fall, that's your 34 mm.

 

[jbriggs444]

This should be a pretty easy experiment to conduct using a typical university building and a laser level, etc., no? Road trip!

100 meters is a 30 story building or thereabouts. Where I matriculated, our physics building was only two or three stories high. I roomed in a dormitory which was 129 feet high at the top of the enclosures above the elevator shafts. Ten stories. Call it 40 meters.

At that height, one would expect a deflection of only 5.5 millimeters.

 

[Baluncore]

As a second approximation, we must realize that during the fall, the direction of the acceleration of gravity is changing. We can decompose the resulting acceleration into a vertical component and a horizontal component.

Correction for changing direction of gravity = 10.97708223127 millimeters
Corrected deflection = 21.95416446254 millimeters

It takes 6 hours for the tangential velocity to become vertical, to directly oppose gravity. I didn't expect a 4.5 second fall to be as significant as you suggest, since the angular change is so close to zero.
360° * 4.5 s / 86400 s/day = 0.01875° = 1.125 arcmin.

 

[berkeman]

100 meters is a 30 story building or thereabouts. Where I matriculated, our physics building was only two or three stories high. I roomed in a dormitory which was 129 feet high at the top of the enclosures above the elevator shafts. Ten stories. Call it 40 meters.

Okay, okay, okay. Jeeze. The road trip needs to go to the city. Whatever...

https://hips.hearstapps.com/edc.h-cdn.co/assets/16/32/1470932555-south-korea-tower.jpg

 

[jbriggs444]

It takes 6 hours for the tangential velocity to become vertical, to directly oppose gravity. I didn't expect a 4.5 second fall to be as significant as you suggest, since the angular change is so close to zero.
360° * 4.5s / 86400s/day = 0.01875° = 1.125 arcmin .

1.125 minutes of arc times 100 meters is... 32 millimeters.

Knowing what to neglect with small angles was one thing that kicked my butt a few times when learning calculus. I wanted to neglect it also in this case, but it just felt wrong to do so.

 

[Baluncore]

Knowing what to neglect with small angles was one thing that kicked my butt a few times when learning calculus. I wanted to neglect it also, but it just felt wrong to do so.

But g is reduced by the cosine of Earth's daily rotation.
You must be missing a unit conversion somewhere.

 

[jbriggs444]

But g is reduced by the cosine of Earth's daily rotation.
You must be missing a unit conversion somewhere.

Cosine of angles near 0 degrees is pretty much a constant. The delta goes as the square of the deflection angle. You need a third approximation to care.

Let us work out the deviation in the vertical component of g for an angle of 1.125 arc minutes...

$5 \times 10^{-7}$ m/s^2.

It changes g from 9.80665 to 9.8066495

 

[Baluncore]

Cosine of angles near 90 degrees is pretty much a constant.

Then your second approximation should be closer to 10 um, not 11 mm.

 

[jbriggs444]

Then your second approximation should be closer to 10 um, not 11 mm.

No. That's your third approximation. My second approximation is based on the sine of small angles.

 

[Vanadium 50]

100 meters is a 30 story building or thereabouts. Where I matriculated, our physics building was only two or three stories high

Ah, but how many basements did it have?

 

[Baluncore]

My second approximation is based on the sine of small angles.

OK, I agree with you. The point of impact of the ball with the bottom of the well will be significantly less than the 32.8 mm.

I use Earth equatorial radius; Re = 6378137 m, with constant g = 9.80665 m/s2 .
The angular velocity of the Earth is; w = 2 * Pi / (24*60*60) rad/s .

I set out to replicate your result numerically, and independently, but first apply a cosine correction for rotation of the Earth. During the time of the trajectory, the rotation of the Earth deepens the well by 345.134 mm, to 100.345134 m.
The first approximation gave a flight time of 4.516008 seconds with a deflection East of 32.841 mm. The deeper well pushes the flight time out to coincidence of the ball with the bottom of the well at 4.523794 seconds, with a first order deflection of 32.936 mm East.

The ball will be decelerating horizontally as it moves away from the centre of the Earth.
At the end of the flight the components of g = 9.80665 m/s2 at the ball will be;
Horizontal; gx = g * Sin( w * t ) = 0.00322619 m/s2
Vertical; gy = g * Cos( w * t ) = 9.80664947 m/s2
If this gx = 0.00322619 was applied for the entire 4.523794 sec time of flight, the deflection would be reduced by; s = 0.5 * gx * t^2 = 33.012 mm .
So it is clearly significant.

The numerical integration would require more work, so at this stage I estimate the actual reduction would be at least one quarter of that; 33.012 / 4 = 8.253 mm
That brings the total deflection to about 32.936 - 8.253 = 24.7 mm .

 

[jbriggs444]

OK, I agree with you. The point of impact of the ball with the bottom of the well will be significantly less than the 32.8 mm.

Good. We seem to be on the same page. I agree with you in the large, though I have some nits to pick.

I use Earth equatorial radius; Re = 6378137 m, with constant g = 9.80665 m/s2 .
The angular velocity of the Earth is; w = 2 * Pi / (24*60*60) rad/s .

I also used g = 9.80665 m/s². However, we were both wrong on that. See below.

I used the sideareal day of 86164.1 seconds rather than the solar day of 86400 seconds.

I set out to replicate your result numerically, and independently, but first apply a cosine correction for rotation of the Earth.

That correction can be applied instead by using apparent $g$ rather than gravitational $g$.

Rather than deepening the well to reflect the centripetal acceleration of the ground beneath our feet, we can reduce the acceleration of gravity to calculate the same arrival time.

I blundered by using apparent $g$ of 9.80665. Google says that 9.78 is correct for the equator. You have blundered by using gravitational $g$ of 9.80665. But 9.80665 is apparent g for some temperate latitude or other. So neither of us are correct on the relevant figure.

Using $\omega^2 r$ as an addition on top of the Googled figure for equatorial apparent g, I get:

Apparent g = 9.78
Centripetal acceleration = 0.0338777555225012
Gravitational g = 9.8138777555225

You get points for using a "g" that is closer to the truth.

During the time of the trajectory, the rotation of the Earth deepens the well by 345.134 mm, to 100.345134 m.
The first approximation gave a flight time of 4.516008 seconds with a deflection East of 32.841 mm. The deeper well pushes the flight time out to coincidence of the ball with the bottom of the well at 4.523794 seconds, with a first order deflection of 32.936 mm East.

I would expect the fall time for a deepened well with gravitational g to match the fall time for a the nominal well with apparent g.

The ball will be decelerating horizontally as it moves [horizontally] away from the centre of the Earth.
At the end of the flight the components of g = 9.80665 m/s2 at the ball will be;
Horizontal; gx = g * Sin( w * t ) = 0.00322619 m/s2
Vertical; gy = g * Cos( w * t ) = 9.80664947 m/s2
If this gx = 0.00322619 was applied for the entire 4.523794 sec time of flight, the deflection would be reduced by; s = 0.5 * gx * t^2 = 33.012 mm .
So it is clearly significant.

Yes. Intuitively and simplistically, one could think of the ball falling through the interval, affected by a force of gravity that remains always in the original "vertical" direction. Or of the ball falling through the interval affected by a force of gravity that remains always in the final "vertical" direction.

That gives you an envelope within with the curve of fall must remain. An envelope which is 33 mm wide at the bottom. Of course, this is considering only the direction of the acceleration of gravity. It does not factor in the relative velocity between launcher and target or the subsequent acceleration of the target. [Those give you a 33 mm deflection for the right hand end of the envelope. The same 33 mm that Coriolis also yields].

The acceleration near the beginning of the fall (vertical) will have a proportionately larger effect than acceleration near the end of the fall (diagonally leftward). You've estimated that as a 3:1 split. My analytic integration puts it at 2:1. It's the 1/3 in the integral of x² dx.

 

[Vanadium 50]

If you can't be accurate, at least be precise. 0.0338777555225012 implies the Earth taking 86400 seconds to spin. In fact, it;s 86,164.1.

 

[Baluncore]

Thanks for the commentary.

I should have used an average sidereal day of 86164.0905 SI seconds. But with a warming climate, I expect that will be increasing to about 86,165.0

Finding a disused equatorial vacuum-well, for an air resistance free experiment to verify the numbers, is going to be somewhat difficult.

I would like to drop the ball, at the same time as hanging a chain next to a plumb line, to see the differences.

 

[jbriggs444]

If you can't be accurate, at least be price. 0.0338777555225012 implies the Earth taking 86400 seconds to spin. In fact, it;s 86,164.1.

Let me check that calculation again. I believe that I used 86,164.1.

my $pi = 3.141592653589;
my $omega = 2 * $pi / 86164.1;
my $r_e = 6371000;
my $centripetal_acceleration = $omega ** 2 * $r_e;
print "Centripetal acceleration = $centripetal_acceleration\n";
Centripetal acceleration = 0.0338777555225012

Edit: Wikipedia gives a better radius figure of 6,378,137.0 meters.The answer I got when I googled for "equatorial radius of the earth" turns out to have been the average radius, not the equatorial radius.

Centripetal acceleration = 0.0339157064785778

 

[DaveC426913]

Y'all fail for not following directions.

There is no need to give an answer accurate to the nanometer.

 

[Trikenstein]

Hello gentlemen,

Was busy since the last 2 days. I am quite intimidated by the solutions you suggested. Need sometime to digest. Here is the answer of the submitter mentioned in the original post.

The submitter introduced by a short visual proof that dead center cannot be a solution. Because a Foucault_pendulum would sit still without swinging if there is no lateral displacement. There is one of such pendulum in the museum of science in Washington DC.

Here is the submitted solution. The base concept is just the conservation of angular momentum. The submitter did mention the Coriolis effects.

Sorry I don't know how to write math symbols. If there is any flaw in the reasoning, I would appreciate your patience to clarify.

// The duration of the free fall, from mvt equation with acceleration:
displacement H = 1/2 * acceleration * t^2 + V0 * t + d0

// Taken referential at the top center with initial V0 = 0 and d0 = 0
Duration free fall T = sqrt(2H/acceleration) = 4.5 seconds

Angular speed of the Earth is ω = 2*Pi/24/60/60 = 7.27 e-5 radian/second

// The linear speed of the top/bottom of the well:
Vt = Rt * ω  (Rt = radius from Earth to the top well in meter)
Vb = Rb * ω  (Rb = radius from Earth to the bottom well)

// The displacement of the those points during the freefall duration
Dt = Vt * T = Vt * 4.5
Db = Vb * T = Vb * 4.5

// Difference in linear displacement
Dt - Db = (Vt - Vb) * T = (Rt - Rb) * ω * T

// Rt - Rb = 100m = height of the well
Dt - Db = 100 * ω * T = 100 * 7.27 e-5 * 4.5 = 0.032 meter = 32mm

// Why off-center on the East side?
Because Earth rotation is West to East.