1. Since the earth is spinning at roughly 1000 mph at the equator, why does a bullet go the same speed no matter what direction you fire it in?

2.

3. Because the bullet starts off already moving with the Earth.
You might as well ask "Since the Earth is spinning at roughly 1000 mph at the Equator, why don't people have to run to stay in one place?".

4. Jump as high as you can on a train (without hurting yourself) when the train is at its fastest. Did you move relative to the floor?

5. While the speed of the bullet (relative to the ground) upon leaving the muzzle is the same no matter which direction you fire it, the distance the bullet will travel relative to the ground before hitting the Earth's surface will not be the same. (assuming all other things are equal.)
A bullet fired East from the equator and one fired West will have different speeds relative to the center of the Earth. This, in turn, alters their trajectory relative to the center of the Earth. If the bullets are fired at the same angle to the ground and at the same height, one will take longer to hit the ground, and hit the ground further from the gun than the other would.
For example, ignoring air resistance, a bullet with a muzzle velocity of 1500 m/s and fired level to the ground, and 1.5 m above the ground, will travel ~120m further if fired East vs. West at the Equator. A drift East or West will occur when firing North and South.
Large artillery guns have to correct for these effects when aiming and ranging.

6. Say I had two targets that are equidistance away to east and west at the equator, and the targets can measure bullet impact, would different impacts be measured?
Im guessing yes. And if the targets were fired at simultaneously, then the bullets would impact at different times, sooner east and later west?

7. Originally Posted by GiantEvil
Say I had two targets that are equidistance away to east and west at the equator, and the targets can measure bullet impact, would different impacts be measured?
Im guessing yes. And if the targets were fired at simultaneously, then the bullets would impact at different times, sooner east and later west?
The impacts would be different, and the bullets would strike at different heights on the targets.
The bullets would arrive at the targets at different times, but which one arrives first depends on the initial muzzle velocity.
I'll use two extremes to illustrate.
First we assume that the muzzle velocity is exactly equal to the tangential speed of the Earth at the equator. The bullet fired West has its tangential speed exactly canceled out by the muzzle velocity and it falls "straight down" (relative to the Earth's center). The time is would take for it to hit the Target would be equal to the time it would take for the target, traveling with the Earth's surface to reach its line of trajectory. It is probably best to think of it in terms of angular velocity relative to the Earth's center the Bullet has zero angular velocity, and the target, which starts at some angular distance from the bullet has an angular velocity equal to that of the Earth.

The bullet fired East enters an orbit with an initial orbital speed of twice the tangential speed of the Earth. It essentially has to "catch up" to the target which has a head start, but is moving slower. The target starts at a certain angular distance away, and is moving at the angular speed of the Earth, while the bullet starts with an initial angular speed twice that of the targets and moving in the same direction. Now if the Eastward bullet were to maintain a constant angular velocity, it would strike the target at the same time as the Westward bullet struck its target. However, it doesn't. As it travels, its orbit takes it closer to the Earth's center*, and in order to maintain a constant angular momentum, it has to gain angular speed. This increase in angular speed means that it will catch up to the target a little sooner and hits its target first. (The Westward bullet must maintain a constant angular velocity also, but since this is 0, the change in distance to the center of the Earth does not require a change in angular velocity)

Now let's consider the scenario where the Eastward bullet is has just enough muzzle velocity to achieve a circular orbit. As before, it is chasing after the target, but now, since it is maintaining a constant height, its angular velocity need not change in order to maintain angular momentum.

The Westward bullet also enters an orbit, but since its initial orbital velocity is equal to the muzzle velocity minus the Earth's tangential speed, it enters an elliptical orbit at that orbit's apogee. Again, if the bullet where to maintain a constant angular velocity, it would hit its target simultaneously with the Eastward bullet. But since it "falls" inward as it travels, it, as the East ward bullet in the last scenario did, must gain angular velocity in order to maintain angular momentum. In this case it has an Westward angular velocity, while the Target, starting at a certain angular distance away to the West, has an Eastward angular velocity. Since the bullet gains angular velocity between being fired and hitting the target, it hits the target sooner than it would have if its angular velocity had remained constant, and thus it hits its target before the Eastward bullet hits its target.

So, which bullet hits first depends on the initial conditions.

*We are assuming the bullets are fired level with the ground.

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