Coriolis force question

[sorgoth]

New question: How much is the penny going to be displaced because of the gravitational pull of the empire state building?

Very, VERY little.

 

[Skeptoid]

davefoc wrote:
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?

This is where davefoc should go woo-woo and defend his result despite evidence to the contrary. Invent fudge factors justified by new and undetectable forces (use of the word quantum will come in handy here). Accuse the others of being muddled in scientific dogmatism. Tell them how Galileo was ridiculed when he dropped things from tall buildings and got the "wrong answers" and won't they be the stupids when you are proven correct.

 

[Walter Wayne]

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.

Virtually identical, but I have southward displacement.

lat     tilt  displacement
 0        0         0
10    0.0337    8.8252
20    0.0634   16.5910
30    0.0854   22.3632
40    0.0972   25.4447
50    0.0972   25.4600
60    0.0856   22.4017
70    0.0635   16.6348
80    0.0338    8.8538
90        0         0

 

[Walter Wayne]

Does anyone have data on the approximate direction of gravity in New York on oblate earth. I think with the small change of height (relative to earth scale) and the fact the coin is prevented from doing a much of an ordit, we could approximate its fall as an elipse. Ideally, I like the value and angle of g at both the top and bottom of the building as a reality check.

So with two values of g one could find the center of gravity as seen from New York, at that point one could calculate the equation of the coin orbit around said center and then calculate it impact point. With one value of g, we would assume constant gravity center, and calculate from their.

My (almost) analytical solution for a spherical went:
1. Calculate equation of elipse in the plane of orbit.
2. Calculate at what point in orbit coin hit sphere.
3. Move co-ordinate system by (latitude) degrees, to bring elipse into "earth" co-ordinate system.
4. Calculate latitude and longitude displacement of coin impact
5. Calculate time it takes coin to hit ground (this is step where I had to resort to numerical methodology)
6. Calculate longitude displacement of building over that time.

I think for an oblate spheroid the changes to this methodology would be: first, a more complicated change in the co-ordinate system as the orbital center is no longer the center of the earth and second, calculate the impact point after changing co-ordinate systems. All lot more intensive, but unless the gravity over the height of the building is more varied than I believe, it is probably doable.

Walt

 

[davefoc]

69dodge said:

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.QUOTE]

And I think he was right.

The 25.701 inches that Ceptimus's simulation determined for the southward motion is almost exactly offset by the northward lean of the building. I got 25.699 inches for the offset of a plumb bob hanging 381 meters above the ground at a latitude of 40.5 degrees based on the vector addition of gravity and centrifugal force.

The key thing that I didn't get was that the coin continues to experience centrifugal force even after it is released.

 

[Walter Wayne]

The key thing that I didn't get was that the coin continues to experience centrifugal force even after it is released.

I'm not sure I understand this. The coin stops experience no centripetal force and thus orbits the middle of the spheroid earth. The building continues to experience a centripetal force, that keeps in orbiting the axis of rotation in the x-y plane.

If this doesn't help you, see if you have a globe that indicates the ecliptic. The ecliptic is a circle centered on the earth's center which goes from the tropic of cancer, down to the tropic of capricorn. Looking at the globe, a building on the intersection of the tropic of cancer and the ecliptic, would travel along the tropic (a latitude line) because the ground exerts a centripetal force on the building. A coin released from the top will try to follow the ecliptic because the only force on it is gravity.

Walt

 

[davefoc]

Walter,
I apologize, but I'm not quite sure I understood your comment.

Do we agree on the following?

1.The building points north relative to a line drawn from the center of the earth.

2. The reason the earth points north is that centrifugal force caused by its rotation around the earth changes the direction slightly of the downward force on the building.

3. If the earth did not rotate the coin would drop straight towards the center of the earth. (I set the rotation period to a very large number in the Ceptimus simulation and in fact this is what happened. There was no offset from the base).

4. The coin follows a path somewhat south of a line that points towards towards the center of the earth.

5. The coin, almost exactly, follows the line of the plumb bob as it falls (excluding the eastward motion caused by the slightly higher speed at the top of the building). My thought here is that there is a slight variation because the coin experiences a slightly different gravitational field as it falls whereas the plumb bob only experiences the gravity field near the earth's surface. Assuming my calculation for the angle of the plumb bob and Ceptimus's simulation are correct, this is a very slight variation.

As to the specifics of what you said about the ecliptic, I put 23.5 degrees in to Ceptimus's simulation and came up with a south displacement of 19.044 inches for the coin. So at least by the results of the Ceptimus simulation the coin does not follow the ecliptic but rather follows a path slightly south of it. I put 23.5 degrees into my vector addition formula. The formula predicted a northward lean of the building of 19.047 inches at a latitude of 23.5 degrees. Once again the northward lean of the building seems to almost exactly match the southward path of the coin as it drops.

I realize that the calculations above are all based on a rigid, spherical earth and the actual results would be somewhat different for the real earth.

 

[Walter Wayne]

2. The reason the earth points north is that centrifugal force caused by its rotation around the earth changes the direction slightly of the downward force on the building.

[pedantic mode]Centrifugal force is an "apparent" force, so when we talk about this, we should talk about centripetal force. When you do a loop on a roller coaster, you don't push down on the seat, the seat pushes up on you. It just so happens having your chair pushed into you, feels like you are being pushed into your chair.[/pedantic mode]

Take a non-rotating earth again. If you are standing a few feet from the north pole and drop a coin it lands directly at your feet.

Again stand a few feet from the north pole on a stationary earth. If one runs east then you in fact run around the north pole in a tight circle. If you drop a coin while running in this circle. It will now go in a straight line, while you continue to run in this eastward circle. This will make the coin appear to turn south, even though you are the one turning. So the coin will not fall to a point directly beneath your head.

Now, when your running around a corner you lean into it, so that the centripetal and gravity forces line up a long your the length of your body. As such you feet are south of your head. So even though the coin doesn't drop directly down, it still lands close tou your feet.

This is exactly the same thing that the empire state building is doing. It is travelling east which is a curved line. So at any giving time the building is turning north. However, to help prevent the building from topling, we have lean into this corner.

Is this a better explaination. I tried to diagram the thought experiment below. It is an overhead view of the pole, with P being the pole, H being the path the head of the eastward runner takes as he leans into his turn, F being the path his feet take and the red line shows the path of a coin dropped from his head.

Walt