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So I'm sure a lot of you have heard the thought experiment where you shoot a bullet parallel to the ground, and drop a bullet at the same time and they hit the ground at the same time. Imagine now you get a very powerful gun that shoots a bullet a quarter or nearly half way around the Earth before hitting the ground. The question arises, does a dropped bullet still hit the ground at the same time? I suppose this question can be a bit more general by asking whether or not the original thought experiment is accurate all the time, or whether it only works assuming the Earth is flat, and any time you introduce a curve the bullets will hit differently. I am not only curious if anyone has a mathematical answer to the question, but also what your initial intuition is before thinking too hard about it. I have asked many people and I would say the answers are around a 60/40 split with the majority saying the bullets hit at the same time.

Here's my qualitative thought process as to why the bullets hit at different times. Or more specifically that the dropped bullet hits earth first. Assume now your starting position is relatively far from Earth's surface, let's say many Earth's diameters. You start so far away that you just have to nudge the bullet a bit to reach a quarter way around Earth. To me, it seems the force of gravity is nearly the same on the two bullets since they have similar distances to the center of Earth, BUT the pushed bullet must travel a much further distance before landing. So the speeds and accelerations are similar, but the distance is not. Therefore the pushed bullet lands later. Another explanation I have is using the same setup, is the moment after you push the bullet it is now further from the center of the Earth as the dropped bullet. So the force of gravity is smaller and it does not accelerate as quickly. Let me know if you all have any other explanations for/against my answer.

Finally, of course everything is ideal (i.e. earth is spherical, gravity is uniform, no air, the usual)

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You are right. The equal time answer assumes a flat earth with uniform gravity. With a curved earth, you could shoot hard enough that the bullet goes into orbit and never lands. Or you could shoot so hard that it flies away from Earth and never returns.

If the Earth was flat, shooting that hard would not make a difference. They would land at the same time.

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For the practical purposes of explaining gravity in high school physics, it is fair to say both bullets land at the same moment. But at a micro-scale of time they do not, and that is for more reasons than just the curvature of the earth. There are forces acting on the fired bullet that affect its orientation and impose non-intuitive forces on the path of the bullet. For instance:

  1. The dropped bullet almost instantly flips tail-down and falls through the drag of air with a much smaller cross sectional area (thus higher cross sectional density). Thus the free-falling bullet falls ever so slightly faster. The fired bullet is spin-stabilized and is falling vertically through the air column exposing its larger side surface, than it’s cross-section.

  2. The fired bullet is moving a lot of air out if it’s way. In free space the resulting pressure from this air is in equilibrium around the longitudinal axis. As the fire bullet nears the ground, however, the air displaced underneath it is resisted by the ground, in the same way and aircraft wing performs differently when in “ground effect.” This would cause a tiny delay in the fired bullet reaching the ground.

  3. The bullet is being acted upon by absolutely enormous gyroscopic forces. Most bullets are spinning at over 150,000 RPM and can often be in the neighborhood of 300,000 RPM. Gyroscopic precession causes the bullet to drift in the direction of its spin, almost always to the right. This precession is actually diverting some of the downward force of gravity into a horizontal force acting parallel to the earth’s surface. This could amount to 6-8 inches over 1,000 yards easily, so it‘ s tiny, but material.

  4. The Eötvös effect tells us that a bullet traveling east essentially “weighs less” and thus falls to earth slightly slower than a bullet moving west. This is why orbital payloads are almost always launched eastward, to exploit the additional eastward motion imparted on the missile by the earth’s rotation. A bullet fired at any azimuth that results in a net eastward displacement compared to its shooter will hit the ground later than a bullet dropped or fired with a net 0 or net westward movement compared to the shooter.

Hope this helps create not just some intuition, but some curiosity to want to learn more.

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  • $\begingroup$ Wow Thanks! Every time I've asked someone this question I've always just prefaced it with "no air" and whatnot since I thought adding all the "non-ideal" effects would make it much harder to think about. I'm glad you could provide all this info on air resistance and the Eotvos effect. I'll be sure to share these ideas! $\endgroup$ Sep 1, 2022 at 15:43
  • $\begingroup$ @John_Mcgowan as far as the second part of your thought experiment, consider this, based on what I described in my answer. You drop a bullet nose-down from a significant height. You also drop a bullet from the same height, nose-down, but a high rate of spin. Both have zero initial velocity. Do they land at the same time? If not, which lands first and why? $\endgroup$
    – Max R
    Sep 2, 2022 at 4:30

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