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Pretend there is a train that has very tall cars, say, 30-40 feet. Also pretend (assume) the train can go very fast. 500 MPH sounds good.
You are in one of the cars. You paint a bulls-eye on the floor of the car. You set-up a ladder that allows you to climb to the top of the car. From there, you position yourself to drop a golf ball exactly over the bulls-eye on the floor below.
Will the ball land exactly below where you drop it, or will it land a little behind the bulls eye, loosing delta while it falls and car is rushing forward at 500 mph?
I think I know that the ball will hit the bulls-eye with no effect from the car moving forward, but common sense tells me it will land behind.
Of course, if it did land behind the mark, then things dropped from any great height would always land a little west of straight down as the earth turned during the fall, wouldn't they?
I owe Nick an answer. Please help me out. Thanks much.
edit typo
[edited by: Leosghost at 10:37 pm (utc) on May 27, 2009]
then things dropped from any great height would always land a little west of straight down as the earth turned during the fall, wouldn't they?
[edited by: LifeinAsia at 10:59 pm (utc) on May 27, 2009]
If the train is moving in a straight line, with uniform speed, on horizontal track, and is airtight, the object will fall vertically relative to an observer on the train.
Kaled.
If the train is moving in a straight line, with uniform speed, on horizontal track, and is airtight, the object will fall vertically relative to an observer on the train.
Rephrased with a tighter condition on the fist three qualifiers - If the train moves at a uniform velocity [en.wikipedia.org], and the container is airtight, the object ill fall vertically relative to an observer on the train (hitting the bullseye)
Leosghost, relativistically you are right. However, at the speeds concerned, air turbulance will have more influence than relativity calculations. At low energies, Newton is more than adequate.
[edited by: lawman at 10:14 am (utc) on May 28, 2009]
[edit reason] Fix Style Code [/edit]
I used a description that someone would understand without knowing the scientific definition of velocity.
Incidentally, whilst relativistic physics is not my forte, the speed of light (in a vacuum) is a universal constant and does not ever change with respect to the observer. i.e. however fast the train is moving, an experiment to measure the speed of light on the train will always yield the same result in every direction, up down, forwards, backwards, etc. (and light will still travel in straight lines).
Kaled.
with respect to the observer.the observer can never be static in relation to the observed phenomena..their masses change and the distance between a dropped object and the "dropper" change ..etc etc ..even simple gravity behaves ( more or less from where we are ..most of the time ) in an inverse square ratio dependant upon the distance between the relative masses ..
...rotation around:
- Earth's own axis
- the Earth-Moon barycentre
- the Sun
- the galaxy
- the local group
- the universe.
[edited by: g1smd at 12:13 am (utc) on May 29, 2009]
Rephrased with a tighter condition on the fist three qualifiers - If the train moves at a uniform velocity, and the container is airtight, the object ill fall vertically relative to an observer on the train (hitting the bullseye)
Just because I am a stickler - the train car must be a perfect vacuum, not just air tight.
The answer here is the same reason jumping upward a moment before a free falling elevator hits the ground will not save you.
The answer here is the same reason jumping upward a moment before a free falling elevator hits the ground will not save you.
But I remember an old Bugs Bunny cartoon where he was in an airplane careening toward earth but was saved when the plane ran out of gas just a few feet before it hit the ground. How do you explain that?
Onya
Woz
[speeliong]
[edited by: Woz at 2:50 am (utc) on May 29, 2009]
Of course, if it did land behind the mark, then things dropped from any great height would always land a little west of straight down as the earth turned during the fall, wouldn't they?
And, they do! Long range artillery fire (as from battleships) must take this Coriolis effect into account.
The Brits failed to adjust for changes moving from northern to southern hemisphere in the Faulkland Island war resulting in some very inaccurate gunnery at the beginning.
Another example would be the Paris gun used by the Germans in WWI (from a range of ~75 miles).
It is the rotation of the earth that causes the phenomon. The projectile is traveling in a straight line while the earth turns beneath it.
A train would be subject to the same effects, since a "straight" line really is following the curve of the earth. The ball's trajectory could be computed via Newtonian physics using a vector matrix. It is a combination of the momentum imparted by the earth's spin (think cetrifugal force) throwing the ball in a straight line away from the earth, the forward momentum of the ball (equal to the train's speed and direction at the moment it is released) and gravity pulling the ball straight towards the center of the earth (at the moment it is released).
The ball would not hit the target, although it is simplistic to say that it would land a bit behind it.
A object will continue in its state of rest or uniform motion unless acted upon by some external force.
There is no horizontal force, therefore there is no horizontal acceleration, therefore the object will travel with precisely the same horizontal velocity as the train.
the observer can never be static in relation to the observed phenomena
Kaled.
Just because I am a stickler - the train car must be a perfect vacuum, not just air tight.
Not true! If the cabin is airtight, then the air will be traveling at the same rate as the train- thus not moving at all relative to the experiment. Conceptually, dropping a ball at a target in a stationary environment (relative to the surface of the Earth), will result in a hit. Same with an airtight cabin.
I used a description that someone would understand
Coriolis effect
With a nod to Leosghost post #3922083 on Heisenburg and cats...
You relativists might as well argue that since you know the precise velocity of the target, you can't possibly know where it will be when the ball is due to hit it. Its simply not the right scale for that argument is to apply.
Of course, if the train was properly sealed, the ball would have both and neither hit and/or missed the target.
There is no horizontal force, therefore there is no horizontal acceleration, therefore the object will travel with precisely the same horizontal velocity as the train.
That would be true if the earth were flat!
The train is following a curved path (the surface of the earth). The object will proceed in a straight path as the train curves away from it.
...the Rate Of Change of latitude of the target is equal to the ROC of latitude of the object.
Again, only if the earth were flat. The "horizontal" vector of the train is in fact following a curved path. Once the object is released its vector will be a straight path.
We are, after all, talking about a three dimensional space.
Once the object is released its vector will be a straight path
If you really want to be that silly and pedantic, the object would travel in a sub-orbital curve after it is released. A curved trajectory would be apparent even to an observer on the train, provided the instrumentation was sufficiently accurate. However, since there is no information available as to speed, latitude, and direction, it is reasonable to ignore all effects that would depend on these variables.
Kaled.
Whats the angular velocity of the fastest train? Negligible. Thus, any impact of any factor relying on this will can be ignored.
But I remember an old Bugs Bunny cartoon where he was in an airplane careening toward earth but was saved when the plane ran out of gas just a few feet before it hit the ground. How do you explain that?
Don't you know that the normal laws of physics don't apply to cartoon characters? Honestly, what were you thinking?
There is no horizontal force, therefore there is no horizontal acceleration, therefore the object will travel with precisely the same horizontal velocity as the train.
Wouldn't the locomotion of the train be considered an external force?
As soon as the object is released, the golf ball will begin horizontal deceleration, losing momentum as it is no longer being directly acted upon by the power of the train. (apart from the air-which is also being acted upon by the locomotion of the train)
Ball hits behind the bulls eye.
I'll also argue that the ball will bounce and roll towards the rear of the train, eventually coming to rest at the back wall.
Unless of course our happy experiment is abruptly halted due to the failure of our myopic engineers to account for that tunnel...
Wouldn't the locomotion of the train be considered an external [b]force[b]
Nope. Velocity (locomotion)* is not a force. Force acts on an object to ACCELERATE it. No (horizontal) acceleration here, so no force present.
As the cabin is defined as airtight, there is no horizontal drag on the ball, thus no horizonal force, thus no horizonal movement. It's straight down, relative to the cabin.
* The train's engine provides sufficient force to overcome the friction of the train through the air. Net force of engine plus drag is zero.
Wouldn't the locomotion of the train be considered an external force?
the golf ball will begin horizontal deceleration
our happy experiment is abruptly halted due to the failure of our myopic engineers to account for that tunnel.
It's straight down, relative to the cabin.
It is straight down relative to the center of the earth's mass.
The ball will miss the target...
We have been asked to assume an initial velocity of 500 MPH (roughly 224 meters per second) and a drop of 40 feet (a bit over a second of free fall). From the moment of decoupling the system the train will traverse a bit over 224 meters of the arc of the earth's diameter while the ball will travel a bit over 224 meters along the tangent of the arc. Been to long since I have done trig, but I would guess a simple solution would prove a miss of several centimeters.
But I have enjoyed the conversation.
