If you fired a bullet in a desert...

[BulgarianPride]

I just wanted to point out one fact. Because of a similar problem (calculating the trajectory of a shell fired by the Destroyer and similar war ships), we have the computer.

 

[Joycee Banercheck]

and there were no obstructions for it to hit, would it eventually just make a steady descent to the floor when it loses all its speed?

YES

I wonder then if thats the case, if there's anyone stupid enough to stand at the point the bullet lands, and see if the second bullet does the exact same thing.

YES

Its a slow work day today, my friends.

YES IT WAS

 

[1.618034]

and there were no obstructions for it to hit, would it eventually just make a steady descent to the floor when it loses all its speed? I wonder then if thats the case, if there's anyone stupid enough to stand at the point the bullet lands, and see if the second bullet does the exact same thing.

Its a slow work day today, my friends.

I am going to go with a more in dept answer.

Existing the gun, the bullet has an initial horizontal velocity. As soon as its out, the only forces acting on it are air resistance and gravity. Ignoring air resistance, we are left with gravity which is acting on the bullet vertically.
But what is the effect of gravity on the bullet? All it does is, every second the vertical velocity is increased by 9.8 meters/second, initially this velocity was zero. Gravity has no effect on the horizontal velocity, so assuming no air resistance this is constant throughout the flight of the bullet.
So how to calculate how far it would travel? All you need is the horizontal velocity and the vertical distance to ground at which the gun was fired, and of course the acceleration due to gravity.

In the above if h is distance to ground and is 1.5 meters, and the bullet was fired at 300m/s, it would travel 165.9 meters.

Ok how about a variation on the question...

What speed would the bullet have to come out of the barrel for it to actually continue in orbit and how long would it take to hit you on the back of the head? Assuming there are no obstructions... Or wind resistance...?

 

[BulgarianPride]

I am going to go with a more in dept answer.

Existing the gun, the bullet has an initial horizontal velocity. As soon as its out, the only forces acting on it are air resistance and gravity. Ignoring air resistance, we are left with gravity which is acting on the bullet vertically.
But what is the effect of gravity on the bullet? All it does is, every second the vertical velocity is increased by 9.8 meters/second, initially this velocity was zero. Gravity has no effect on the horizontal velocity, so assuming no air resistance this is constant throughout the flight of the bullet.
So how to calculate how far it would travel? All you need is the horizontal velocity and the vertical distance to ground at which the gun was fired, and of course the acceleration due to gravity.

In the above if h is distance to ground and is 1.5 meters, and the bullet was fired at 300m/s, it would travel 165.9 meters.

Ok how about a variation on the question...

What speed would the bullet have to come out of the barrel for it to actually continue in orbit and how long would it take to hit you on the back of the head? Assuming there are no obstructions... Or wind resistance...?

Interesting...
I will need give a thought or two.

 

[1.618034]

I reckon there are tiny provisos.

A bullet travelling through the air does not always produce exactly equal aerodynamic lift and down force.

The spin imparted on the bullet may help create an imbalance in the pressures over and underneath it. If a bullet is spinning to the right, when the wind comes from that direction, a slightly lower pressure is created above, holding the projectile in the air for slightly longer than would be the case if it had been dropped dead, fired without any spin... and slightly longer still than would be the case if it was spinning 'with' the wind (which would create lower pressure below the projectile).

I mention it, because this so called 'magnus' effect has something to do with the wierd and wonderful things that happen to footballs in mid-air. With bullets, I imagine it to have a larger effect on lateral movement, they probably 'skid' sideways more than they do 'hang' or 'dip'... as they fly 'straight ahead' and don't spin repeatedly on their vertical axis.


<a class="postlink" href="http://en.wikipedia.org/wiki/Magnus_effect" onclick="window.open(this.href);return false;">http://en.wikipedia.org/wiki/Magnus_effect</a>

Balls are 'dynamic', they change profile, and so do bullets (to a much, much smaller degree).

There is also the coreolis effect, the effect of the geometry of the earth and it's spin, as perceived by a viewer spinning with it. the bullet travels in free space whilst the earth, and viewer moves beneath it. imagine standing on halfway out on a rotating turntable... in order to throw a ball to a guy standing parallel to you, you have to aim ahead of, and beyond him. Now imagine your standing on a sphere.... not really relevant with bullets but hey we're doing the whole lot here.

There is probably even a touch of ground effect at play as well, at very low heights, part of the wave-front of air deflected by the bullet travels back off the earth, creating lift.

Even, at the tiniest imaginable level, 'frame dragging', which is the effect of space time itself being pulled out of shape by the body travelling through it, as described by General Relativity. For one thing, the inertia of the air around the bullet increases. More bizarrely, the two bullets may hit the earth within at the same time by your clock, but measured by its own clock, the one that travelled at twice the speed of sound fell in a fractionally shorter time. (I hate relativity, I've probably got that wrong). The effect is so small, don't expect to measure it.

But that's all bollocks anyway; the important bits are demonstrated beautifully here;

<a class="postlink" href="http://www.phy.hk/wiki/englishhtm/ThrowABall.htm" onclick="window.open(this.href);return false;">http://www.phy.hk/wiki/englishhtm/ThrowABall.htm</a>

Ah. Chaos. Where would we be without it?

That applet won't work... Grr.

 

[persvenality cult]

Ok how about a variation on the question...

What speed would the bullet have to come out of the barrel for it to actually continue in orbit and how long would it take to hit you on the back of the head? Assuming there are no obstructions... Or wind resistance...?

Interesting...
I will need give a thought or two.

isn't it around 11km/s? escape velocity, in other words. the usual 300km altitude 'stable' orbits are slower as gravity obeys the law of inverse squares. but as the surface of the earth is what, 5000km from the core, another 300km isn't actually all that big a difference? standard orbits are around 8k/ms I think.<br /><br />-- Sat Nov 06, 2010 1:05 am --<br /><br />

I reckon there are tiny provisos.

A bullet travelling through the air does not always produce exactly equal aerodynamic lift and down force.

The spin imparted on the bullet may help create an imbalance in the pressures over and underneath it. If a bullet is spinning to the right, when the wind comes from that direction, a slightly lower pressure is created above, holding the projectile in the air for slightly longer than would be the case if it had been dropped dead, fired without any spin... and slightly longer still than would be the case if it was spinning 'with' the wind (which would create lower pressure below the projectile).

I mention it, because this so called 'magnus' effect has something to do with the wierd and wonderful things that happen to footballs in mid-air. With bullets, I imagine it to have a larger effect on lateral movement, they probably 'skid' sideways more than they do 'hang' or 'dip'... as they fly 'straight ahead' and don't spin repeatedly on their vertical axis.


<a class="postlink" href="http://en.wikipedia.org/wiki/Magnus_effect" onclick="window.open(this.href);return false;">http://en.wikipedia.org/wiki/Magnus_effect</a>

Balls are 'dynamic', they change profile, and so do bullets (to a much, much smaller degree).

There is also the coreolis effect, the effect of the geometry of the earth and it's spin, as perceived by a viewer spinning with it. the bullet travels in free space whilst the earth, and viewer moves beneath it. imagine standing on halfway out on a rotating turntable... in order to throw a ball to a guy standing parallel to you, you have to aim ahead of, and beyond him. Now imagine your standing on a sphere.... not really relevant with bullets but hey we're doing the whole lot here.

There is probably even a touch of ground effect at play as well, at very low heights, part of the wave-front of air deflected by the bullet travels back off the earth, creating lift.

Even, at the tiniest imaginable level, 'frame dragging', which is the effect of space time itself being pulled out of shape by the body travelling through it, as described by General Relativity. For one thing, the inertia of the air around the bullet increases. More bizarrely, the two bullets may hit the earth within at the same time by your clock, but measured by its own clock, the one that travelled at twice the speed of sound fell in a fractionally shorter time. (I hate relativity, I've probably got that wrong). The effect is so small, don't expect to measure it.

But that's all bollocks anyway; the important bits are demonstrated beautifully here;

<a class="postlink" href="http://www.phy.hk/wiki/englishhtm/ThrowABall.htm" onclick="window.open(this.href);return false;">http://www.phy.hk/wiki/englishhtm/ThrowABall.htm</a>

Ah. Chaos. Where would we be without it?

That applet won't work... Grr.

it's not that special, it's just lets you set the height, attitude and velocity and visualises the parabolas in 'real time'.

 

[BulgarianPride]

Got it.
In order to always stay in orbit, the centripetal force acting on the bullet must the force of gravity.

F=m*v^2/r

where m is the mass of the bullet, v is it's velocity and r is the distance from the center of the earth (approximately 6 378 101.5 meters if i fire from 1.5 meters above the ground). Now for the the bullet to be always in orbit, it must travel at

v=(9.8*6378101.5)^0.5=7906.035 m/s

This is the limit, any faster and it will leave earth for good.

In order for the bullet to comeback and hit me in the back it must travel 2*pi*r meters.

edit: This stuff below is wrong. Everything above sounds right.

so the speed that it must come out is,

v=2*pi*r/0.553 = 72 467 981.25 m/s

Sadly if this bullet is fired from 1.5 meters above the ground , it would leave the earth.
I've probably made a mistake...

 

[yerwot?]

way out of my shallow here but fascinated by ballistics so here goes,how does a round ball change profile? and in the equation where the acceleration is given a value of 0 ,instead of a deceleration value.(brain hurts now).

 

[King Geedorah]

Ok how about a variation on the question...

What speed would the bullet have to come out of the barrel for it to actually continue in orbit and how long would it take to hit you on the back of the head? Assuming there are no obstructions... Or wind resistance...?

The acceleration of a body in circular motion is calculated by a^2/r, the force as a result of gravity by F=GMm/r^2. So a bit of a rejig and v=root(GM/r) where G is the universal gravitational constant, M the mass of the earth and r the distance between the earth and your bullet. M= 3x10^24 I think, G is something silly like 6.02 x 10^23 and r about 7000km? Then again I'm no physicist, it's quite late and I've had a few, god knows why I'm on here doing this.

ha I made v=a Silly error. Can't be bothered changing it.




-- Sat Nov 06, 2010 1:18 am --

Got it.
In order to always stay in orbit, the centripetal force acting on the bullet must the force of gravity.

F=m*v^2/r

where m is the mass of the bullet, v is it's velocity and r is the distance from the center of the earth (approximately 6 378 101.5 meters if i fire from 1.5 meters above the ground). Now for the the bullet to be always in orbit, it must travel at

v=(9.8*6378101.5)^0.5=7906.035 m/s

This is the limit, any faster and it will leave earth for good.

In order for the bullet to comeback and hit me in the back it must travel 2*pi*r meters.
so the speed that it must come out is,

v=2*pi*r/0.553 = 72 467 981.25 m/s

Sadly if this bullet is fired from 1.5 meters above the ground , it would leave the earth.
I've probably made a mistake...

I don't follow that but it may well be right. All I know is orbital speed/distance is only dependent on those two, the mass of the orbiting object is irrelevant.

-- Sat Nov 06, 2010 1:18 am --

way out of my shallow here but fascinated by ballistics so here goes,how does a round ball change profile? and in the equation where the acceleration is given a value of 0 ,instead of a deceleration value.(brain hurts now).

What does it mean for a round ball to change profile?

I might be able to answer the question.

 

[Gelsons Dad]

If u fired the bullet and dropped one at the same time from the same height theyd hit the ground at the same time

Not exactly true.

Both rounds are subject to the same gravitational force but the one that is fired is also accelerating away from the earth at a rate proportional to its velocity and elevation. Even if fired parallel to the surface at the point of discharge this will still equal a certain elevation due to the earths curvature.