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Coriolis force question

Walter Wayne's south displacement estimate: 25.63639 inches
Davefoc's south displacement estimate: 25.6512 inches

Using
40,000 km circumference
9.8067 gravitational acceleration
40.5 degrees latitude
381 meter drop
23 hours 56 minute sidereal day

Hmm, the last time I had a result this close to somebody else's we were both wrong. Still it is nice to see a result that corresponds roughly with one of mine since my 7.33 inch eastward displacement seems to be getting battered by a raft of results that don't agree with it.

I'm going to spend some time looking at what's going on here with respect to the eastward displacement. I don't understand why the 7.33 inch estimate is so wrong. But now there have been two simulations and a theoretical calculation by Hamish that suggest its about 4.8 inches.
 
Did you notice that my program came up with a z displacement (relative to the release point) of 247.936664 metres? My model worked with a tower 381 metres high ata latitude of 40.5, perpendicular to a spherical Earth, so that accounts for 247.4397 metres. The difference is 0.4969 metres or 19.564 inches, and indicates that the weight would land that far to the south of such a tower.

I've thoroughly confused myself reading about oblate spheroids, geoids and so on. Apparently, local gravity to an oblate spheroid is not perpendicular to the surface, unless you count centrifugal force as part of the gravity. Neither does local gravity point through the centre of the Earth when the rotation effect is ignored - the bulge of material around the equator causes a 'towards the equator' component, that produces a torque on the orbits of satellites in low inclined orbits, causing the orbits to precess. I don't understand how all these effects combine, and they are not easy to model.

I confess I have no idea now whether the weight would land north or south of the tower, or dead in line with it.
 
Originally posted by ceptimus
Apparently, local gravity to an oblate spheroid is not perpendicular to the surface, unless you count centrifugal force as part of the gravity.
Click to expand...
Right. So count it.

I think the easiest approach here is to work in a rotating coordinate system fixed to the earth; all we need to do is include in our calculations the Coriolis force and centrifugal force. (These are simply fictional forces that are due to the use of a rotating coordinate system; they have nothing to do with the shape of the earth, just its rotation rate.) So the three forces on our falling object are gravity, centrifugal force, and the Coriolis force. But the sum of gravity and centrifugal force has a known direction: it's the local vertical, based on which buildings are built, and relative to which we wish to measure the displacement of the falling object.

So I'm pretty sure that the object will not be deflected north or south of the building's base, except for an exceedingly small southward Coriolis force due to the object's small velocity eastward, which in turn is caused by the Coriolis force due to its significant velocity toward the earth's axis.

The only thing missing that we need to know is the angle of the local vertical relative to the plane of the equator. Then we can calculate the component of the object's velocity toward the axis, which, together with the earth's rotation rate, determines the eastward Coriolis force. On a spherical earth, this angle is just the latitude. On an ellipsoidal earth, a quick Google search yielded this page, which says, I think, that we want the "geodetic latitude" of the point. I don't know whether that is the same as the usual latitude. (This is just to calculate the size of the eastward deflection. It's not related to the question of the existence of a northward or southward deflection.)
 
I think the difference between local gravity, and gravity towards the center, will be a significant difference at this level of precision. So we've probably gone about as far as we can without modelling that.

I'm still intrigued by the difference between the two methods. It does seem like the difference in speed times the fall time would give a pretty darn precise answer, but it's quite a bit off from the numerical approaches. I'd like to understand this better.
 
Although the net force, due to gravity and rotation is normal to the surface, at the surface this effect varies with height above the surface. This means that a very tall building, with each floor accurately lined up to plumb bob, would actually have a slight curve!

It is easy to calculate the centrifugal component of the force - this can be subtracted from the normal-to-the-surface total force to give the local gravitational component. This has a slight southward-pulling component in New York, due to the extra 'bulge' of material around the equator.

Once you are far enough away from the surface (say as far as the moon) the gravitational component does point through the centre of the Earth, and the normal to the point on the Earth that happens to be closest to the moon, even after the centrifugal component is subtracted, will point in a slightly different direction.

I don't know how fast the effect falls off with height, and as we are only considering small deflections, this may be important. A true simulation would need to cater for how the Earth's mass is distributed, taking into account not just the non-spherical shape, but also the differing density with depth.

I don't think this affects the eastward offset, as symmetry cancels out most of these awkward effects, but to get an accurate number for the north/south offset, maybe such complications need to be considered?

I'm going to abandon any further changes to my simulation unless someone comes up with an accurate model that is still simple enough to work with.
 
Getting my geek on

I did a calculation of the plumb bob displacement on a spherical planet, and interestingly I got 25.52" south displacement. Not far from the 25.64" I got for the dropped ball using a numerical approach.

The method I used to come up with the formula was fairly simply. I calculated the net force given that the plumb line must see given that it is following a latitude circle and subtract the force of gravity. The result must be the force applied by the string upon which we hang the plumb bob. Calculate the angle of that. Note that the result for the angle were identical to about several decimals wether I looked at the top or bottom of the tower, so I didn't need to integrate over the height of the plumb bob.

The equation below is my result, where:
g = acceleration due to gravity at the spheres surface (m/s<sup>2</sup>)
R = radius of sphere (m)
h = height of plumb bob above surface (m)
omega = radial velocity (radians/second)
phi = latitude (radians)
theta = deviation from vertical of plumb line (radians)

Walt
 
Walter it is surprising how different your formula looks to the way I got my answer, but the results seem quite similar.

Could you try a few different latitudes and heights. I'd like to put them into my calculation and see if we continue to get the same results.

Dave
 
Suppose you scrap the plumb bob entirely and replace it with a laser that is leveled is such a way as to point to the Earth's center of mass. Drop the penny from the point where the laser light is emitted. Where does the penny strike the sidewalk in relation to the red dot on the sidewalk?
 
Skeptoid said:
Suppose you scrap the plumb bob entirely and replace it with a laser that is leveled is such a way as to point to the Earth's center of mass. Drop the penny from the point where the laser light is emitted. Where does the penny strike the sidewalk in relation to the red dot on the sidewalk?
Click to expand...
According to my model, which simulated a spherical Earth, the penny strikes the ground 19.564 inches closer to the equatorial plane than the laser. Measured along the ground, this figure must be divided by cos(latitude) and as I used 40.5 for my latitude figure, this gives a southward offset of 25.728 inches from the laser.

With a spherical Earth, the density distribution doesn't matter, but as soon as you consider a non-sphere, you need to take into account the Earth's density variations with depth.
 
What is the force that is pulling the weight south?

If the earth stopped rotating and all we were dealing with was earth gravity wouldn't the weight be drawn exactly towards the center in parallel with the laser beam?

Does the weight only have eastward motion when it is released? If so what force is making it deviate from its eastward path?

Do we agree that building are leaning to the north relative to the laser beam?

Sorry, several questions asking basically the same thing, but I just don't get why weight would end up south of drop point.

edited to add: I just noticed that I have been using the words south displacement for what we are talking about. I meant that the weight is displaced north of the building base in my thinking. Hmm, is Walter saying he thinks the weight is displaced south of the building base? We end up with nearly the same quantitative estimate but Walters is South and mine is north?
 
My model assumes a spherical Earth, with the building perpendicular to the surface. So the axis of the building points through the centre of the Earth.

The reason for the 25.7 inch southward deviation of the penny (from the base of the building, or the laser in Skeptoid's version) is that the penny starts out on an orbit that would eventually take it around the centre of the Earth (if the Earth didn't get in the way). The furthest north the penny would ever get on it's orbit is the release point. The plane of the orbit is at the same angle to the equatorial plane as the latitude of the building, 40.5 degrees in my version.

Now the axis of the building is in this same plane when the penny is released, but when the penny reaches the ground, some 8.8 seconds later, the Earth has rotated carrying the building out of (north of) this plane. The base of the building is now 25.7 inches north of the penny's impact point.

[edit to add: This only applies to the model. I've still no idea on the north/south offset of a real coin with real building, on a real Earth. I think the 4.898 inches eastward offset is about right though.]
 
Same goes for a plumb bob. Anything on a sphere moving in a straight line follows a great circle route (circle centered on center of sphere with radius r). If one walks along a line of latitude one must continually turn away from the equator. Thus New York turns to the north, due to lateral forces the spinning earth places upon it.


On an analytical solution:
I am getting close to an analytical solution for a spherical planet. I have found Keplers Laws somewhat anoying for finding east-west displacement. With them one can precisely define where the elipse intersects the planet. Thus I have an analytical solution for south displacement of the coin (25.63").

However, to find the east displacement is pretty difficult. Even finding the solution for where the coin lands doesn't help, if you don't know when it landed. Using Kepler's second law, I end up with a very difficult integral to t. If I approximate the integral to find t, I can bound the time of impact to within about 1.1 milliseconds. However, in 1.1 milliseconds, the earth surface travels about 14". Not very accurate.

The ironic thing is, my previous approximations were more accurate on this, as any error in time ended up effecting both the building movement and coin impact. However, my "exacting" solution is less accurate because I can determine the impact point precisely, but the amount the building has moved has a significant error due to approximations in impact time.

If I have time a might post a bit more about it later.

If anybody is really bored, you can try the integral below. :D
I'll feel real stupid if someone does this quickly.
 
Ok, enough theorising. I say we submit a grant request for a 382 metre bell jar, a bloody huge vacuum pump, some lasers, a plumb bob, a cunning scaffold to fix to the roof and the exclusive use of the Empire Sate Building for a couple of days.

Oh, and a penny too.

That ought to settle the issue.
 
ceptimus,
If the spherical earth you are assuming the buildings and the plumb bob will not point toward the center.

Spin the earth faster so the effect is more pronounced.

What happens? Centrifugal force pulls at the weight.in the plane of the rotation. So the angle the plumb bob hangs at is the sum of two vectors, one the gravity vector and the other the centrifugal force vector.

The above is true even if the earth is a perfect sphere.

Eventually as the earth spins faster and faster the weight at 381 meters won't fall at all. 381 meters will become high enough for a geosynchronous orbit.

Wayne,
I'd really like to see you try your equation out on a different latitudes so I can see how what it predicts compared to my calculations.

My numbers
displacement (inches)
Code: 
latitude   nortward tilt(degrees) displacement from base of 381 meter drop
0              0                 0
10            .0340              8.9 
20            .0639             16.74
30            .0861             22.55
40            .0978             25.63
50            .0977             25.61
60            .0859             22.51
70            .0637             16.7
80            .0339              8.824
90             0                 0
 
davefoc,

Yes, I agree the buildings in New York would lean to the north on a spherical Earth. Maybe I was the first to mention that? The problem, as I see it, is that there are other effects to take into account:

1. The Earth isn't round. It's approximately an oblate spheroid (the shape generated by rotating an ellipse around its minor axis).

2. The distribution of mass within the Earth isn't constant - it has a dense nickel-iron core, a mantel and a lighter crust etc.

3. Gravity direction close to an oblate spheroid (even a constant density one) varies. Close to the spheroid, there is a 'towards the equator' component of gravity. This component decreases with height.

4. Because of these effects, the direction of 'up' is not a straight line, so buildings with each floor accurately aligned to 'up' will be slightly curved.

I can't see the point of allowing just for the centrifugal component, while neglecting the others, until someone works out, or finds a reference on, the relative magnitude of each effect.
 
Ceptimus, thank you for your response. I think I understand better what you are saying now.

My thoughts:
I think to a very high degree of accuracy one could assume that the earth is a sphere plus some material that causes it to take the shape it does. The sphere will pull towards the center. It doesn't matter that the sphere is made up of different density material. The material is arranged concentrically so when a mass is outside the sphere the gravitational field it experiences is just like all the mass was at center.

The material that is outside the sphere also contribute to the local gravity. Since there is more of it towards the equator there will be a net southward pull (in northern latitudes) from this material so buildings will lean somewhat more north to compensate for that. In addition there are local gravitational variations caused by variations in local densities and large masses near by that will effect the direction of gravity.

However, I think these variations in gravity will effect the plumb bob and the dropped weight equally whereas the centrifugal force on the weight is removed once it is let free.

So my thought here is that the shape of the earth will not cause a significant displacement between where the plumb bob points and where the weight lands. The shape of the earth causes a variation in the gravitational force. This is a minor effect and would only affect the calculation slightly in that the drop time is changed sligthly as a result of the gravitational variation.
 
Originally posted by davefoc
But now there have been two simulations and a theoretical calculation by Hamish that suggest its about 4.8 inches.
Click to expand...
Add the formula at the bottom of this page to the list.
 
I plugged the numbers into equation referenced by 69dodge.

Using their value for sidereal rotation I got 4.8878 inches. Using value for sidereal rotation we have been using I got 4.8887 inches.

Either way the estimates are very close to the Hamish calculation and the Ceptimus and Walter Wayne simulations and very far from the davefoc number.

Walter Wayne 4.88549 inches
Ceptimus 4.8982 inches
Hamish 4.8819 inches (.124 meters)

So at this point what is davefoc to conclude? That he is wrong and the rest of the world is right or that he is right and the rest of the world just doesn't see this quite correctly?

Well, davefoc is going to think about this some more but I sense a lack of confidence in his original estimate.
 
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