Could a .50 calibre bullet reach escape velocity if there were no atmosphere?
No. A .50BMG round has a muzzle velocity of about 1200 m/s. Escape velocity from the Earth is about 11000 m/s, minus about 465 m/s for the Earth’s rotation at the equator.
A tank round is about the maximum chemical muzzle velocity, at about 1800 m/s.
The maximum muzzle velocity of a railgun that has supposedly been tested is about 7000 m/s.
Now, if you mean can a .50BMG bullet (just the lead and copper part) can be accelerated to escape velocity, the answer is a pretty obvious yes, but I think that’s not what you were asking.
I looked at a table of muzzle velocities for various bullets and the highest I saw was about 3000 feet per second. According to Wikipedia, tank guns can have muzzle velocities as high as 5900 feet per second, close to the limit for chemical propellants. Escape velocity at the Earth’s surface is about 37,000 feet per second, so the answer to your question is a resounding “No!”
So is the escape velocity figure not counting air drag?
Escape speed does not take air resistance into account, mostly because there’s really no way that you can take it into account without specifying a boatload of other parameters into account.
Note, by the way, that I said “escape speed”, not “escape velocity”. Velocity is a vector and thus has a particular direction, but direction is actually completely irrelevant for escape speed, so long as you don’t slam into the surface…
Correct. Escape velocity is figured from some basis physics; roughly speaking how fast does something have to start out moving so that when gravity reduces its velocity to zero, the object is infinitely far away from the body pulling on it. Any slower than that and gravity would stop it at some finite distance and thus it’d start pulling the projectile back. The math is really simple and it depends on the gravitational constant G, the mass of the earth (or whatever body you start from) and the distance of your projectile from the center of mass of the earth (so assuming you start at sea level that’s basically the radius of the earth).
Here’s a Wiki article:
Anyhow that gives you the theoretical escape velocity under ideal conditions. If the projectile is subject to drag forces that will increase the escape velocity (it’s a minimum in a vacuum, higher in our atmosphere, still higher if our atmosphere was maple syrup).
Are you confusing the speed needed to maintain orbit? It’s not speed that is required to leave the planet, it’s the energy necessary to maintain a given speed until gravity is no longer a factor. If you fire straight up, with no atmosphere, it should take less speed to escape the pull of gravity.
No, I’m not confusing anything. A bullet does not accelerate once it leaves the muzzle of a gun. I don’t know what definition of escape velocity you are using. Wikipedia’s definition is as good as any:
For a given gravitational potential energy at a given position, the escape velocity is the minimum speed an object without propulsion needs to have sufficient energy to be able to “escape” from the gravity, i.e. so that gravity will never manage to pull it back.
If you fire a round straight up with a velocity of 11,000 m/s, it will never come back. If you shoot the projectile tangent to the surface aiming north/south at 11,000 m/s, it will never come back. If you managed to fire straight down without the ground getting in the way, you would still have the same escape velocity.
However, if you fire it in the direction of the rotation of the Earth at the equator, you gain 465 m/s, so you only need to fire the round at 10,545 m/s to have escape velocity. If you fire it against the rotation, you lose 465. This is why we launch rockets E/W from as close to the equator as possible – to take advantage of the free velocity.
Thanks guys. This was in reference to the whole myth on mythbusters where a bullet shot absolutely straight up wasn’t harmful, but one shot at anything less than a 90 degree angle could. My wife said “If it was in a vacuum the bullet would reach the same velocity on the way back down” and I was under the assumption in a vacuum a bullet would keep going.
You can break it to your wife that she’s wrong there as well… A bullet when it is fired by a weapon is moving at the speed of the explosion that propelled it out of the barrel in the first place (roughly).
This is probably going to be faster than the actual speed of free fall, which is what a fired round is doing as it comes back to earth.
Nope. In a vacuum, as she stated, the speed at which it lands is the same as it was when it left the barrel. There is no terminal velocity in a vacuum.
Possibly the first man-made object to achieve escape velocity: a shaft cover during a 1957 nuclear device test.
As Baracus said, your wife is correct (although you are right to raise the question of escape velocity)
Would not the gravity of the earth pull at (essentially) a constant, that being the maximum speed of fall?
Maybe I’m not being clear.
Round is fired into a vaccuum cylinder. The bullet is launched forward at a speed determined by the pressure wave of expanding gas from the gunpowder explosion.
The round travels up, until all of the … uh… inertia(?) is expended, at which point it turns around and falls back towards the earth, at a speed determined by the pull of gravity. It will accelerate at a constant rate until it is falling at a speed determined by the gravitational pull of the earth.
My statement is that the round will travel faster up, due to being propelled by an explosion.
What I understand you to be saying is that the bullet would travel up, then come back down reaching the exact same speed it was ejected out of the cartridge via gravity alone?
If the gravitational pull is strong enough to slow down and stop a bullet traveling, say, 8,000 feet per second initially, it will also be strong enough to pull it back in at the same speed by the time the bullet has traveled the same distance back.
The first part is true. The earth keeps pulling with a force F. According to Newton’s laws of motion, the bullet will undergo an acceleration, g, equal to F/m, where m is the mass. If F and m stay constant, then the acceleration will be constant and the bullet will continue to speed up until it hits the earth.
In the real world, the bullet would reach some maximum speed, its terminal velocity, but that is due to air resistance, not because it catches up with gravity.
There is no such speed. The gravitational force of the Earth produces an acceleration in the bullet that depends only on how far the bullet is from the Earth. In a vacuum, the falling bullet will never stop increasing its speed until it hits the ground.
When the bullet is fired straight up, it has a certain initial velocity that it receives from the chemical propellant. The key point is that during the entire trip, the bullet is always accelerating downwards. It just takes time for that acceleration to consume all of the bullet’s upward speed. The same force continues to act on it on the way down, so the downward trip is an exact mirror of the upward trip, and it has the same speed when it reaches the bottom. (Assuming no atmosphere, of course.)
Does a bullet stop accelerating the instant it leaves a gun barrel, particularly in a vacuum? There will be some expanding gasses coming out of the muzzle that might give it just a little extra push. But apart from that, and some other absolutely miniscule effects, the trip down is a mirror of the trip up and the speed comes out the same.