# Being shot by a bullet in space. Damaging? Why/why not?

1. Apr 15, 2010

### rflxion

Hi guys,

Just a simple question about space and bullets. I realise this topic comes up a lot, but specifically the only answer I can find is that of the shooter moving back with an equal and opposite force to the bullet, by the conservation of momentum.

Rather, (and ignoring for a moment the fact that the gun wouldn't fire due to the lack of oxygen) IF you were to fire a gun from space, at another person also in space, would the target get hurt/damaged?

Myself and a friend were having this debate, and to me it seems as though the net result would be the target being moved in the same direction as the bullet, but overall coming away unscathed because of the absence of any external opposing forces.

Is there something obvious I'm missing here, or would that, in fact, be the case?

2. Apr 15, 2010

### Staff: Mentor

First, I believe the gun will fire in space. The round does not use O2 from outside the casing.

Second, Force and Momentum are different things. You are correct that momentum is conserved. Do you know the definition of momentum, in terms of mass and velocity? You can find it on wikipedia.org or other similar websites.

Using the momentum equation, and the fact that the total momentum of the bullet and the target person will be the same before and after impact, calculate the final velocity of the combined person+bullet. Does that tell you something about the damage?

3. Apr 15, 2010

### mate0

yes, the gun will fire in space because gunpowder contains its own oxidizer, and yes it will still damage the target. If the bullet does not penetrate all the way through the target all of the momentum will be transferred, but this transfer will still damage tissues.

the best comparison to this is meteorite collisions with earth. as we know these do a large amount of damage to the planets surface.

4. Apr 15, 2010

### Staff: Mentor

When doing these kinds of problems you need to consider conservation of momentum and energy. Momentum will be conserved. This may result in a change of kinetic energy, all of the kinetic energy that was lost will go into deformation energy and heat, which is what causes the damage. So the damage is related to the change in kinetic energy.

5. Apr 15, 2010

### Noesis

It might also help to intuitively consider in what manner would the unlucky target avoid damage. Only if the bullet fully stopped (before impact) or if the target were moving equal to or quicker than the bullet could this occur.

Now consider that the target is originally stationary, while the bullet is zipping through space without air resistance, and then consider how, if at all, the above two non-damaging scenarios are possible.

Last edited: Apr 15, 2010
6. Apr 15, 2010

### sophiecentaur

The force multiplied by the time it is acting is called the Impulse. The impulse is the amount by which the momentum of the bullet and victim changes (same amount of force - just opposite direction- and same length of time for each). This means the bullet's loss in momentum is the same as the victim's gain in momentum.
The same velocity change will be experienced, in fact, by the victim as that of the person who fired the shot (assuming they are both of similar mass and both are 'detatched'.

But this is an inelastic collision; Kinetic Energy is not conserved. The final velocity of the victim (plus embedded bullet) will be a very small fraction of the bullet's initial velocity. The Kinetic Energy of the combination will be a lot less than the KE of the bullet so the balance of the Energy will have been transferred into damage to the victim. It will make very little difference whether the victim is in space or standing on the ground. Most of the damage will have been done before the victim has reached his final velocity. It's not like they show in the films, where people get knocked over by bullets!

7. Apr 15, 2010

### SpectraCat

Minor point, but people get "knocked over" by bullets in real life too. It's called stopping power, and many different firearms were developed with that in mind. Back in the day, an 0.50 inch musket ball would certainly knock you over, and the 0.45 caliber pistols were initially developed to knock over tribal warriors in the Phillipines during the Moro rebellion. They would get all hopped up on stimulants, wrap themselves in thick hempen rope, and attack the US soldiers by running at them with machetes. The 0.38 caliber side-arms of the time simply didn't deliver enough impulse to knock them over when they struck the rope, and they would shrug off any indirect hits, get in close, and hack the soldiers up with machetes. The 0.45 caliber rounds delivered enough stopping power to knock them over, and (much like the kid brother in "A Christmas Story") they couldn't get up easily because they were wrapped up in rope.

8. Apr 16, 2010

### sophiecentaur

I did some simple sums, using momentum conservation.
Assume, first, that it happens in space. If a 80kg body is hit by a 0.2 kg 45 bullet (about the largest used), traveling at 300m/s and the bullet doesn't travel through the body, the two will move off at 0.6m/s or 1.6mph. This would be the speed change of a warrior, running at at least 2m/s. It would 'slow him down' but not 'stop' him, in the simple mechanical sense. He would feel a bit unwell after the impact and probably fall down, of course, due to his injuries. The rope must have acted as a very effective padding against the 38 rounds but was not sufficient to deal with the 45 shots. The fact that the 45 rounds worked does not constitute an explanation of why.
The same thing applies to the guy firing the gun - if the 'recoil' momentum (which you can't eliminate except by anchoring yourself to the ground) were any more, the gun would not be usable. This would be a real problem with continuous fire!
The Cinema shots of people being thrown backwards are just not real. I saw a sequence on some trashy Science TV prog in which they hung a pig(?) carcass on a rope and fired various guns at it. The result was never more than a small movement backwards. That was reality - not a special effect.

Last edited: Apr 16, 2010
9. Apr 16, 2010

### mate0

no spectra bullets will not knock you over. when they said the .45 could knock over tribal warriors they just meant that it killed them very quickly. because of newtons 3rd law if they person being shot is knocked back the person firing the gun will be sent back an equal distance.

10. Apr 16, 2010

### SpectraCat

Please note here that I am not saying Hollywood gets it right ... I was only saying that bullet impacts could knock over a human in real life as well. I believe this is more a question of imparting torque to a person's body who is standing/running on solid ground in a gravitational field (and thus not so applicable to the OP's question). I do agree that a bullet could not lift a person off their feet and propel them through a wall, as is sometimes shown in the movies.

However, it looks like I will have to do some calculations of my own ... I am sure the historical anecdotes are accurate, but sophiecentaur is correct, I don't really know why the .45 was more effective. Perhaps it just caused the warriors to be knocked off balance and fell down. I will also look up the historical references with regards to large caliber musket balls, I have definitely seen eyewitness accounts from revolutionary war era soldiers who recounted their comrades being knocked off their feet by bullet impacts.

One thing I can perhaps shed some light on is the issue of why recoil doesn't knock the shooter backwards. Several factors contribute to the physics there. First, the chamber gases are vented, allowing much of the recoil force to be directed along vectors normal to the bullet flight path. Second, the shooter usually has a firm stance and is expecting the recoil, allowing them to direct the recoil force upwards in a (hopefully) controlled manner.

11. Apr 16, 2010

### sophiecentaur

Whatever you do with gases, you cannot change the Momentum situation due to the bullet. One thing you can do is to make sure that the force on the shooter is spread over as long a time as possible -so it can be less for the same value of impulse. Some resilience between the barrel and your shoulder can achieve this.
There will be a finite amount of momentum in the emerging gases. If they can be directed sideways (equal and opposite) instead of forwards then there is less momentum imparted to the shooter.

12. Apr 16, 2010

### mate0

actually in a handgun the majority of the gasses are vented forward, which does nothing to reduce recoil, there are no handguns in which the gases go anywhere but forwards. in some rifles gases are vented to the sides in what is called a flash guard, this is to keep the muzzle flash from temporarily blinding the shooter, and has little effect on recoil.
The only gun that could possibly knock someone over is the .50 cal rifle, which fires an 800 grain bullet at 2900fps with force of 14,895 ft lbs(20195 J).

13. Apr 16, 2010

### sophiecentaur

"force of 14,895 ft lbs(20195 J). "
That would be the Kinetic Energy, not force.
I can't be naffed to work out what those other units mean. I only speak SI, as a rule.

14. Apr 16, 2010

### cyberfish99

Just to point out, many modern guns rely on gravity in some form or another to reload or eject a cartridge, so there would have to be compensation for that or the gun would jam after only one or two shots

15. Apr 16, 2010

### sophiecentaur

You mean you can't shoot whilst you're laying on your back? That seems a bit of a limitation.

16. Apr 16, 2010

### AndyK

Going back to the original post, in this collision the bullet may enter the 80kg body and travel with it or just pass through it. In this case the kinetic energy of the bullet is being applied to a very small area on the human body, specifically its tip. The pressure is applied to the skin and organs and it does not change because it is happening in a vacuum, laws of Physics are the same everywhere. If you thought of them bouncing away in opposite directions, or the bullet dragging the person without any damage also consider that the bullet has a mass which is a whole lot less than the human being but most of its energy is due its speed not it's mass. At the 1/2mv^2 we can see kinetic energy depends more on speed than mass however the effect it applies depends on the surface area that the force is being applied to, which in this case is less than 0.1mm^2 (tip of the bullet, I am not sure of the exact size, but it is pretty small.) If what you said was correct a sharp knife wouldn't cut anything that is not being held in space by an external force. And you would be right if you tried it using a kitchen knife and a leaf of lettuce. However if you took a sharp knife and wield it against a water melon floating in space with enough velocity, you just committed to the murder of a melon. :)

17. Apr 16, 2010

### pantaz

I can't think of any modern firearm requiring gravity to assist its function. It's essentially all springs and levers. The only thing that comes to mind is when inserting/removing shells from a revolver's cylinder, but even in that case, gravity assist is optional.

18. Apr 16, 2010

### ruko

I fired my modern 9 MM Luger autoloader several times holding it upside down and it worked just fine. What modern fire arm uses gravity to eject and reload a cartridge?

19. Apr 17, 2010