Coriolis force and the Earth's rotation (1 Viewer)

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1. The problem statement, all variables and given/known data
A body is thrown as shown in the picture (0°<x<90°). In what direction the body will the body move in relation to the point it was thrown from - east or west (assume the distance between the point the body was thrown from and the point it lands at is no more than a few meters).
Fry3pbo.png


2. Relevant equations
maR=maI-2mωxuR-mω2r

3. The attempt at a solution
I got really confused, I know which way the earth will rotate but I can't understand how to find the direction. Can anyone give me an explanation? I know which way Coriolis force will act but still can't figure this out.
 

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haruspex

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I know which way Coriolis force will act
Then say what that is.

Just think about the EW component of velocity relative to the Earth's surface. As it moves NE, how will that change?
However, I note the instruction to assume it is only a few metres, and that it is being thrown at the equator. This makes me wonder if it is a trick question. If you consider the magnitude of the effect, what do you notice?
 
Then say what that is.

Just think about the EW component of velocity relative to the Earth's surface. As it moves NE, how will that change?
However, I note the instruction to assume it is only a few metres, and that it is being thrown at the equator. This makes me wonder if it is a trick question. If you consider the magnitude of the effect, what do you notice?
The answer is supposed to be west do I doubt it is a trick question. Coriolis force will act inwards the picture I believe. I still can't understand the answer though.
 

haruspex

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Ah, just noticed this detail:
in relation to the point it was thrown from
That makes it trickier. Need to think some more.

Ok, I've thought some more, and it only seems to make it more certain that the answer is East, and no need for Coriolis. E.g. consider x nearly 90 degrees.
I am assuming that the launch direction shown is relative to the moving surface of the earth.
 
Ah, just noticed this detail:

That makes it trickier. Need to think some more.
I don't believe it is supposed to be really tricky because this was our introduction to the topic, and the professor just said the answer right away and moved on.
 

haruspex

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I don't believe it is supposed to be really tricky because this was our introduction to the topic, and the professor just said the answer right away and moved on.
Please see my edits...
Seems to me your prof has been rather careless, both in specifying the problem and in determining the answer.
 

haruspex

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One more thought... I see it is a dashed line out as far as the horizon, then solid. Is this indicating the launch point is on the far side?!
 
One more thought... I see it is a dashed line out as far as the horizon, then solid. Is this indicating the launch point is on the far side?!
The launch is from the surface of the earth, the dashed line is for indicating the angle.

In the problem simply stated that the body is thrown with some velocity in direction shown in the picture.
 

haruspex

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The launch is from the surface of the earth, the dashed line is for indicating the angle.
Sure, but why is it a dashed line at first, then becoming a solid line? Is it possible the launch point is on the opposite side of the earth from the viewer? Ok, it's extremely unlikely - just looking for a way to make sense of the answer.
Another trick way to make the answer right is to take the launch direction shown as being in an inertial frame moving with the Earth's or Sun's centre. In that case it will be West unless the launch speed is hundreds of kph.

You do understand that in the question as posed Coriolis is irrelevant, yes? It was thrown, it would seem, NE relative to the motion of its launch point. If it only goes a short distance, it will therefore land NE relative to where its launch point is at that time, and even further East relative to where its launch point was when it was thrown.
If the question is intended to be about Coriolis, it should be asking about EW deflection relative to the original launch direction. The answer would still be East, but extremely small. If thrown 10m N and in the air for 1 second the deflection would be about 1 nanometre.
 
Sure, but why is it a dashed line at first, then becoming a solid line? Is it possible the launch point is on the opposite side of the earth from the viewer? Ok, it's extremely unlikely - just looking for a way to make sense of the answer.
Another trick way to make the answer right is to take the launch direction shown as being in an inertial frame moving with the Earth's or Sun's centre. In that case it will be West unless the launch speed is hundreds of kph.

You do understand that in the question as posed Coriolis is irrelevant, yes? It was thrown, it would seem, NE relative to the motion of its launch point. If it only goes a short distance, it will therefore land NE relative to where its launch point is at that time, and even further East relative to where its launch point was when it was thrown.
If the question is intended to be about Coriolis, it should be asking about EW deflection relative to the original launch direction. The answer would still be East, but extremely small. If thrown 10m N and in the air for 1 second the deflection would be about 1 nanometre.
I have found a few sources online that say the same as my professor - that it will deflect to the west. I still have no idea how to reach that.
 

haruspex

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As I noted above, the question as you posted it is not about deflection. Did you word it accurately?

Please post the links you found.
 
If you are standing on a toilet in a hurricane in the southeastern united states, and you throw a wiffle ball to the north, it will curve around to the east since the world turns towards new york.

That's what I remember from science class 50 years ago.

Have a nice day, and don't stand on toilets during hurricanes.
 
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@haruspex are you imagining the ball being thrown from the center of the diagram?

I had a different interpretation. The ball is being thrown vertically upwards from the location where the dashed line becomes solid (in other words, the angle x does not give the direction of motion but rather the location of launch). Then the problem makes sense.

I do agree that the verbal statement of the problem was ambiguous. “As shown in the picture” is not saying enough.

(Sometimes a picture says a thousand wrong words!)

@Eitan Levy are you familiar with the cross product formulation of the coriolis force? I think that is the most powerful way to understand the effect.
 

haruspex

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The ball is being thrown vertically upwards from the location where the dashed line becomes solid
Aha! That's why it is a dashed line! Thanks.
 
@haruspex are you imagining the ball being thrown from the center of the diagram?

I had a different interpretation. The ball is being thrown vertically upwards from the location where the dashed line becomes solid (in other words, the angle x does not give the direction of motion but rather the location of launch). Then the problem makes sense.

I do agree that the verbal statement of the problem was ambiguous. “As shown in the picture” is not saying enough.

(Sometimes a picture says a thousand wrong words!)

@Eitan Levy are you familiar with the cross product formulation of the coriolis force? I think that is the most powerful way to understand the effect.
I understand how to calculate the diversion and even its direction, but only in relation to the picture itself. Say I know the diversion would be outwards (or inwards, I can't remember) in relation to the picture, what is the relation between the outwards-inwards axis of the picture and the east-west axis of Earth? Why are they parallel as my professor mentioned? Why inwards=east and outwards=west? (or the opposite, again I can't remember, but that's not my point).

How do I make that transformation?
 

haruspex

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the east-west axis of Earth?
Other than in geopolitical terms, the earth does not have an east-west axis. Wherever you stand, except at the poles, if you face north then east is to your right and west to your left.
At the point where the body is thrown straight up in the diagram, west is out of the page and east is into the page.
 
At the point where the body is thrown straight up in the diagram, west is out of the page and east is into the page.
How did you reach that conclusion? That's what I can't understand.
 

jbriggs444

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How did you reach that conclusion? That's what I can't understand.
As I think we are reading the diagram...

The circle is the globe viewed from the equatorial plane, high above the prime meridian. (i.e. we are high above the ocean just off the west coast of Africa).

The object is launched from the horizon 90 degrees east longitude and maybe 60 degrees north latitude (somewhere in Russia).

The dotted line represents a backwards projection of its launch trajectory to emphasize that the launch is vertical.

The solid line represents the forward projection into the air, i.e. the object is launched vertically.

The Earth is rotating from west to east. So the right hand globe edge is rotating eastward, away from us and into the page. The left hand edge is also rotating eastward, toward us and out of the page.

The object was launched at the right hand horizon. So it is rotating eastward and into the page.
 
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How did you reach that conclusion? That's what I can't understand.
Think about walking east along the equator (or any line of constant latitude). When you reach the point on the globe which is on the right side of the image, what direction are you walking, in to the page or out of the page? What if you were walking west?
 

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