# Dropping a bullet in a non-vacuum

Cecil says that, in a vacuum, a bullet fired from a gun in a horizontal direction will hit the ground after one dropped from the same height due to curvature of the earth (http://www.straightdope.com/columns/read/1373). No argument there.

But we don’t live in a vacuum. What happens in real life?

I’d say the bullet fired horizontally at sea level will still fall to Earth after a bullet dropped from the hand. The fired bullet will still be cruising along for a second or two, at least, after the dropped bullet reaches the ground.

I heard someone (a hunter, not a physicist) once say that the bullet simply dropped will hit the ground first because the spinning can make bullets rise somewhat, or at least (I presume) give it some small amount of “lift” to forestall the effects of gravity. Apparently that was not scientifically accurate.

In real life without a vacuum, wind will play a role. Is the wind blowing up or down? Headwind or tailwind?

Air resistance will slow the bullet from it’s foward trajectory, making it fall to the Earth closer than if there were no resistance to slow it down, so it will reach the ground slightly sooner than no air because of the same curvature effect.

And of course a lump of ground, tree, or building, etc in front of the bullet will ruin the experiment.

I don’t think spinning will provide any aerodynamic lift. It will stabilize the flight, and keep the bullet from tumbling, which will reduce drag and make the bullet go farther and straighter.

A bullet fired from a gun be hot, and thus slightly less dense than a bullet that was just dropped (cite). It would probably also have a slightly higher drag coefficient because of the scratches from engaging the rifling. Well, all in all probably negligible effects.

Why would you say that?

If the scenario is one where I am stood on a beach say, holding a bullet level with my shoulder in one hand, and in the other, pointing a handgun along the beach, if I was able to fire the pistol and drop the bullet at exactly the same time, I’d fully expect the bullet I dropped, to hit the ground first. So, would I be wrong then?

Why would you expect that?

Assuming a vacuum and a flat, non-curving earth, both a horizonally fired bullet and a bullet dropped from the same height will hit the ground at the same time. Think Galileo.

Add in air resistance. This would have a serious forward motion effect to the fired bullet, but in the downward direction wouldn’t have much effect for either.

Add in curvature of the earth and other factors like trajectory alteration due to spin, tumble, bullet shape and birds in the way for the fired one to get the final answer.

I’d just expect the dropped bullet to have already hit the earth by the time gravity has overtaken the velocity of the bullet that is fired. Is that wrong then?

If you fire it exactly horizontally, then gravity isn’t overtaking anything and it will be pulled down at the same rate as the one you drop. This is simple vectors.

I haven’t got the maths to deal with this, but my logic says that the speed of the bullet overcomes gravity upon leaving the barrel, otherwise it would just drop straight to the floor. What am I missing?

It drops to the floor just as quickly as the bullet you just let go of. But that takes some time, and a bullet that’s also moving forward at 300 m/s can do a lot of other interesting things in that time, like hitting a target, since the forward speed is a lot greater than any speed it’ll reach from falling.

If it helps, think of it this way: Suppose I aim the gun straight down, and both shoot and drop. Obviously, the fired bullet will hit the ground much sooner than the dropped one. Now, suppose that I aim the gun straight up. Obviously, in this case, the dropped one will hit first. Clearly, then, there’s some direction I could aim the gun, somewhere between straight up and straight down, such that the two would hit at the same time. What do you suppose that angle is?
OK, that’s the vacuum version of the problem. What about air resistance? Let’s suppose we have a perfectly flat Earth, perfectly still air, and perfectly spherical projectiles. In this case, the fired bullet actually does hit the ground slightly later, because air resistance is proportional to speed squared. So the fact that the bullet is moving rapidly horizontally actually makes the resistance against its vertical motion stronger.

Halfway between directly down, and directly up, ie. 90 degrees?

So does it fall quicker or slower? It sounds like you are saying quicker? Sorry if I’m coming over somewhat dense!

In practical terms, there are confounding factors such as air currents, birds in the way and the curvature of the earth.

In the pure scenario of a flat earth, constant gravity etc, both bullets hit the ground at the same time. The bullet’s forward motion doesn’t overcome gravity because it has no vector component in opposition to gravity.

Using the term “vector” isn’t going to help someone who doesn’t understand physics already.

The simple answer is that the pull of gravity is independent of the forward motion of the bullet, so all other things being equal the two bullets fall at the same rate and hit the earh at the same time.

You’re missing that the forward speed of the bullet has no effect on gravity’s downward pull on that bullet. Forward speed does not impart any lift to the bullet; it doesn’t have wings where is where a plane’s lift comes from.

Y’know that scene in Speed where the bus drives off the end of an unfinished freeway and magically rises in the air over the gap to the next intact part of the road? Total fantasy. Think about it; if it was driving along and the road underneath it vanished, it would immediately start falling. Well the same would happen in the situation depicted in the movie; the sudden lack of road support does not magically impart lift to the bus. Forward speed by itself generates no lift at all, it takes a wing to generate that lift.

No amount of forward speed gives any lift to the bullet (or to anything else). It falls to the ground at the same time as the stationary one, all else being equal.

All the force that causes the bullet to accelerate does is cause the bullet to travel in an arc. If you are on a building and you fall off you will fall straight down. Now get a running start and run straight off (no jumping). You will still fall at the same rate (9.8 m/s[sup]2[/sup]), but the rate at which you are running will impart a distance away from the building, exactly how far contingent upon how fast you’re traveling. In theory, although impossible, if you can go fast enough you can orbit the Earth (that’s exactly what an orbit is, you’re falling at the same rate the Earth curves away from you).

The factors that matter here are whether or not the spin imparts a lift component and how fast the bullet is traveling, because if the Earth curves away in any measurable degree from a right angle it will necessarily take longer to impact than if you drop the other bullet perpendicular to the Earth. Wind resistance, barring lift, will just cause the fired bullet to impact closer to you than if there were no wind. A tailwind might cause it to go far enough that curvature of the Earth matters, but that is unlikely.

How so?

I disagree with this. Drag, or air resistance, is still a force, and therefore a vector quantity. It is resolvable into horizontal and verical components (or any other components you like) that add like vectors to make the resultant force. If we just concern ourselves with the drag that affects the bullet’s speed of falling, we see that it only proportional to the square of that downward speed. The vertical component of the drag force is unaffected by the bullet’s forward velocity, even though the net drag force is much different. Put another way, the bullet’s significant horizontal velocity makes for significant horizontal drag, but this doesn’t affect its motion or the forces it experiences in the vertical direction.

In short, my point is that air resistance doesn’t change the main result of the experiment; the bullet dropped and the bullet fired horizontally take the same amount of time to hit the ground regardless of whether there is air or not (also ignoring the curvature of the earth). The only difference between being in a vacuum and being in the atmosphere is that the bullet fired into air doesn’t go as far downrange.

The bullet is encountering a lot more air molecules on all sides - I wonder if this is equivalent, from the bullet’s perspective, as the surrounding air being more viscous (meaning it would take longer for the bullet to sink through the air).