Just wondering, as oft I do, what would happen if an Astronaut fired a .45 acp pistol in space? Would the bullet go 850fps in one direction and the Astronaut 850fps in the other direction or would a gun even shoot in space? Since there is no resistance to bullet travel would a gun that shot a bullet 850fps here on Earth shoot the same bullet at a much greater velocity?I just had the idea that maybe shooting in space would be a way to get around up there. Just a dumb idea but heck who knows. I have never heard of it being tried. Y'all know how us Texicans are. Always wondering about this or that.
Shooting in Space
- Thread starter Lucky Jim
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Dregan
Member
Neither
The energy of the charge would be equally split by both.
But each wouldn't go flying off at the same speed. Since the bullet is much smaller than the astronaut, and has much less mass, the bullet would go faster. The same amoung of force would be applied to the astronaut, but he would move much slower. The force expended to each would be the same, although their speeds would be radically different becasue of the massive difference of mass between the two.
This is of couse assuming that someone can get a cartridge to go bang in space. No air. No O2 for the gunpowder to burn. Wouldn't even hear a click, b/c no air for the sound to move through.
2 cents.
-D.
The energy of the charge would be equally split by both.
But each wouldn't go flying off at the same speed. Since the bullet is much smaller than the astronaut, and has much less mass, the bullet would go faster. The same amoung of force would be applied to the astronaut, but he would move much slower. The force expended to each would be the same, although their speeds would be radically different becasue of the massive difference of mass between the two.
This is of couse assuming that someone can get a cartridge to go bang in space. No air. No O2 for the gunpowder to burn. Wouldn't even hear a click, b/c no air for the sound to move through.
2 cents.
-D.
Since gunpowder makes it's own oxygen when it burns it would fire in space.
As for recoil propeling the astronaut in the opposite direction, it would, but to a much smaller degree. You just have to figure out the differnces in mass between the bullet weight and the astronaut weight and do the calculus.
As for recoil propeling the astronaut in the opposite direction, it would, but to a much smaller degree. You just have to figure out the differnces in mass between the bullet weight and the astronaut weight and do the calculus.
Dregan
Member
didn't know that gunpoder made oxygen as it burns... gonna have to look into that...
the formula's easy = force (newtons of energy released by the gunpowder) times mass = acceleration. F x M = A.
You need to know exactly how many newtons are released by the gunpowder, and divide that number by two. For the bullet, it's easy. You'd express it like this F/2 X M (bullet) = A.
The astronaut is a little trickier, because you also have to figure in the recoil action of the gun (i'm going to assume that he had his arms locked rigid for the shot, because figuring his strength to absorb kinetic energy in his arm and shoulder is impossible without using a real person and doing some real tests) - Anyway. If he's using a 18 pound recoil spring, that spring will absorb the square of that amount of energy (during compression and release) and that figure needs to be removed from the force actually propelling the astronaut backwards. In this case, "r" is the strength of the recoil spring, and don't forget to convert to metric! [F/2 - (F/2r^2)] X M = A.
Easy!
the formula's easy = force (newtons of energy released by the gunpowder) times mass = acceleration. F x M = A.
You need to know exactly how many newtons are released by the gunpowder, and divide that number by two. For the bullet, it's easy. You'd express it like this F/2 X M (bullet) = A.
The astronaut is a little trickier, because you also have to figure in the recoil action of the gun (i'm going to assume that he had his arms locked rigid for the shot, because figuring his strength to absorb kinetic energy in his arm and shoulder is impossible without using a real person and doing some real tests) - Anyway. If he's using a 18 pound recoil spring, that spring will absorb the square of that amount of energy (during compression and release) and that figure needs to be removed from the force actually propelling the astronaut backwards. In this case, "r" is the strength of the recoil spring, and don't forget to convert to metric! [F/2 - (F/2r^2)] X M = A.
Easy!
Back in an old shooting magazine they were shooting a 1911 .45ACP in a swimming pool underwater. They went on to explain that as long as there was no air bubble in the barrel of the gun when it was fired that it would not blow up. Cool pics of the bullets going about 7 ft in the water. Went on to explain that gunpowder makes it's own oxygen as it burns.
Dregan is correct that the bullet will go much faster than the astronaut due to large diff in mass. Bullet would still go about 850 fps just as in the earth's atmosphere (actually a tiny amount faster due to no air to be pushed out of barrel). Big difference is that it would continue to go 850 fps as long as it remains in space. Lots of different definitions of "space" though. Near earth, there is still a fairly large amount of gases. Interplanetary, less. Interstellar, lots less. Intergalactic, practically none. So depending where it is fired, it can slow down a little to not at all. Let's not even think about how gravity will affect it.
kudo is sort of correct about why the round will fire in space. The gunpowder (or smokeless powder) doesn't make it's own oxygen, but the oxygen is contained in the chemicals making up the powder. No outside oxygen is required.
kudo is sort of correct about why the round will fire in space. The gunpowder (or smokeless powder) doesn't make it's own oxygen, but the oxygen is contained in the chemicals making up the powder. No outside oxygen is required.
rrader
member
To clear up one thing:
Newtons Law; Force = Mass* Acceleration
so: A = F/M
As far as the motion of the bullet and astronaut:
This is assumed to be a system where energy is conserved, i.e., is the same at all times, so the initial potential energy stored in the gunpowder is converted perfectly to an equal amount of kinetic energy of the bodies after the gun is fired.
As for the motion of the bodies before and after the gun is fired, making the usual assumptions that the system exists in perfect isolation with no net energy input or friction, and that the bodies travel in pure linear motion:
Start with: momentum = mass * velocity
Momentum initial = momentum final = zero with respect to a local frame of reference
Let:
Mb = mass bullet
Ma = mass Astronaut
Vbi = initial velocity of bullet
Vbf = final velocity of bullet
Vai = initial velocity of astronaut
Vaf = final velocity of astronaut
so we get:
MbVbi + MaVai = MbVbf + MaVaf = 0 total momentum = 0 always
or simply:
MbVbf + MaVaf = 0
MbVbf = -MaVaf
So the bullet and astronaut have equal and opposite momenta after gun is fired, so the less massive bullet goes away with much higher velocity than the more massive astronaut, and in the opposite direction.
(This neglects the momentum of the gas produced)
Newtons Law; Force = Mass* Acceleration
so: A = F/M
As far as the motion of the bullet and astronaut:
This is assumed to be a system where energy is conserved, i.e., is the same at all times, so the initial potential energy stored in the gunpowder is converted perfectly to an equal amount of kinetic energy of the bodies after the gun is fired.
As for the motion of the bodies before and after the gun is fired, making the usual assumptions that the system exists in perfect isolation with no net energy input or friction, and that the bodies travel in pure linear motion:
Start with: momentum = mass * velocity
Momentum initial = momentum final = zero with respect to a local frame of reference
Let:
Mb = mass bullet
Ma = mass Astronaut
Vbi = initial velocity of bullet
Vbf = final velocity of bullet
Vai = initial velocity of astronaut
Vaf = final velocity of astronaut
so we get:
MbVbi + MaVai = MbVbf + MaVaf = 0 total momentum = 0 always
or simply:
MbVbf + MaVaf = 0
MbVbf = -MaVaf
So the bullet and astronaut have equal and opposite momenta after gun is fired, so the less massive bullet goes away with much higher velocity than the more massive astronaut, and in the opposite direction.
(This neglects the momentum of the gas produced)
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Mike Irwin
Member
Oxygen is NOT required for gunpower to ignite and burn.
Think about it.
Given the ability to push a bullet from zero to several hundred, or even thousand, feet per second, you'd need a LOT more oxygen than you'd find in an enclosed cartridge case.
Black powder contains 3 ingredients, Sulphur, which is an element, Charcoal, which is carbon with some oxygen, and Potassium Nitrate, Kn03.
Potassium Nitrate in gun powder is called the oxidizer -- it provides the free oxygen needed for the rest of the reaction to take place.
To provide enough oxygen, the formula for black powder contains roughly 75% Potassium Nitrate.
During combustion, the sulphur and carbon in the charcoal act as fuels, and the KnO3 gives up its oxygen to fuel the process.
The effect is roughly similar in nitrocellulose, although the chemical formulae are much different.
Mal
"Big difference is that it would continue to go 850 fps as long as it remains in space."
Wouldn't a bullet fired in space be subject to very very slow, very long term gravitational slowing?
Think about it.
Given the ability to push a bullet from zero to several hundred, or even thousand, feet per second, you'd need a LOT more oxygen than you'd find in an enclosed cartridge case.
Black powder contains 3 ingredients, Sulphur, which is an element, Charcoal, which is carbon with some oxygen, and Potassium Nitrate, Kn03.
Potassium Nitrate in gun powder is called the oxidizer -- it provides the free oxygen needed for the rest of the reaction to take place.
To provide enough oxygen, the formula for black powder contains roughly 75% Potassium Nitrate.
During combustion, the sulphur and carbon in the charcoal act as fuels, and the KnO3 gives up its oxygen to fuel the process.
The effect is roughly similar in nitrocellulose, although the chemical formulae are much different.
Mal
"Big difference is that it would continue to go 850 fps as long as it remains in space."
Wouldn't a bullet fired in space be subject to very very slow, very long term gravitational slowing?
rrader
member
Mike Irwin
Maybe it had escape velocity from our solar system, in which case it would begin to gain velocity by gravitational attraction to the next nearest star system.
Maybe it had escape velocity from our solar system, in which case it would begin to gain velocity by gravitational attraction to the next nearest star system.
C.R.Sam
Moderator Emeritus
Mal got it covered.
Robert Goddard worked it out bout 90 years ago, even makin a really big bell jar so he could fire a handgun in a vacuum. He measured velocity, energy, recoil effect etc.
Gravitational deceleration ....not necessairily...could be gravitational acceleration ....difference being the the direction fired in relation to the most influential mass in the area.
Sam
Robert Goddard worked it out bout 90 years ago, even makin a really big bell jar so he could fire a handgun in a vacuum. He measured velocity, energy, recoil effect etc.
Gravitational deceleration ....not necessairily...could be gravitational acceleration ....difference being the the direction fired in relation to the most influential mass in the area.
Sam
In outer space there is no up, there is no down, there is no left, there is no right. Off the bullet would go in a straight line until it is pulled into a gravitational path of a larger object. And wouldn't the bullet be traveling slo compared to the speed at which the Shuttle and other crafts soar through space?
C.R.Sam
Moderator Emeritus
Again, depends on direction of fire in relation to the shuttle or other craft it is fired from.
Sam
Moparmike
Member
Fly me to the moon, let me play among the stars...Oh, sorry.
As the bullet travelled out of our solar system, it would be slowed down by the gravity of our sun assuming that no planets' gravity wells affected its path. What would be really cool is the top velocity of the bullet on its way to a black hole.
Or the velocity of an unladen swallow on its way to a black hole.
As the bullet travelled out of our solar system, it would be slowed down by the gravity of our sun assuming that no planets' gravity wells affected its path. What would be really cool is the top velocity of the bullet on its way to a black hole.
Or the velocity of an unladen swallow on its way to a black hole.
Mike Irwin
Member
Just shoot a .45 ACP at that black hole.
That'll stop it! A .45 will stop ANYTHING!
That'll stop it! A .45 will stop ANYTHING!
Orthonym
Member
MOMENTUM IS CONSERVED!
E= 1/2 Mv^2!
Momentum is a vector, energy a scalar.
No more stupid questions, please?
I thought they taught this in the 8th grade.
There was an excellent article on this subject in Astounding in about 1940.
Oh, and F=Ma.
E= 1/2 Mv^2!
Momentum is a vector, energy a scalar.
No more stupid questions, please?
I thought they taught this in the 8th grade.
There was an excellent article on this subject in Astounding in about 1940.
Oh, and F=Ma.
Wow. That is way to deep for me. Think I will go back to work with the old shovel. I know how fast it will dig here on Earth in 105 degree heat or myself X 105 degree heat X shovel X 8 hours = pooped. Thanks for clearing that thought up for me guys.
Mike:
On escape velocities: A .45 at 850 FPS will not be able to leave the earth's gravitational field. The escape vel required is around 37,000 FPS. The escape velocity of the Solar System is much much higher. I don't think Man has ever built a gun that could fire a bullet that would escape from the Solar System. The energy required would be astronomical, so to speak.
Yes. No matter where in "space" (see my list of spaces) you are, gravity will have an effect on the velocity and trajectory of the bullet. The effect might not be negative though, it could increase the velocity just as easily as it decreases it. That's why I said, "Let's not even think about how gravity will affect it."
On escape velocities: A .45 at 850 FPS will not be able to leave the earth's gravitational field. The escape vel required is around 37,000 FPS. The escape velocity of the Solar System is much much higher. I don't think Man has ever built a gun that could fire a bullet that would escape from the Solar System. The energy required would be astronomical, so to speak.
rrader
member
Mal H:
That would depend on the distance the gun was from earth when the bullet was fired.
For escape velocity the bullet's kenetic energy must be at least greater than its gravitational potential energy at all times:
Let:
Mb = mass of bullet
Me = mass of earth
Ve = escape velocity
G = universal gravitational constant
R = distance from earth's center
Kenetic Energy must be > Gravitational Porential Energy
1/2Mb(square of Ve) > GMeMb/(R)
so Ve > square root of (2GMe/R)
so as R increases Ve decreases.
That would depend on the distance the gun was from earth when the bullet was fired.
For escape velocity the bullet's kenetic energy must be at least greater than its gravitational potential energy at all times:
Let:
Mb = mass of bullet
Me = mass of earth
Ve = escape velocity
G = universal gravitational constant
R = distance from earth's center
Kenetic Energy must be > Gravitational Porential Energy
1/2Mb(square of Ve) > GMeMb/(R)
so Ve > square root of (2GMe/R)
so as R increases Ve decreases.
bedlamite
Member
Physics 101
Orthonym is right, momentum is conserved. Disregarding any gravitational pull and the ejected case, the sum of mass times velocity before will equal the sum of mass times velocity after, and this is vector addition.
MI x VI = (MB x VB) + (MA x VA)
Assuming initial velocity is 0, the astronaut weighs 250 lbs with gear, the bullet is 230 gr, and the bullets velocity is 850 fps we have:
0 = (0.00102 slugs x 850 fps) + (7.764 slugs x VA)
VA = -.112 fps
This also assumes the barrel is lined up exactly with the COM of the astronaut. Not exactly the best method of propulsion.
Orthonym is right, momentum is conserved. Disregarding any gravitational pull and the ejected case, the sum of mass times velocity before will equal the sum of mass times velocity after, and this is vector addition.
MI x VI = (MB x VB) + (MA x VA)
Assuming initial velocity is 0, the astronaut weighs 250 lbs with gear, the bullet is 230 gr, and the bullets velocity is 850 fps we have:
0 = (0.00102 slugs x 850 fps) + (7.764 slugs x VA)
VA = -.112 fps
This also assumes the barrel is lined up exactly with the COM of the astronaut. Not exactly the best method of propulsion.
rrader
member
Bedlamite:
I stated several times that momentum doesn't change here (remains at zero), i.e., is conserved. :banghead:
The formulation I gave assumed linear motion, so that it's unnecessary to use vector addition to resolve the motion ot two bodies relative to one another, i.e., along a straight line. A simple + or - suffices to account for directionality.
Systems may be conservative with respect to scalar quantities like energy, I mentioned that to obviate use of mass energy equivalency which seems kind of unnecessary given these very slow speeds and heavy objects
I stated several times that momentum doesn't change here (remains at zero), i.e., is conserved. :banghead:
The formulation I gave assumed linear motion, so that it's unnecessary to use vector addition to resolve the motion ot two bodies relative to one another, i.e., along a straight line. A simple + or - suffices to account for directionality.
Systems may be conservative with respect to scalar quantities like energy, I mentioned that to obviate use of mass energy equivalency which seems kind of unnecessary given these very slow speeds and heavy objects
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