Coriolis force question

[QuarkChild]

This is a really dumb question, but it came up in the lab I teach today and I couldn't think of the answer.

If you let a ball drop from some height above the surface of the Earth, it's deflected easterly if you're in the Northern hemisphere, right? And if you throw a ball up in the air and let it come down again, you get a net westerly displacement. (At least that's how it came out when I did the calculation.) Conceptually, though, I can't figure out why the deflection as it travels upward doesn't cancel the deflection as it travels downward.

Tez? Anyone?

(I tried googling it but my google skills failed me.)

 

[Tez]

This is a really dumb question, but it came up in the lab I teach today and I couldn't think of the answer.

If you let a ball drop from some height above the surface of the Earth, it's deflected easterly if you're in the Northern hemisphere, right? And if you throw a ball up in the air and let it come down again, you get a net westerly displacement. (At least that's how it came out when I did the calculation.) Conceptually, though, I can't figure out why the deflection as it travels upward doesn't cancel the deflection as it travels downward.

Tez? Anyone?

(I tried googling it but my google skills failed me.)

Hmm - I think I only understand coriolis force for firing projectiles tangentially (horizontally) not for vertical dropping etc. i.e. if a fire a missile northwards it'll be deflected east, if I fire it eastward it'll go south and so on. If I just fire a bullet straight up and down does it get deflected? If so, I'll have to think about it...

 

[Hamish]

It sounds crazy but it's true.

To solve this, just return to the tried and trusted method of solving all these types of problems: conservation of angular momentum.

In the case where you drop the ball from a height, the ball starts off with more angular momentum than the ball which starts on the ground. I'm assuming that you're dropping this from a tall building and not from an aircraft with motion relative to the earth. As you drop the ball, it's angular velocity increases to conserve angular momentum, so it speeds up relative to the surface of the Earth. Result: easterly deflection. If you throw the ball upwards, the angular velocity decreases so it slows down relative to the earth. It returns with the same angular velocity with which it left, but the avarage angular velocity over the throw is less than the angular velocity of the Earth's surface. Result: westerly deflection.

The fundamental difference, therefore, is that a drop is not the same as half a throw since in preparing for the drop, you have given the ball more angular momentum by raising it.

Hope this answers your question.

 

[Hamish]

Just a quick note to add that this is not, strictly speaking, a coriolis effect question. As I understand it, the coriolis effect deals with the difference in tangential velocitites of the earth's surface at different latitiudes. There is no change in latitude in this question. If you were throwing the ball north or south, you'd get a coriolis effect but since the ball falls to the same lattitude each time, there is no coriolis effect.

Somebody please correct me if I'm wrong.

 

[Tez]

Both your posts sound reasonable to me Hamish... Thanks.

 

[xouper]

Hamish: In the case where you drop the ball from a height, the ball starts off with more angular momentum than the ball which starts on the ground. I'm assuming that you're dropping this from a tall building and not from an aircraft with motion relative to the earth. As you drop the ball, it's angular velocity increases to conserve angular momentum, so it speeds up relative to the surface of the Earth. Result: easterly deflection.

So, if I drop a U.S. penny from the Empire State Building (in New York) how much westerly offset should I use if I'm trying to hit a specific target on the ground (in addition to any wind drift corrections that might be needed)?

 

[Abdul Alhazred]

So, if I drop a U.S. penny from the Empire State Building (in New York) how much westerly offset should I use if I'm trying to hit a specific target on the ground (in addition to any wind drift corrections that might be needed)?

The ESB has setbacks rather than being straight up the whole way. Also the top floor is an enclosed observatory.

So lots of luck hitting the ground at all, unless you're on the 85th floor terrace (or perched on top like King Kong) with a strong wind behind you.

IIRC the first setback is somewhere around the twentieth floor.

 

[ceptimus]

So, if I drop a U.S. penny from the Empire State Building (in New York) how much westerly offset should I use if I'm trying to hit a specific target on the ground (in addition to any wind drift corrections that might be needed)?

I could work it out for a straight sided building in a vacuum. I'll post a guess though first to see how bad my guessing is.

I guess 7.33 inches.

 

[xouper]

Abdul Alhazred: The ESB has setbacks rather than being straight up the whole way.

It's always someone.

OK, how about Sears Tower in Chicago? One of the sides doesn't have any setbacks, but I don't recall which one.

 

[CurtC]

Hamish wrote:
Just a quick note to add that this is not, strictly speaking, a coriolis effect question.

Yes, it is the coriolis effect. The coriolis effect, in its simplest terms, describes the bahavior of the forces you feel on a merry-go-round. If you're on the outside and try to walk to the inside, you'll perceive a force pushing you in the direction of rotation. If the carousel is turning anticlockwise (viewed from above), the force will always appear to be pushing you to the right, whether you're headed in or out.

The reason the question about dropping objects is about coriolis is because what makes for the slightly strange behavior is that the top of the building is closer to the outside of the merry-go-round, going faster, than the bottom.

And my guess is much much less than 7.33 inches.

 

[CurtC]

Well, my own back-of-the-envelope calculations give something like three or four inches. Much more than I guessed.

 

[QuarkChild]

It sounds crazy but it's true.

To solve this, just return to the tried and trusted method of solving all these types of problems: conservation of angular momentum.

In the case where you drop the ball from a height, the ball starts off with more angular momentum than the ball which starts on the ground. I'm assuming that you're dropping this from a tall building and not from an aircraft with motion relative to the earth. As you drop the ball, it's angular velocity increases to conserve angular momentum, so it speeds up relative to the surface of the Earth. Result: easterly deflection. If you throw the ball upwards, the angular velocity decreases so it slows down relative to the earth. It returns with the same angular velocity with which it left, but the avarage angular velocity over the throw is less than the angular velocity of the Earth's surface. Result: westerly deflection.

Wow. That makes sense.

Thank you Hamish.

 

[Hamish]

Curt C wrote:

Yes, it is the coriolis effect

You are, of course, correct. I could only remember the applications where latitude lines were crossed but this is just s special case. Thanks for pointing that out.

My calculations give a deflection of 20.3 cm (about 8 inches) for a penny dropped from the Sears Tower and 16.7 (about 6.6 inches) for the Empire State Building.

 

[davefoc]

Results of my calculations (dimensions are in meters)

height      at equator
10	0.001
100	0.033
442.8	0.306

height      at NYC latitude
10	0.001
100	0.023
442.8	0.215

Assumptions
gravitational acceleration - 9.8 meters/per sec./per sec
circumference of earth - 40075.0 meters
NYC latitude - 44.7 degrees
empire state bldg height - 442.8 meters

The predicted displacement of a coin dropped in a vacuum from height of the top of the antenna on the empire state building at the latitude of New York City is about 8.5 inches from point on the ground pointed to by a plumb bob suspended from the location the coin was dropped from. The contact point will be east of the point marked by the end of the plumb bob.

 

[davefoc]

I just noticed that Hamish and I came up with somewhat different answers.

When I subtracted the height of the antenna from the height of the overall height of the Empire State Building I got 379.7 meters. Using this height I got 6.7 inches of displacement. Perhaps the Observation deck is a little below this in which case my calculations would agree with Hamish's exactly. Which of course means that we're both right or that we both screwed up in similar ways.

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

 

[Hamish]

I just noticed that Hamish and I came up with somewhat different answers.

Yep, and the Empire State Building is 381m according to my sources. The Sears Tower is 442m. I did just assume that I was dropping this from the top of the antenna. I can redo the calculation really easily with whatever parameters you'd like.

I will admit to having forgotten about latitude. It sorts itself out in the earlier stage of my working since the angular deflection (longitudinal) works out the same wherever you are on the earth. Of course, this doesn't equate to the same displacement.

I've done some revised and ever so slightly more accurate calculations. I'll take 381 metres as the ESB height and assume 44.7 degrees latitude.

I get 0.123m (5.1 inches) as the displacement. So now we completely disagree again.

Shall we compare methods?

 

[ceptimus]

I just googled on NYC latitude and got 40.5 degrees

You also need to take into account how far the top of the tower is from the earth's axis. Remember the tower is leaning at an angle of approximately (90 - latitude) to the axis.

I will keep adding quibbles till you get nearer to my guess.

 

[Hamish]

I just googled on NYC latitude and got 40.5 degrees

That's what you get for trusting people. Yes, NYC is at about 40.5 latitude. However, due to a mistake I've just spotted in my calculation, my new figure happens to be 0.124m, so not much change.

You also need to take into account how far the top of the tower is from the earth's axis. Remember the tower is leaning at an angle of approximately (90 - latitude) to the axis.

No, this was implicit in my calculation.

Any more quibbles?

 

[Abdul Alhazred]

I just googled on NYC latitude and got 40.5 degrees

You also need to take into account how far the top of the tower is from the earth's axis. Remember the tower is leaning at an angle of approximately (90 - latitude) to the axis.

I will keep adding quibbles till you get nearer to my guess.

There's a Foucault's pendulum at the United Nations HQ, in but not of NYC.

It's been a stationary hanging weight every time I've seen it. The last time was August of 2001, the first time was sometime in the 1960s. Any updates on that?

 

[ceptimus]

Any more quibbles?

Err, Err... Ah!

You do realise, I take it, that the earth spins in a sidereal day, not a mean solar one? About 23 hours 56 minutes per rev.

Type 'radius of earth in inches' into google, and it will say: 251 106 299 inches

What's the latest figure? Are we getting near to 7.33 inches? I'm running out of ideas.