How does frame dragging affect the trajectory of a falling bullet?

[15123]

A friend of mine asked me this question out of curiosity:

If you shoot a bullet straight up into the air, how far would the bullet land from the point it went up?

Possible factors:
- Earth's rotation
- air resistance

What is the answer?

 

[Nugatory]

In theory:
It wll depend on a number of other variables. For example, the effect of the rotation and curvature of the Earth depends on how high the bullet goes and how long it is in flight; you worry about these things when you're aiming an ICBM but not when you're aiming a bottle rocket.

In practice:
Only two things matter: crosswinds and the difficulty of really aiming exactly straight up, not the least bit sideways.

 

[256bits]

Hello 15123
If the Earth was not rotating then if the bullet was shot straight up, it should fall back to Earth at the exact same spot. That is of course assuming there is 'no wind' in the atmosphere.

With a rotating Earth and the same 'no wind' assumption, the only place the bullet would fall back straight down to the same spot is at one of the poles.

With a rotating Earth and the same 'no wind' assumption, at the equator the bullet would fall someplace behind the spot where the bullet was shot up, meaning to the west of that location. The bullet would keep a sideways motion of 1000mph ( the velocity of the Earth's surface at the equator ) as it is ascending and descending. The Earth would rotate under the bullet in its trajectory, and if you shot it up high enough the Earth could make one complete revolution, plus a part of a revolution ( remember the bullet keeps the 1000mph velocity to the east ) so that the bullet could fall back to the same spot.

'no wind' would mean no atmosphere and therefor no drag to slow the velocity of ascent or descent due to terminal velocity, nor any drag to slow the 1000mph velocity to the east. the bullet would be subject to gravity only, and one could calculate the fall back location somewhat easily.

At any other latitude, someone might come along to explain that.

Also, with an atmosphere, it becomes more difficult since a few more assumptions would have to be made on the wind velocities and the does the bullet end up being carried along with the air due to drag.

 

[15123]

Let us suppose that the bullet is aimed 100% exactly straight up (through some device perhaps) and that the bullet is a standard 9 mm bullet with a symmetric body (like all bullets).
Let us also suppose that this happens on a day where there is no wind resistance at all.

Other factors:
- bullet's rotation around its own axis

I tend to think that the bullet's nose might be pulled sideways by gravity when the bullet is on its way back down to earth. I find it hard to believe that the bullet would fall straight down the same way that it went up (with its nose pointing forwards 100% of the time). It does not sound realistic to me.

 

[Nugatory]

I find it hard to believe that the bullet would fall straight down the same way that it went up (with its nose pointing forwards 100% of the time). It does not sound realistic to me.

You're right, it's not realistic. But the thing that's not realistic is the part where we get everything exactly perfectly right: 100% straight up, no wind deflection at all, we'll do the experiment at one of the Earth's poles so that the Earth isn't moving sideways while bullet is in flight, not even the tiniest microscopic flaw in the symmetry of the bullet., ... If we got all of those things right, the bullet would go straight up into the air and fall right back into the muzzle of the gun. Of course that's absurd, but the absurdity just goes to show that it's absurd to expect to get everything spot on right in the real world.

 

[Damo ET]

Please correct me if I'm wrong, but why would the Earth's rotation have any effect on where the bullet (or any other free falling object) would land? We don't get pummeled by falling hail at 1600km/h from the east at the equator, because the atmosphere is also rotating with the earth. If I drop a ball from the east side of a tall building, it doesn't hit the side of the building because the Earth moves from under the ball. This is the same case for the bullet, the rotating Earth has nothing to do with the frame of reference that the bullet is moving in.Damo

 

[Nugatory]

If I drop a ball from the east side of a tall building, it doesn't hit the side of the building because the Earth moves from under the ball. This is the same case for the bullet, the rotating Earth has nothing to do with the frame of reference that the bullet is moving in.

The corrections are too small to worry about, but they are non-zero. The dropped object is moving in free-fall in a straight line while the building is moving on a circle of radius 6400 kilometers (if at the equator). If the building were tall enough, and if other effects such as crosswinds did not dominate, you would see that the trajectory of the ball does not remain parallel to the wall of the building throughout the fall.

We don't get pummeled by falling hail at 1600km/h from the east at the equator, because the atmosphere is also rotating with the earth.

That is indeed correct, but it works that way only because the distances involved are small enough that we can ignore the rotational effects. We (rightly) approximate the behavior of a falling hailstone as if the surface of the Earth were flat and stationary, and this works fine over distances of five or ten kilometers, especially if we aren't demanding the unreasonable precision required to drop a bullet back into the muzzle of the gun it was fired from.

Try rolling that hailstone from the north pole to a point on the equator, and you will have to consider Coriolis effects.

 

[A.T.]

the rotating Earth has nothing to do with the frame of reference that the bullet is moving in.

The bullet is not moving in a specific reference frame. If we chose to describe the bullets movement in the rotating rest frame of the earth, there will be a Coriolis acceleration. In an inertial frame the bullet will stay behind the Earth's rotation when it increases its distance from the center. Either way the bullet doesn't land, where it was shot up (except at the poles), but the deviation is small.

 

[technician]

The effect is not insignificant. It needs to be taken into account when firing artillery shells.
They may not go 'vertically upwards' but they do have a vertical component.
The effect is known as the Coriolis effect

 

[Damo ET]

Roger roger, that makes sense. :)

I can remember my father telling me when I was quite young, that when constructing large buildings these effects could be expected also. Although they would be miniscule.Damo

 

[ViolentCorpse]

The corrections are too small to worry about, but they are non-zero. The dropped object is moving in free-fall in a straight line while the building is moving on a circle of radius 6400 kilometers (if at the equator). If the building were tall enough, and if other effects such as crosswinds did not dominate, you would see that the trajectory of the ball does not remain parallel to the wall of the building throughout the fall.

When the object is resting in a building floor, it's rotating with the Earth. I don't understand why that should change during free-fall? Shouldn't the object's horizontal motion still be same as that of the Earth during the entire time it is falling (as a coin tossed in a steadily moving car moves with the car)?

 

[A.T.]

... it's rotating with the Earth...the object's horizontal motion ...

You are confusing angular and linear velocity here. They are related via the distance from the center, which changes during a fall.

 

[ViolentCorpse]

.. which changes during a fall.

How would that change if the object is falling in a straight line?

 

[Nugatory]

When the object is resting in a building floor, it's rotating with the Earth. I don't understand why that should change during free-fall? Shouldn't
the object's horizontal motion still be same as that of the Earth during the entire free-fall

The object, once in free-fall, keeps its horizontal motion just fine; drop it from the top of the building and it will start out with the exact same horizontal speed as the top of the building and keep it the whole way down. When it hits the ground it will still be moving sideways at the same speed as the top of the building.

But the top of the building is moving sideways faster than the bottom of the building. It has to, because the top and the bottom both make a full circle every 24 hours, but the two circles have different radiuses (they differ by the height of the building), hence different circumferences, hence different distance traveled in 24 hours, hence different speeds. Thus, the object will be moving horizontally relative to the ground when it hits. The effect is too small to notice or care about unless you're aiming long-range artillery, but it's there.

 

[ViolentCorpse]

Such a lucid explanation! I understand it now.

Thanks so much guys.

P.S: If I remember correctly, non-believers of the heliocentric model of the solar system actually used this observation (that Earth doesn't move out underneath a falling object) as an objection to it. Galileo pointed out that need not be so if Earth is uniformly rotating. Turns out they were both wrong, hehe.

 

[Nugatory]

Galileo pointed out that need not be so if Earth is uniformly rotating. Turns out they were both wrong, hehe.

Yes, but Galileo was much less wrong .

To quote Isaac Asimov:

when people thought the Earth was flat, they were wrong. When people thought the Earth was spherical, they were wrong. But if you think that thinking the Earth is spherical is just as wrong as thinking the Earth is flat, then your view is wronger than both of them put together

 

[voko]

The trajectory of a bullet "shot upward" will be an ellipse. The inclination of orbital plane for a bullet "shot upward" depends solely on the latitude of the firing location, but the important thing is that it will be some non-zero (except when fired on the equator). Because the trajectory of the firing location itself is a circle parallel to the equatorial plane, the trajectory of the bullet and the trajectory of the firing location may intersect in at most two points, one being the initial firing location, and the other one "on the other side of the Earth", which is impossible for an ordinary bullet.

 

[ViolentCorpse]

Yes, but Galileo was much less wrong .

To quote Isaac Asimov:

Well-said. I love that entire essay written by Asimov!

Edit: Nurgatory, could you please elucidate how the top of the building would trace out a larger circle than the bottom of it? I can't quite visualize it.

Thank you!

 

[Damo ET]

The top of the building will travel 2(pi)meters further than the base based on the height of the building. If it is 100mtrs tall, the top will travel 628.3mtrs further in one day than the base.Damo

 

[Nugatory]

Edit: Nugatory, could you please elucidate how the top of the building would trace out a larger circle than the bottom of it? I can't quite visualize it.

It might be easier to visualize if you imagine yourself floating in space above the south pole, watching the Earth rotating below you like a giant ball. Put a building at the equator so that it's sticking out from the surface of the Earth - from your vantage point above the poles the building is sticking out sideways.

The radius of the Earth is about 6400 kilometers, so when the Earth turns one full rotation, a point on the equator such as the base of the building, will have gone around a circle of circumference $2\pi{r} = 6400\times{2}\pi$.

However, the top of the building is further away from the center of the Earth (6400 km plus the height of the building) so it traces a slightly larger circle: $2\pi(r+h)$.

 

[CraigH]

Sorry but I don not understand why the rotating Earth would affect where the bullet lands. Surely if the angular velocity of the bullet stays the same it will land in the same place it was fired?

The Earth's surface has an angular velocity of ω=7.3x10^-5rads-1 and a horizontal velocity of 465ms-1 at the equator (for a stationary observer outside of the earth). If a bullet is taken to the top of a very high building it will have a larger horizontal velocity because the radius of the circular path it is traveling will be larger.

However when the bullet is dropped it's horizontal velocity will decrease and approach 465ms-1 as it approaches the Earth's surface. This is because it's angular velocity will remain constant. Therefore the bullet will stay above the same spot it was fired.
v=rw
As r decreases, v must also decrease so that w stays constant.

The reason I think w (angular velocity) will stay constant is because there is no torque acting on the bullet.
angular acceleration = T/I
w1=w0 +(T/I)t
if T=0 angular velocity will not change

I have created an image to explain what I am talking about


Can someone please tell me where I am going wrong? I believe you that the bullet would fall behind its firing point but I can't see why. According to my math it will fall where it was fired.

Thanks

 

[cjl]

However when the bullet is dropped it's horizontal velocity will decrease and approach 465ms-1 as it approaches the Earth's surface. This is because it's angular velocity will remain constant. Therefore the bullet will stay above the same spot it was fired.

Why do you think the angular velocity would be constant? What force would be slowing down its tangential velocity as it fell? There would have to be some net force acting on it for this to happen - where would it come from?

 

[Nugatory]

This is because it's angular velocity will remain constant. Therefore the bullet will stay above the same spot it was fired...
The reason I think w (angular velocity) will stay constant is because there is no torque acting on the bullet.

Something has to be wrong with that argument, because there's also no horizontal force acting on the bullet so no reason for its horizontal speed to change.

The bullet's angular velocity relative to the center of the Earth does not remain constant. You are right that there is no torque acting on the bullet and therefore that its angular momentum does not change. However, constant angular momentum implies constant angular velocity only if we're doing uniform circular motion so the radius is kept constant - and in this case as soon as the bullet is released it's no longer moving along a circle. Mathematically, you'll have to use the more general definition of angular momentum L=rxp where r and p are the position and linear momentum vectors. The angular velocity equations you used come from the constant-radius special case of this equation.

 

[CraigH]

Why do you think the angular velocity would be constant? What force would be slowing down its tangential velocity as it fell? There would have to be some net force acting on it for this to happen - where would it come from?

ah okay this makes sense, I forgot about f=ma. If the tangential velocity changes there must be a tangential force on it. But I don't get why the angular velocity changes when there is no torque on it.

Something has to be wrong with that argument, because there's also no horizontal force acting on the bullet so no reason for its horizontal speed to change.

The bullet's angular velocity relative to the center of the Earth does not remain constant. You are right that there is no torque acting on the bullet and therefore that its angular momentum does not change. However, constant angular momentum implies constant angular velocity only if we're doing uniform circular motion so the radius is kept constant - and in this case as soon as the bullet is released it's no longer moving along a circle. Mathematically, you'll have to use the more general definition of angular momentum L=rxp where r and p are the position and linear momentum vectors. The angular velocity equations you used come from the constant-radius special case of this equation.

sorry this has confused me. Are you saying that the equation:

(net torque) = (moment of inertia) * (angular acceleration)

is only true for constant radius circles?

 

[Nugatory]

Are you saying that the equation:

(net torque) = (moment of inertia) * (angular acceleration)

is only true for constant radius circles?

Quoting wikipedia, with emphasis mine: "Moment of inertia is a property of a body that defines its resistance to a change in angular velocity about an axis of rotation". The Earth and everything on it is rotating about the Earth's axis, but the bullet, once released, is not so you can't use this equation to relate net torque to the change in its angular velocity.

An example of angular velocity changing even though there is no net torque and therefore no change in angular momentum: You are sitting at the point (0,0) in the Cartesian plane, and a particle moves past you from left to right at a constant speed along the line $y=1$. It's angular velocity gradually increases as it nears you, reaches a maximum as it moves through the point (0,1) and then diminishes to zero as it moves away.

 

[A.T.]

Are you saying that the equation:

(net torque) = (moment of inertia) * (angular acceleration)

is only true for constant radius circles?

This equation tells you that angular velocity changes when the radius changes under zero torque.

 

[voko]

Sorry but I don not understand why the rotating Earth would affect where the bullet lands.

See #17.

Surely if the angular velocity of the bullet stays the same it will land in the same place it was fired?

Nope. Angular velocity stays constant only with an object that is constrained to rotate about a point. A bullet is not.

 

[jbriggs444]

But I don't get why the angular velocity changes when there is no torque on it.

For a simple example, walk to the street outside your house and turn your head to track a car as it passes. When the car is approaching from far away its angular velocity is small. As it passes close by its angular velocity is large. As it recedes into the distance, its angular velocity is small again. All without any torque or acceleration at all.

 

[MikeLacroix]

Sorry but I don not understand why the rotating Earth would affect where the bullet lands. Surely if the angular velocity of the bullet stays the same it will land in the same place it was fired?

The Earth's surface has an angular velocity of ω=7.3x10^-5rads-1 and a horizontal velocity of 465ms-1 at the equator (for a stationary observer outside of the earth). If a bullet is taken to the top of a very high building it will have a larger horizontal velocity because the radius of the circular path it is traveling will be larger.

However when the bullet is dropped it's horizontal velocity will decrease and approach 465ms-1 as it approaches the Earth's surface. This is because it's angular velocity will remain constant. Therefore the bullet will stay above the same spot it was fired.
v=rw
As r decreases, v must also decrease so that w stays constant.

The reason I think w (angular velocity) will stay constant is because there is no torque acting on the bullet.
angular acceleration = T/I
w1=w0 +(T/I)t
if T=0 angular velocity will not change

I have created an image to explain what I am talking about
http://img824.imageshack.us/img824/4900/bdss.png

Can someone please tell me where I am going wrong? I believe you that the bullet would fall behind its firing point but I can't see why. According to my math it will fall where it was fired.

Thanks

add in frame dragging ... similar to the gps adjustment for time.
The curve of the trajectory would need to be considered by a receiver to shove the bullet back down the barrel. Factor in the rotation of the barrel to the incoming projectile by some amount by a function of the velocity of the incoming mass.