Effects of Earth's rotataion on a simple experiment

[amitSingh95]

I am not sure if this question even makes any sense, it just popped in my mind while reading a book on special relativity.
Suppose an observer throws a ball upwards, now at the time of throwing, both the observer and the ball have a horizontal velocity same as the instantaneous velocity of Earth at that moment (considering Earth's rotation only).
Now the ball, after falling back will have no vertical displacement (relative to the observer), but if the ball is thrown with sufficient velocity, shouldn't it land ahead of the observer, as the path of the observer is somewhat an arc, due to Earth's rotation, so its displacement should be less than the horizontal displacement of the ball?

 

[256bits]

You have to ask yourself - What is the horizontal velocity due to Earth's rotation of the thrower and the ball before the throw? What is the horizontal velocity of the ball of the ball as it moves upwards? At the peak? Has the horizontal velocity of the ball changed? Knowing the time to the peak, how much hozizontal distance has the thrower and the ball moved? With these positions in mind, are the two radial angles wrt the axis of Earth rotation, for ball and thrower from initial to peak position the same? If not, which angle is less than the other?

 

[ChrisVer]

Coriolis force is what you're asking for?

 

[amitSingh95]

You have to ask yourself - What is the horizontal velocity due to Earth's rotation of the thrower and the ball before the throw? What is the horizontal velocity of the ball of the ball as it moves upwards? At the peak? Has the horizontal velocity of the ball changed? Knowing the time to the peak, how much hozizontal distance has the thrower and the ball moved? With these positions in mind, are the two radial angles wrt the axis of Earth rotation, for ball and thrower from initial to peak position the same? If not, which angle is less than the other?

You are absolutely right and considering what you said, I think the ball SHOULD land ahead of the thrower as at the time of throwing, both have the same horizontal velocity, which doesn't change, but the distance of the ball from the axis of rotation is, on an average, greater than that of the thrower as it goes up. So as in an annular disc rotating about an axis passing through its center and perpendicular to its plane, displacement of outer edge is always greater than that of inner edge, similarly, the ball should traverse a greater arc than the thrower.

 

[Orodruin]

If the ball goes out with maintained horizontal velocity, what happens to its angular velocity?

 

[Orodruin]

The relation is v = ωr. What does this tell you about the angular velocity if you keep v fixed and increase r?

 

[amitSingh95]

The relation is v = ωr. What does this tell you about the angular velocity if you keep v fixed and increase r?

Sorry, I actually intended to write ω=v/r, the same as you said.
v is not fixed here, my last message was wrong, ω is fixed, just as it is for different points on a rotating disk, but v increases with r, so as the ball goes higher, its velocity increases, and as it lands back, it decreases, but even its lowest velocity, given by v=ωr, is equal to that of the thrower, which is constant, so shouldn't it cover a greater path?

 

[voko]

the ball goes higher, its velocity increases

The ball goes higher and its velocity increases?

Have you heard about conservation of energy?

 

[Orodruin]

But it is not a question of a fixed disk. It is a ball that you throw up in the air. If it was a fixed disk the angular velocity would be the same. The angular momentum of the ball is conserved and given by mvr, where v is the horizontal component of the velocity. What does this tell you about the horizontal velocity as r increases? What does it tell you about the angular velocity?

 

[amitSingh95]

The ball goes higher and its velocity increases?

Have you heard about conservation of energy?

I am not aware of much physics terminology, but let me explain you what I mean with an example. Suppose you are standing on a huge merry go round rotating rapidly. Now if you are right at the center, nothing will happen, but if move towards corner, you will be ultimately thrown away, because your velocity increased.
Now I don't know if velocity is the appropriate term to use here.

 

[amitSingh95]

But it is not a question of a fixed disk. It is a ball that you throw up in the air. If it was a fixed disk the angular velocity would be the same. The angular momentum of the ball is conserved and given by mvr, where v is the horizontal component of the velocity. What does this tell you about the horizontal velocity as r increases? What does it tell you about the angular velocity?

So things will not work here as they do in case of a fixed rotating disk?

 

[voko]

I am not aware of much physics terminology, but let me explain you what I mean with an example. Suppose you are standing on a huge merry go round rotating rapidly.

That is true for the observer, fixed at the surface if the Earth. But that is not true for the ball, which is not fixed to anything rotating.

 

[Nugatory]

So things will not work here as they do in case of a fixed rotating disk?

That is correct, they do not. Google for "Coriolis force", read what you find, come back with more questions if necessary.

 

[amitSingh95]

That is correct, they do not. Google for "Coriolis force", read what you find, come back with more questions if necessary.

Understood. Thanks everyone for helping.

 

[Orodruin]

I think it is not really necessary to go into Coriolis force in order to understand this. It could be quite confusing without some basic level of mechanics. I would say the easiest way to understand it is from a non-rotating frame using conservation of angular momentum and the realization that the ball's angular velocity will be smaller than that of the observer on Earth.

 

[256bits]

I am not aware of much physics terminology, but let me explain you what I mean with an example. Suppose you are standing on a huge merry go round rotating rapidly. Now if you are right at the center, nothing will happen, but if move towards corner, you will be ultimately thrown away, because your velocity increased.
Now I don't know if velocity is the appropriate term to use here.

Pondering that, and relating to the ball being thown upward we have 2 particular cases:
Case 1. ball is thrown upwards in a vacuum - no atmosphere
Case 2. ball is throwing upwards in an atmosphere, and is carried with the atmosphere, which rotates with the Earth as a unit. Subsequentally, the earth, atmosphere, ball all have the same angular velocity.

Case 1 you will not be able to catch the ball at the same spot.
Case 2 you will be able to catch the ball at the same spot.

 

[voko]

I do not see why we need to consider the Coriolis force. I find it much more straightforward to treat this as an orbital motion problem. The ball follows an elliptic trajectory. Find the intersection of the trajectory with the surface of the Earth and estimate the elapsed time.

 

[amitSingh95]

Pondering that, and relating to the ball being thown upward we have 2 particular cases:
Case 1. ball is thrown upwards in a vacuum - no atmosphere
Case 2. ball is throwing upwards in an atmosphere, and is carried with the atmosphere, which rotates with the Earth as a unit. Subsequentally, the earth, atmosphere, ball all have the same angular velocity.

Case 1 you will not be able to catch the ball at the same spot.
Case 2 you will be able to catch the ball at the same spot.

"Case 1" - that's what I wanted to say though my reasoning was wrong.

 

[mfb]

I do not see why we need to consider the Coriolis force. I find it much more straightforward to treat this as an orbital motion problem. The ball follows an elliptic trajectory. Find the intersection of the trajectory with the surface of the Earth and estimate the elapsed time.

I don't think calculating orbital elements and the intersection of an ellipse with circle is easier than finding the direction of the Coriolis force.
The calculation via the unchanged v should be even easier, as long as the velocity is small compared to the orbital velocity.

@amitSingh95: This effect can be observed in free-fall towers (those for science, not those for humans), but it is small.

 

[Nugatory]

I don't think calculating orbital elements and the intersection of an ellipse with the trajectory of an observer moving along a circle is easier than finding the direction of the Coriolis force.

Bolded text just supports mfb's point about the difficulty of the orbital calculation, and we haven't even started to consider what happens if the thrown ball has any north-south velocity component.

 

[voko]

I don't think calculating orbital elements and the intersection of an ellipse with circle is easier than finding the direction of the Coriolis force.

Do we actually need to compute anything to answer the original question? Clearly an ellipse and a (different) circle can only intersect in four locations at most; what does that alone imply w.r.t. the possibility of catching the ball thrown up with an arbitrary velocity?

And what would one say upon considering the fact that the motions happen in different planes (save for a couple of degenerate cases)?

 

[olivermsun]

I would say you can start with motion on the equator, and then work your way up.

If you want to know the detailed trajectory of an object as viewed by the observer at a given (but arbitrary) latitude, then this might be easier to compute using the Coriolis force.

 

[Orodruin]

Do we actually need to compute anything to answer the original question?

No, I completely agree with this. The only thing we need to answer is whether or not the angular velocity of the ball will increase, decrease, or stay the same with height and it is trivial to deduce which by just using conservation of angular momentum.

 

[D H]

Out of order response:

You are absolutely right and considering what you said, I think the ball SHOULD land ahead of the thrower as at the time of throwing, both have the same horizontal velocity, which doesn't change, but the distance of the ball from the axis of rotation is, on an average, greater than that of the thrower as it goes up. So as in an annular disc rotating about an axis passing through its center and perpendicular to its plane, displacement of outer edge is always greater than that of inner edge, similarly, the ball should traverse a greater arc than the thrower.

The ball will land behind (to the west of) the thrower.

Google for "Coriolis force", read what you find, come back with more questions if necessary.

This is by far the easiest way to understand what happens to a ball that is thrown straight up.

I think it is not really necessary to go into Coriolis force in order to understand this. It could be quite confusing without some basic level of mechanics. I would say the easiest way to understand it is from a non-rotating frame using conservation of angular momentum and the realization that the ball's angular velocity will be smaller than that of the observer on Earth.

From the perspective of a non-rotating frame, the ball's initial angular velocity will be identical to that of an observer on the Earth. So how does this help? You need to supply more information to make this explanation work.

I do not see why we need to consider the Coriolis force. I find it much more straightforward to treat this as an orbital motion problem. The ball follows an elliptic trajectory. Find the intersection of the trajectory with the surface of the Earth and estimate the elapsed time.

Seriously? You're going to have to solve for the eccentricity and the semi-major axis. Then you need to solve for initial and final true anomalies. Then you need to convert those to eccentric anomalies and then mean anomalies. Then you need to convert the change in mean anomaly to Δt. Then you need to compute how much the Earth has rotated during that interval. Then you need to compare that to the change in true anomaly. This is anything but straightforward.


Explaining this westward drift using the Coriolis effect is simple. The Coriolis effect on an object moving vertically is a westward acceleration proportional to velocity when the object is moving up and an eastward acceleration proportional to velocity when the object is moving down. This appears to cancel, but it doesn't. The westward acceleration while the ball is on the upward part of the trajectory will be strongest when the ball is released. The ball will drift westward throughout the trajectory, reaching a maximum at the peak of the trajectory. That westward drift velocity starts diminishing when the ball starts falling, but it will still have a non-zero westward velocity component throughout the fall. It is only when the ball hits the ground that the east-west component of velocity once again reaches zero.

 

[voko]

Seriously?

Seriously. See #21.

 

[Orodruin]

From the perspective of a non-rotating frame, the ball's initial angular velocity will be identical to that of an observer on the Earth. So how does this help? You need to supply more information to make this explanation work.

And I did. The only other information you need is how angular momentum relates to angular velocity.

Instead you want to start laying out non-inertial frames to a novice in mechanics? (I can vouch that there are students not understanding non-inertial frames even after passing their basic mechanics courses) I find this non-pedagogical. Even if some people would consider it an easier explanation, it is definitely not the universal truth that you seem to think it is.

 

[olivermsun]

Seriously. See #21.

So how would you like to explain #21 to the OP?

 

[voko]

So how would you like to explain #21 to the OP?

I thougt that was self-explanatory :)

Kepler's first law states that the trajectory will be an ellipse, with the centre of the Earth in its focus. The shape of the ellipse in the given setup depends on the magnitude of upward velocity (that is the only variable parameter) and so it is obvious from very general principles that the ball has a chance to meet the observer only if the initial upward velocity is "just right", so it is clear that in the general case that won't happen.

Kepler's second law (your argument essentially) means that the ball's angular velocity will be almost everywhere less than the observer's, thus falling behind; so no initial (upward) velocity is "just right".

 

[olivermsun]

I thougt that was self-explanatory :)

Kepler's first law states that the trajectory will be an ellipse, with the centre of the Earth in its focus...

Kepler's second law (your argument essentially) means that the ball's angular velocity...

That's fine. I was just hoping you'd thought of a simpler or more self-explanatory way of explaining the result to the poster than appealing to Kepler's laws and angular velocity…

 

[D H]

And I did. The only other information you need is how angular momentum relates to angular velocity.

And you didn't put that information on the table. That's what I was getting at.

Instead you want to start laying out non-inertial frames to a novice in mechanics?

Yes. It's done quite often as a matter of fact. The derivation? No. The formula? Yes. High school physics is replete with apparently unrelated formulae. That's the way it goes. They don't have the math to derive those formulae. They can use them, however. Try explaining a hurricane from the perspective of a non-rotating frame. Explaining them using the concept of the Coriolis effect is understandable at the high school physics level.Qualitatively, yes, your explanation does explain why there's a westward drift. Quantitatively, good luck. That's voko's approach. Invoking the concept of the Coriolis effect and assuming gravity doesn't vary with height and it's just a matter of high school level calculus to arrive at the quantitative result that the ball thrown straight up at the equator will land a distance $d=\frac 4 3 \frac {\Omega v^3}{g^2}$ to the west.

 

[Orodruin]

And you didn't put that information on the table. That's what I was getting at.

I beg to differ:

The relation is v = ωr. What does this tell you about the angular velocity if you keep v fixed and increase r?

The angular momentum of the ball is conserved and given by mvr, where v is the horizontal component of the velocity. What does this tell you about the horizontal velocity as r increases? What does it tell you about the angular velocity?

I gave all of the information necessary for the OP to work this out for himself/herself rather than simply providing the answer.

Qualitatively, yes, your explanation does explain why there's a westward drift. Quantitatively, good luck. That's voko's approach. Invoking the concept of the Coriolis effect and assuming gravity doesn't vary with height and it's just a matter of high school level calculus to arrive at the quantitative result that the ball thrown straight up at the equator will land a distance $d=\frac 4 3 \frac {\Omega v^3}{g}$ to the west.

If you reread the OP, you will see that it is not necessarily asking for a quantitative solution. I find it counter productive to meet a qualitative question of what happens and why with "there is this fictitious force in a rotating system", rather than trying to explain what is actually going on.

In addition, within your assumptions that the ball is not thrown to heights that require consideration of a varying g, it is also very easy to integrate the equations I supplied and arrive at the same result (assuming that your $g$ really should be $g^2$, which is necessary to make the dimensions consistent):

Conservation of angular momentum gives:
$$
\omega(r) = \omega_0 \frac{r_0^2}{r^2} \simeq \omega_0 - 2 \omega_0 \frac {h(t)}{r_0}
\quad \Rightarrow \quad
\frac{d\phi}{dt} = \omega(r) - \omega_0 \simeq - 2 \omega_0 \frac {h(t)}{r_0}.
$$
Given that $g$ is constant
$$
h(t) = v t - \frac{gt^2}2.
$$
Integrating from 0 to $2v/g$ yields
$$
\phi = - \frac{4\omega_0 v^3}{3r_0 g^2}.
$$
Multiply with $r_0$ to get the distance rather than angle:
$$
d = - \frac{4\omega_0 v^3}{3 g^2}.
$$

 

[Orodruin]

And just for curiosity reasons as to the size of the effect. Modern rifles can have a muzzle velocity of about 1200 m/s (source http://en.wikipedia.org/wiki/Muzzle_velocity). The resulting effect is of the order of a few hundred meters.

However, in order to have the effect as the dominant one, you must be able to aim the rifle to the vertical direction within some arcminutes. If not, your aim will be determining whether the bullet lands to the east or west.

 

[voko]

And just for curiosity reasons as to the size of the effect. Modern rifles can have a muzzle velocity of about 1200 m/s (source http://en.wikipedia.org/wiki/Muzzle_velocity). The resulting effect is of the order of a few hundred meters.

However, in order to have the effect as the dominant one, you must be able to aim the rifle to the vertical direction within some arcminutes. If not, your aim will be determining whether the bullet lands to the east or west.

Does that take the atmosphere into account?

 

[Orodruin]

Of course not, neither does anything else we have said. :)

 

[D H]

And just for curiosity reasons as to the size of the effect. Modern rifles can have a muzzle velocity of about 1200 m/s (source http://en.wikipedia.org/wiki/Muzzle_velocity). The resulting effect is of the order of a few hundred meters.

It's almost eighteen hundred meters if the bullet is fired from the equator, not just a few hundred. (I get (1787 meters using an orbital calculation, 1762 meters using $\frac 4 3 \frac {\Omega v^3}{g^2}$ as an estimate.)

Does that take the atmosphere into account?

Of course not. A drag-free bullet fired straight up at 1200 m/s will climb to almost 75 kilometers before falling back to Earth. A real 50 caliber bullet? I doubt it will even climb to 5 kilometers. A bullet shot straight up will most likely be tumbling on the way down, making for a lowish terminal velocity (a few hundred meters/second). The randomness that results from this tumbling will overwhelm the westward drift.