# Effects of Earth's rotataion on a simple experiment

1. Aug 10, 2014

### 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?

2. Aug 10, 2014

### 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?

3. Aug 10, 2014

### ChrisVer

Coriolis force is what you're asking for?

4. Aug 10, 2014

### amitSingh95

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.

5. Aug 10, 2014

### Orodruin

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

6. Aug 10, 2014

### Orodruin

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

7. Aug 10, 2014

### amitSingh95

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?

8. Aug 10, 2014

### voko

The ball goes higher and its velocity increases?

Have you heard about conservation of energy?

9. Aug 10, 2014

### Orodruin

Staff Emeritus
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?

10. Aug 10, 2014

### amitSingh95

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.

11. Aug 10, 2014

### amitSingh95

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

12. Aug 10, 2014

### voko

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.

13. Aug 10, 2014

### Staff: Mentor

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

14. Aug 10, 2014

### amitSingh95

Understood. Thanks everyone for helping.

15. Aug 10, 2014

### Orodruin

Staff Emeritus
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.

16. Aug 10, 2014

### 256bits

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.

17. Aug 10, 2014

### 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.

18. Aug 10, 2014

### amitSingh95

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

19. Aug 10, 2014

### Staff: Mentor

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.

20. Aug 10, 2014

### Staff: Mentor

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.