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Falling bullet scenario

  1. Aug 10, 2013 #1
    A friend of mine asked me this question out of curiosity:

    If you shoot a bullet straight up into the air, how far would the bullet land from the point it went up?

    Possible factors:
    - earth's rotation
    - air resistance

    What is the answer?
     
  2. jcsd
  3. Aug 10, 2013 #2

    Nugatory

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    In theory:
    It wll depend on a number of other variables. For example, the effect of the rotation and curvature of the earth depends on how high the bullet goes and how long it is in flight; you worry about these things when you're aiming an ICBM but not when you're aiming a bottle rocket.

    In practice:
    Only two things matter: crosswinds and the difficulty of really aiming exactly straight up, not the least bit sideways.
     
  4. Aug 10, 2013 #3
    Hello 15123
    If the earth was not rotating then if the bullet was shot straight up, it should fall back to earth at the exact same spot. That is of course assuming there is 'no wind' in the atmosphere.

    With a rotating earth and the same 'no wind' assumption, the only place the bullet would fall back straight down to the same spot is at one of the poles.

    With a rotating earth and the same 'no wind' assumption, at the equator the bullet would fall someplace behind the spot where the bullet was shot up, meaning to the west of that location. The bullet would keep a sideways motion of 1000mph ( the velocity of the earth's surface at the equator ) as it is ascending and descending. The earth would rotate under the bullet in its trajectory, and if you shot it up high enough the earth could make one complete revolution, plus a part of a revolution ( remember the bullet keeps the 1000mph velocity to the east ) so that the bullet could fall back to the same spot.

    'no wind' would mean no atmosphere and therefor no drag to slow the velocity of ascent or descent due to terminal velocity, nor any drag to slow the 1000mph velocity to the east. the bullet would be subject to gravity only, and one could calculate the fall back location somewhat easily.

    At any other latitude, someone might come along to explain that.

    Also, with an atmosphere, it becomes more difficult since a few more assumptions would have to be made on the wind velocities and the does the bullet end up being carried along with the air due to drag.
     
  5. Aug 10, 2013 #4
    Let us suppose that the bullet is aimed 100% exactly straight up (through some device perhaps) and that the bullet is a standard 9 mm bullet with a symmetric body (like all bullets).
    Let us also suppose that this happens on a day where there is no wind resistance at all.

    Other factors:
    - bullet's rotation around its own axis

    I tend to think that the bullet's nose might be pulled sideways by gravity when the bullet is on its way back down to earth. I find it hard to believe that the bullet would fall straight down the same way that it went up (with its nose pointing forwards 100% of the time). It does not sound realistic to me.
     
  6. Aug 10, 2013 #5

    Nugatory

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    You're right, it's not realistic. But the thing that's not realistic is the part where we get everything exactly perfectly right: 100% straight up, no wind deflection at all, we'll do the experiment at one of the earth's poles so that the earth isn't moving sideways while bullet is in flight, not even the tiniest microscopic flaw in the symmetry of the bullet., ... If we got all of those things right, the bullet would go straight up into the air and fall right back into the muzzle of the gun. Of course that's absurd, but the absurdity just goes to show that it's absurd to expect to get everything spot on right in the real world.
     
  7. Aug 10, 2013 #6
    Please correct me if I'm wrong, but why would the earth's rotation have any effect on where the bullet (or any other free falling object) would land? We don't get pummeled by falling hail at 1600km/h from the east at the equator, because the atmosphere is also rotating with the earth. If I drop a ball from the east side of a tall building, it doesn't hit the side of the building because the earth moves from under the ball. This is the same case for the bullet, the rotating earth has nothing to do with the frame of reference that the bullet is moving in.


    Damo
     
  8. Aug 10, 2013 #7

    Nugatory

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    The corrections are too small to worry about, but they are non-zero. The dropped object is moving in free-fall in a straight line while the building is moving on a circle of radius 6400 kilometers (if at the equator). If the building were tall enough, and if other effects such as crosswinds did not dominate, you would see that the trajectory of the ball does not remain parallel to the wall of the building throughout the fall.

    That is indeed correct, but it works that way only because the distances involved are small enough that we can ignore the rotational effects. We (rightly) approximate the behavior of a falling hailstone as if the surface of the earth were flat and stationary, and this works fine over distances of five or ten kilometers, especially if we aren't demanding the unreasonable precision required to drop a bullet back into the muzzle of the gun it was fired from.

    Try rolling that hailstone from the north pole to a point on the equator, and you will have to consider Coriolis effects.
     
    Last edited: Aug 10, 2013
  9. Aug 10, 2013 #8

    A.T.

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    The bullet is not moving in a specific reference frame. If we chose to describe the bullets movement in the rotating rest frame of the earth, there will be a Coriolis acceleration. In an inertial frame the bullet will stay behind the earths rotation when it increases its distance from the center. Either way the bullet doesn't land, where it was shot up (except at the poles), but the deviation is small.
     
  10. Aug 10, 2013 #9
    The effect is not insignificant. It needs to be taken into account when firing artillery shells.
    They may not go 'vertically upwards' but they do have a vertical component.
    The effect is known as the Coriolis effect
     
  11. Aug 11, 2013 #10
    Roger roger, that makes sense. :)

    I can remember my father telling me when I was quite young, that when constructing large buildings these effects could be expected also. Although they would be miniscule.


    Damo
     
  12. Aug 11, 2013 #11
    When the object is resting in a building floor, it's rotating with the Earth. I don't understand why that should change during free-fall? Shouldn't the object's horizontal motion still be same as that of the Earth during the entire time it is falling (as a coin tossed in a steadily moving car moves with the car)?
     
  13. Aug 11, 2013 #12

    A.T.

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    You are confusing angular and linear velocity here. They are related via the distance from the center, which changes during a fall.
     
  14. Aug 11, 2013 #13
    How would that change if the object is falling in a straight line?
     
  15. Aug 11, 2013 #14

    Nugatory

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    The object, once in free-fall, keeps its horizontal motion just fine; drop it from the top of the building and it will start out with the exact same horizontal speed as the top of the building and keep it the whole way down. When it hits the ground it will still be moving sideways at the same speed as the top of the building.

    But the top of the building is moving sideways faster than the bottom of the building. It has to, because the top and the bottom both make a full circle every 24 hours, but the two circles have different radiuses (they differ by the height of the building), hence different circumferences, hence different distance travelled in 24 hours, hence different speeds. Thus, the object will be moving horizontally relative to the ground when it hits. The effect is too small to notice or care about unless you're aiming long-range artillery, but it's there.
     
  16. Aug 11, 2013 #15
    Such a lucid explanation! I understand it now.

    Thanks so much guys.

    P.S: If I remember correctly, non-believers of the heliocentric model of the solar system actually used this observation (that Earth doesn't move out underneath a falling object) as an objection to it. Galileo pointed out that need not be so if Earth is uniformly rotating. Turns out they were both wrong, hehe.
     
  17. Aug 11, 2013 #16

    Nugatory

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    Yes, but Galileo was much less wrong :smile:.

    To quote Isaac Asimov:
     
  18. Aug 11, 2013 #17
    The trajectory of a bullet "shot upward" will be an ellipse. The inclination of orbital plane for a bullet "shot upward" depends solely on the latitude of the firing location, but the important thing is that it will be some non-zero (except when fired on the equator). Because the trajectory of the firing location itself is a circle parallel to the equatorial plane, the trajectory of the bullet and the trajectory of the firing location may intersect in at most two points, one being the initial firing location, and the other one "on the other side of the Earth", which is impossible for an ordinary bullet.
     
  19. Aug 12, 2013 #18
    Well-said. I love that entire essay written by Asimov!

    Edit: Nurgatory, could you please elucidate how the top of the building would trace out a larger circle than the bottom of it? I can't quite visualize it.

    Thank you!
     
    Last edited: Aug 12, 2013
  20. Aug 12, 2013 #19
    The top of the building will travel 2(pi)meters further than the base based on the height of the building. If it is 100mtrs tall, the top will travel 628.3mtrs further in one day than the base.


    Damo
     
  21. Aug 12, 2013 #20

    Nugatory

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    It might be easier to visualize if you imagine yourself floating in space above the south pole, watching the earth rotating below you like a giant ball. Put a building at the equator so that it's sticking out from the surface of the earth - from your vantage point above the poles the building is sticking out sideways.

    The radius of the earth is about 6400 kilometers, so when the earth turns one full rotation, a point on the equator such as the base of the building, will have gone around a circle of circumference ##2\pi{r} = 6400\times{2}\pi##.

    However, the top of the building is further away from the center of the earth (6400 km plus the height of the building) so it traces a slightly larger circle: ##2\pi(r+h)##.
     
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