# The Ball in the Train

1. Jan 12, 2006

### Bugzmom200

Hi everyone. I have a question that my friends and I have debated for several years. I think I'm right and everyone else is wrong....

Let's say I was in a train. This train is moving forward at a constant speed, maybe 200 mph. The ceiling is very high and there are no windows, doors, or other way a draft could get in. I am sitting on a chair in the middle of the train. I thow a ball as high as it can go. When it lands, does it land in my hand or behind me?

2. Jan 12, 2006

### ZapperZ

Staff Emeritus
I'll answer that by asking you to think about this: Our earth is moving a few thousand miles per hour according to some celestial body. So the earth is your train. I'm sure you have thrown a ball up in the air as high as you can. Where did it land?

Zz.

3. Jan 12, 2006

### robphy

During a steady ride on an airplane, do you have to adjust the way you usually pour soda from a can into a cup at rest on your table?

4. Jan 13, 2006

### pervect

Staff Emeritus
If you threw it straight up, it would probably land to your west due to coriolis forces, just as it would if you were standing still. Assuming you're not too far north of the equator

Some unusual conditions of very high latitude or very high speed might be able to make this statement untrue.

You probably wouldn't be able to throw it accurately enough to measure the effect, though, even if you were at the equator.

5. Jan 13, 2006

### arildno

I always got a big bump in my head when I did that sort of thing.
That's probably why I'm so bad at sports.

6. Jan 13, 2006

### daniel_i_l

Before you threw it up the ball was traveling at the same speed as you. So according to Newtons first law the ball will continue with that speed even after you throw it cause of inertia. If you throw it out the window for example, you'll see it fly backwards but thats only due to air resistants, but in the train there's no resistants cause also the air is going at the speed of the train.

7. Jan 13, 2006

### LURCH

Speaking of effects too small to measure;
1) What about the fact that the ball is going into an orbital path around the Earth that is more eleptical than the train's? The ball would orbit faster than the train while climbing, slower at appoge, and speed up again after appoge. Would the sum of these changes in relative volocity average out to lateral progression equal to that of the train, or would it be slightly behind? I know that the arch traced out by the ball will have the same area as that of an object in a freefall, more circular orbit at the train's altitude, but the fact that the train is not in freefall orbit, but under power makes me unsure.

2) Also, if the train is travelling along a track that runs East and West (and not on the Equator), wouldn't orbital dynamics cause the ball to come down slighty closer to the Equator than the train?

8. Jan 13, 2006

### DaveC426913

So, in summary:

It will land in your hand. To any reasonable degree of measurement, it will go straight up and straight down.

However, as others are pointing out, if you want to be a stickler and do a controlled experiment, there are some factors that can be accounted for that will make very, very small changes in exactly how far from straight up and down the ball will travel. You'd need a very steady train, very accurate (laser) instruments and a robotic tossing mechanism.

9. Jan 13, 2006

### Taviii

If you are facing in the direction that the train moves in then the ball lands behind you. My physics teacher explained this to my class 2 days ago.

10. Jan 13, 2006

### Homer Simpson

So what force is causing it to go backwards??

11. Jan 13, 2006

### Taviii

You are getting it all wrong, the ball isnt going backwards.

The ball tends to slow down because of its weight, and the speed of the train remains constant at 200 mph. So if you think of a point under the train in the moment when you trow the ball, the ball travels a smaller distance from that point to the place were it lands than the hand of the trower (which has the speed of the train at all time)

Last edited: Jan 13, 2006
12. Jan 13, 2006

### Homer Simpson

But the train is not accelerating, it has constant velocity. The ball is going 200 mph when you release it. With no outside forces to slow it down, it will continue to move at 200 mph for ever. (F=ma)

13. Jan 13, 2006

### Orefa

Wow, this is so wrong. If you're lucky you have simply misunderstood your teacher. If you're not lucky then you are stuck with a lousy teacher.

The ball's inertia is proportional to its weight (or mass, more accurately). No matter how much it weighs, if it is in motion then it retains this motion until some force is applied to it.

14. Jan 13, 2006

### ZapperZ

Staff Emeritus
As has been pointed out, you are strongly suggested to go back to your teacher, and MAKE SURE you understand this correctly, because this is seriously wrong.

Again, look at my FIRST response to this thread. WE, the earth, are the "train" according to someone on alpha centauri. Now toss a ball vertically in the air. Do you see it landing behind you? In front of you? To the side of you?

Zz.

15. Jan 14, 2006

### Taviii

Is it posible that if you trow the ball from the top of the train it will not land in the same place because of the friction with the air?? If yes then I made a mistake because in the train there is no friction with the air on the horizontal movement because the air also travels at 200 mph. I made this mistake because of my poor english. I didnt understand what "draft" means, from the initial post in this thread.

16. Jan 14, 2006

### DaveC426913

Yes that would be true. But the OP carefully described the interior of the train for good reason.

17. Jan 14, 2006

### pervect

Staff Emeritus
This is using alternate language to talk about the same physics. In fact, that's the mental picture I used to figure out that the ball dropped to the west. (I hope I didn't make a sign error somewhere along the way!).

Lets adopt an earth-centered inertial (non-roating) frame of reference. Then the angular momentum of the ball relative to the center of the Earth will be constant. It's angular velocity will thus be slightly lower at higher altitude than it will be at a lower altitude. So it will appear to move west from the POV of someone on the ground.

This physics is explainable in terms of the coriolis pseudo-force by someone who adopts a coordinate system fixed to the Earth.

My main comment is based on the fact that at the equator, one's veocity in the ECI (Earth-centered-inertial) frame is 460 meters/second, or about 1040 miles/hour. This is due to the rotation of the Earth. A few hundred miles per hour due to the trains motion is probably not going to be that important compared to the effect of one's velocity due to the Earth's rotation.

Of course this part of the analysis fails completely at the North pole, which is why I remarked that it wouldn't always be true.

Last edited: Jan 14, 2006
18. Jan 14, 2006

### pervect

Staff Emeritus
For the brave of heart, here is what I'm getting for the complete equations of motion. I won't describe how I got them in detail, hopefully it's right :-).

Coordinates:

$\theta$ is zero at the equator, and $pi/2$ at the north pole. It's similar to a latitude.

$\phi$ is zero on the Grenwich meridian, similar to a longitude.

r is of course the distance from the center of the Earth.

Differentiation with respect to time is indicated by a dot above the variable, i.e. $\dot{r} = dr/dt$

$$\ddot{r} - r \, \dot{\theta}^2 - r \, sin^2(\theta) \, \dot{\phi}^2 = -gM/r$$
$$\ddot{\theta} + \frac{2}{r}\,\dot{\theta}\dot{r} - \frac{1}{2} \, sin(2\theta) \, \dot{\phi}^2 = 0$$
$$\ddot{\phi} + \frac{2}{r} \, \dot{\phi}\dot{r} + 2 \, tan(\theta) \, \dot{\theta}\dot{\phi}= 0$$

What this means is that if someone is standing still on the equator, $\dot{\phi}$ is 2 pi radians / 24 hours due to the Earth's rotation while $\dot{\theta}$ is zero. This gives us

$$\ddot{\phi} + \frac{2}{r} \, \dot{\phi}\dot{r} = 0$$

which means that if someone throws a ball up, making $\dot{r}$ nonzero (positive), $d^2 \phi / dt^2$ becomes nonzero (negative). The magnitude of the force is proportional to the radial velocity, as one expects of a coriolis force. Thus the ball accelerates west immediately when it is thrown up, stops accelerating at the peak of its trajectory, and accelerates east on its downward trip, eventually returning to an east/west velocity of zero. However, while the east/west velocity is zero, the east/west position is negative (west).

A trouble spot - the above equations are for a circular Earth. It can be seen that the second equation doesn't have a stationary solution for $\theta$ :-(, thus it's not quite right.