Is Newton's Cradle True? Investigating Bullet Experiments in Physics

[been_jamin]

i have a question. you can call me stupid if you want but i need to confirm. i remember talking about an experiment in physics where if you shoot a bullet perfectly level from a gun and you just drop a bullet at the same time would both of the bullets hit the ground at the same time? i'd swear that was true but i need proof. thanks.

 

[gregmead]

yes it is true, because the velocity (of anything moving in 2D) is a vector with components sideways and upwards (x and y axes).

the bullet if it has been fired perfectly level will start off with zero velocity in the y-axis, which is the same velocity that a bullet you just drop from the same height will have.

Becasue they both start from the same height they both fall at the same velocity (accelerate over the same distance), so they both hit the ground at once.

The bullet will obviously hit the ground a long way away though, becuase it also has velocity in the x-axis.

Hope this helps, and I've written it clearly :)

 

[Andrew Mason]

i have a question. you can call me stupid if you want but i need to confirm. i remember talking about an experiment in physics where if you shoot a bullet perfectly level from a gun and you just drop a bullet at the same time would both of the bullets hit the ground at the same time? i'd swear that was true but i need proof. thanks.

Rifled bullets resist lateral aerodynamic forces very well so there is virtually 0 lift. So they fall at the same rate as a bullet dropped from rest. A non-rifled bullet, such as a muzzleloader round ball, will behave eratically due to aerodynamic forces and could either fall faster or slower than the same ball dropped from rest.

AM

 

[DaveC426913]

Rifled bullets or no, in principle, and on average, they will still fall at the same rate (the drag forces ought to average out).

The point here, is that the horizontal forces (detonation) acting on the bullet and the vertical forces (gravity) acting on the bullet are completely independent of each other.

So, yes.

 

[Andrew Mason]

Rifled bullets or no, in principle, and on average, they will still fall at the same rate (the drag forces ought to average out).

The point here, is that the horizontal forces (detonation) acting on the bullet and the vertical forces (gravity) acting on the bullet are completely independent of each other.

Well, it depends on the shape of the projectile. A golf ball, for example, generates substantial lift, not just drag. A muzzleloader round can also generate lift. A properly rifled bullet (ie. a spinning bullet) is stable and flies nose-first generating virtually no lift. However, if it starts tumbling it will move in a spiral (helical) path with ever increasing radius of spiral, so it does not drop at a steady rate.

On the moon, they would all fall at the same rate (although the fired bullet would have farther to fall due to curvature of the moon surface over its much longer trajectory range).

AM

 

[Danger]

On the moon, they would all fall at the same rate (although the fired bullet would have farther to fall due to curvature of the moon surface over its much longer trajectory range).

This brings up a point of curiosity. If you fired it fast enough, at least in the same direction as the planet's rotation, wouldn't it continue straight as the planet 'falls away' beneath it and escape the gravity well?

 

[Andrew Mason]

This brings up a point of curiosity. If you fired it fast enough, at least in the same direction as the planet's rotation, wouldn't it continue straight as the planet 'falls away' beneath it and escape the gravity well?

Direction of rotation wouldn't matter much (it just adds/subtracts to the required velocity. If the moon was all by itself in the solar system, and if you gave it escape velocity (.5mv^2 = GmM/r) it would keep going and escape. But if you gave it anything less, I think it would return and could smack you in the back of the head (if you waited there long enough). To put it into orbit, you would have to have it change direction after launch.

But the moon is not that far from other bodies. If your shot was aimed at the earth, the bullet would just have to go to a point between Earth and moon to escape the moon (and be caught by Earth gravity).

AM

 

[Huckleberry]

This brings up a point of curiosity. If you fired it fast enough, at least in the same direction as the planet's rotation, wouldn't it continue straight as the planet 'falls away' beneath it and escape the gravity well?

Hmm, the fastest bullets I know of have a velocity of about one mile per second. Escape velocity from the surface of the Earth is about 7 miles per second. If you could shoot a bullet that fast straight up then it would escape the Earth, if anything remained of the bullet at all.

The spinning motion of a bullet would reduce the lateral shifting of the wind. The high density of lead compared to air helps reduce the effects of air resistance in any direction. I think the rotation of the Earth would have almost no effect. The rotation of the Earth would be added to the velocity of the bullet because it was fired by a person who was on the Earth's surface.

The curvature of the Earth would have some small effect. If a bullet that travels 1 mile/second is fired level with the ground then it will reach the level of the ground from the point it was fired in about half a second. Since it will be half a mile away by that time the curvature of the Earth for half a mile will need to be included. Anyone know what that would average?

 

[krab]

Rifled bullets or no, in principle, and on average, they will still fall at the same rate (the drag forces ought to average out).

The drag forces won't average out because the flow of air around the bullet is turbulent so the air resistance is proportional to v^2. Your statement would be correct if the air resistance were proportional to v. The dropped bullet hits the ground first.

 

[DaveC426913]

Well, it depends on the shape of the projectile. A golf ball, for example, generates substantial lift, not just drag.

True, but that is due to the spin (imparted by the club) altering the aerodynamics. If you used a club that spun the ball the other way, you'd get a 'grounder'.

However, if it starts tumbling it will move in a spiral (helical) path with ever increasing radius of spiral, so it does not drop at a steady rate.

Also true, which is why I said 'on average'. If you did the test 100 times, 50 would land sooner and 50 would land later. Effectively, the tumbling is contaminating the experiment (designed to determine whether a 0 verlocity bullet fell sooner than a fired bullet).

This brings up a point of curiosity. If you fired it fast enough, at least in the same direction as the planet's rotation, wouldn't it continue straight as the planet 'falls away' beneath it and escape the gravity well?

Absolutely! And brilliant deduction I might say. Exactly how Newton conjectured the possibilties of artificial satellites! Look at http://www.space.gc.ca/asc/eng/educators/resources/orbital/earth_orbit.asp .

The drag forces won't average out because the flow of air around the bullet is turbulent so the air resistance is proportional to v^2. Your statement would be correct if the air resistance were proportional to v. The dropped bullet hits the ground first.

Please clarify. Drag forces, big or small, will not - on average (say, over 100 shots) - affect the rate at which the bullet drops. Why would the dropped bullet hit the ground sooner?
.

 

[Huckleberry]

Absolutely! And brilliant deduction I might say. Exactly how Newton conjectured the possibilties of artificial satellites!

I certainly believe that the bullet would curve around the Earth as it fell. If air resistance were eliminated and the bullet had enough velocity it would orbit the Earth. I think this is because the Earth is curved and not because the Earth is rotating. It wouldn't matter if you fired east or west.(besides any effects wind caused by the Earth's rotation would have) If I'm wrong then let me know.

 

[DaveC426913]

I certainly believe that the bullet would curve around the Earth as it fell. If air resistance were eliminated and the bullet had enough velocity it would orbit the Earth. I think this is because the Earth is curved and not because the Earth is rotating. It wouldn't matter if you fired east or west.(besides any effects wind caused by the Earth's rotation would have) If I'm wrong then let me know.

You are 100% correct. That's what orbital motion is. Satellites are all trying to fall towards the Earth. But they are moving 'horizontally' so fast that the Earth falls away at just the right rate so that the satellite misses it.



(A bit of detail.) You are correct that you could fire East or West (or North or South in fact). But if you fire East, you don't need as big a gun, since you start off with (and thus get for free) the rotation of the Earth (1000mph at the equator) If you fire West, you'll need a gun that can fire 2000mph faster. This is why all rockets from Canaveral and elsewhere rise Eastward.


Note that you are still correct - the speed of the orbit is independent of the direction in which the Earth is rotating underneath it - I am only talking about how much effort it takes to get it to the correct orbital speed.

 

[Huckleberry]

Okay, I'm just typing out loud now to get a better understanding of this.

From the northern hemisphere the Earth rotates in a counterclockwise direction. It turns towards the east. If the Earth were in a vacuum and a bullet were fired east then it would appear to move less distance because the point at which it was fired would approach it as the Earth rotated. If a bullet were fired west then it would travel more distance on the sphere because the point from which it was fired is moving further away?

Or would the velocity of the sphere's rotation not be important? It would be added to the velocity of the bullet, but it would also be added to the velocity of the point from which the bullet was fired. It would be as if the sphere were not rotating at all.

I'm leaning towards the second option, but to be honest I've rather confused myself now. Can someone explain this better?

 

[Danger]

Man, it's great how thorough the responses are here. Unfortunately, I don't think that I expressed myself clearly enough the first time. I'll take another shot at it (pardon the expression), and if it's still obscure I'll post a diagram. What I had in mind is:
Initially, let's say that instead of a gun, you have an impossibly long solid rod. If Venus is right on the horizon and you poke it with the rod, the end of the rod is outside of Earth's gravitational field. If Venus is rising at the time, the rod will alter it's angle relative to Earth and rise with it. If it's setting, the rod will follow and hit the Earth.
Now replace the rod with a laser. If you fire a pulse at Venus on the horizon, it will be on its way and hit where Venus was when it was fired.
Lastly, replace the laser with a very high-velocity gun (say 90% of c). Would it not hit somewhere even farther away from Venus' current position, but still outside of Earth's influence? On the assumption that Venus is setting when it's fired, at some point on the lowering speed scale it would go into orbit, and even lower on the scale it would hit somewhere on Earth. If Venus is rising instead, then...
I just saw where I screwed up the rotation part of it. I was thinking of the trajectory as a solid line that follows the target, and of course it isn't. Anyhow, the escape velocity question is still what I was wondering about.

 

[krab]

Please clarify. Drag forces, big or small, will not - on average (say, over 100 shots) - affect the rate at which the bullet drops. Why would the dropped bullet hit the ground sooner?

Let's say x is horizontal, z is vertical, up being positive. Then
[tex]\vec{F}=m{d\vec{v}\over dt}=-mg\vec{k}-bv^2\vec{u}[/tex]
where [itex]\vec{u}[/itex] is the unit vector in the velocity direction and [itex]v=\sqrt{v_x^2+v_z^2}[/itex]. We care about the vertical motion:
[tex]m{dv_z\over dt}=-mg-b\sqrt{v_x^2+v_z^2}\,v_z[/tex]
In the dropped case, v_x is zero so there is less air resistance on v_z than in the fired case where v_x is very large. QED

 

[Huckleberry]

I wish I had a head for math. It would really help in cases like this. I'm interested in learning how the drag of a bullet would reduce its downward velocity, but I don't understand the meaning of even these relatively simple mathematics. I could do the problems and come up with a solution and still not understand why it works. What creates the drag? What makes it different than the dropped bullet?

Here's an interesting thought that might explain a rise in the altitude of a bullet after it leaves the barrell of a gun. The position from which the gun is fired is on a rotating sphere. It's motion is an arc that forms a circle. The bullet is moving in an arc that is less curved initially because of its greater velocity relative to the firing position. From the point of view of a person firing the gun their arc curves more than the bullets arc. The difference in the arcs of the bullet and the firing position could appear as a rise in the altitude of the bullet. This would be most pronounced when the bullet is at its highest forward velocity and lowest downward velocity, the moment it exits the barrell. Would this work? In any direction compared to the rotation of the sphere?

In any case, the fired bullet would hit after the dropped bullet because of the curvature of the surface of the sphere. It would have a few more inches or feet to drop depending on the velocity of the bullet, radius of the sphere and the gravitational force.

 

[Integral]

I hate to dampen this party, but aren't you all way over complicating the question. Seems to me the simplest most direct answer is yes, with the in a vacuum qualification in place. We must understand that things get a bit more uncertain in air.

 

[DaveC426913]

Okay, I'm just typing out loud now to get a better understanding of this.

From the northern hemisphere the Earth rotates in a counterclockwise direction. It turns towards the east. If the Earth were in a vacuum and a bullet were fired east then it would appear to move less distance because the point at which it was fired would approach it as the Earth rotated. If a bullet were fired west then it would travel more distance on the sphere because the point from which it was fired is moving further away?

Or would the velocity of the sphere's rotation not be important? It would be added to the velocity of the bullet, but it would also be added to the velocity of the point from which the bullet was fired. It would be as if the sphere were not rotating at all.

I'm leaning towards the second option, but to be honest I've rather confused myself now. Can someone explain this better?

Whatever you do, if you fire your bullets from the surface, you need to factor in the rotation of the Earth.

If you want your two bullets orbiting at the same altitude, they will need to be traveling at the same (if opposite) velocities. To do that, you'll fire the Eastward one at speed X-1000mph, and the Westward one at speed X+1000mph. The Earth's rotation will add/subtract, leaving both bullets traveling at speed x (from an objective point of view). From the POV of someone on Earth, the Westward bullet will cross the sky 2000mph faster than the Eastward bullet.

If you do not bother to correct for the Earth's rotation, your Eastward bullet will have an (objective) speed 1000mph faster, while the Westward bullet will have an (objective) speed 1000mph slower, thoiugh from Earth, they will appear to cross the sky in the same times.

However, the Eastward bullet will be in a higher orbit, while the Westward bullet might not even achieve orbital speed and fall back to Earth.

 

[DaveC426913]

I wish I had a head for math. It would really help in cases like this. I'm interested in learning how the drag of a bullet would reduce its downward velocity, but I don't understand the meaning of even these relatively simple mathematics. I could do the problems and come up with a solution and still not understand why it works. What creates the drag? What makes it different than the dropped bullet?

Here's an interesting thought that might explain a rise in the altitude of a bullet after it leaves the barrell of a gun. The position from which the gun is fired is on a rotating sphere. It's motion is an arc that forms a circle. The bullet is moving in an arc that is less curved initially because of its greater velocity relative to the firing position. From the point of view of a person firing the gun their arc curves more than the bullets arc. The difference in the arcs of the bullet and the firing position could appear as a rise in the altitude of the bullet. This would be most pronounced when the bullet is at its highest forward velocity and lowest downward velocity, the moment it exits the barrell. Would this work? In any direction compared to the rotation of the sphere?

In any case, the fired bullet would hit after the dropped bullet because of the curvature of the surface of the sphere. It would have a few more inches or feet to drop depending on the velocity of the bullet, radius of the sphere and the gravitational force.

Ah well, you are adding a completely unnecessary element to this - that of the curvature of the Earth.

The question of a fired bullet vs. a dropped bullet only has to do with their fall under the influence of gravity. It is assumed that the curvature of the Earth is insignificant.

 

[DaveC426913]

I hate to dampen this party, but aren't you all way over complicating the question. Seems to me the simplest most direct answer is yes, with the in a vacuum qualification in place. We must understand that things get a bit more uncertain in air.

Yes, things are more uncertain in air. I am trying to get across that, while air turbulence will add mess up the meaurements (by adding a deviation), it won't bias them up or down.

i.e:

If you fired one hundred bullets and dropped 100 bullets, you would find a theoretically 0 deviation in times for the dropped bullet, while you would find a wide deviation in times for the fired bullets. All the devation would be due to air turbulence; some would hit sooner, and some would hit later. But they would average the same time as the dropped bullets.

 

[Danger]

I don't think that I expressed myself clearly enough the first time. I'll take another shot at it (pardon the expression), and if it's still obscure I'll post a diagram. What I had in mind is:

http://img203.echo.cx/my.php?image=bulletpf1zr.jpg

 

[DaveC426913]

Did you look at the diagram at http://www.space.gc.ca/asc/eng/educators/resources/orbital/earth_orbit.asp ? (The one near the bottom.)

It's pretty much what you've got, except it should have one more dotted line going off at a gentle curve into space.

In your diagram, the line to Venus won't be straight, it will be a gentle curve. You cannot fire a gun directly at Venus and hope to hit it no matter how fast it is (at least a projectile gun i.e. with a bullet that has mass). Even if you spirit Venus and Earth away to intergalactic space, where no other bodies (such as the Sun) will interfere, the object is still (sort of) in orbit, it's just a very loooong one.

(In reality, the curve will flip over and curve the other way (i.e. forming a long S shape), once it nears Venus, since Venus' gravity will become dominant- but that's over-complicating the experiment.)

see diagram

 

[Danger]

Did you look at the diagram at http://www.space.gc.ca/asc/eng/educators/resources/orbital/earth_orbit.asp ? (The one near the bottom.)

Yeah, I did. The reason that I drew my own is that the linked one still showed an elevated gun, whereas what I had in mind was one firing as close to horizontal as you can get on a curved surface. I like your diagram better.
I hadn't bothered considering the effect of solar and other extraterrestrial gravitational effects because I was in fact only concerned with whether or not I was right in believing that even a horizontal firing can result in escape from Earth. Everyone kept talking about pointing it upward, so I wasn't sure that they were following my thoughts. Venus was just used as a reference point as opposed to a specific trajectory goal. I could as likely have said Betelgeuse instead, except that I'm not sure how to spell it.

 

[krab]

If you fired one hundred bullets and dropped 100 bullets, you would find a theoretically 0 deviation in times for the dropped bullet, while you would find a wide deviation in times for the fired bullets. All the devation would be due to air turbulence; some would hit sooner, and some would hit later. But they would average the same time as the dropped bullets.

The last statement is incorrect. Whassa matter Dave? Didn't read my post? Didn't understand it? Don't agree with the air resistance formula? What? It WAS a proof, you know.