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I think that when you shoot it, it will land later because since the earth is round wherever it land will be slightly lower than were the dropped one landed. So why do they say they will land at the same time?

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    $\begingroup$ Bullets don't typically go far enough that the curvature of the earth becomes significant. $\endgroup$ – Chris Nov 18 '17 at 21:37
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    $\begingroup$ The effect of air drag will be far greater than any curvature effects. What Chris said. But if you ignore both effects the landing time will be the same, assuming you shoot horizontally. Vertical and horizontal movement are perfectly independent, you see. $\endgroup$ – Gert Nov 18 '17 at 21:55
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    $\begingroup$ Mythbusters tested this; it's easily found on YouTube. $\endgroup$ – Kyle Kanos Nov 18 '17 at 22:05
  • $\begingroup$ @KyleKanos: you don't seriously watch MB, do you? 99 % of their myths can be busted with a piece of paper and a pencil! LOL. But they SO love blowing things up, these cr*tins. $\endgroup$ – Gert Nov 18 '17 at 22:08
  • $\begingroup$ @Gert There's more to life (and physics) than paper and pencil. :-) $\endgroup$ – StephenG Nov 18 '17 at 23:41
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Edited to fix a mistake.

You raise a very good point. As everyone has said, it is a very small effect for bullets. But if you go faster than a bullet, it becomes important. And if air resistance is a problem, you can go above the atmosphere to try it.

Without air resistance, the bullet follows a parabolic path in a uniform gravitational field.

But if we are considering a region large enough that the curvature of the earth matters, the field is not uniform. The force is toward the center of the earth. The trajectory is an ellipse.

If you drop the bullet, the "ellipse" is very skinny - a line. You only get part way through the ellipse before you hit the earth.

If you shoot sideways with a low velocity, you go part way around the earth before the ellipse hits the earth. It is a little like shooting from the top of a hill. The bullet fall farther before hitting, and takes longer than on level ground.

If you shoot fast enough, it would go halfway around the world before it landed. At this point, it would be half way through the ellipse. It would be traveling horizontally, and just graze the earth.

If you shoot faster still, it would not hit the earth. It would follow an elliptical orbit all the way around and come back at the same altitude as it started.

Faster still, and the orbit would become circular.

Faster still, and the orbit would be elliptical, but the point where you shoot would be the lowest point of the orbit.

If you shot extremely fast, faster than the escape velocity of the earth, it would fly away into space and never return.

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I assume you mean "shot horizontally". A bullet shot vertically upwards will certainly land later!

You are technically correct: due to the curvature of the earth the bullet has further to fall and hence will take longer to land. In practice it will be very hard to tell, though.

If the bullet travels 1 km, the earth's curvature adds 8 cm to the height. If the range is 5 km, then the earth drops 196 m over that range. See this page for calculations. Those extra heights may seem significant.

However, in most cases the bullet cannot travel that far as it will be falling. If we fire from a 1 m height, then the bullet will drop all of that 1 m in only .44 s. Assuming a typical bullet speed of 500 m/s, in 0.44 s it has travelled only 220 m. The earth's curvature over that distance adds only 3.8 mm to the drop height.

In other words @Chris's comment is correct: you'd be hard pressed to measure the difference.

The measurement is also affected by various possible errors, including the following:

  • the exact levelling of the gun
  • the flatness of the land
  • timing errors

All of these can have effects that are greater than the difference caused by the curvature.

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This is a typical question of the style illustrated in the image.

enter image description here

Assuming air resistance is neglected, the key point is that there is only the force of gravity acting vertically in both situations. As a result of $\vec F=m\vec a$, there is no acceleration horizontally in either situations, so that ball 2 with merrily continue its horizontal motion, unaffected by the vertical motion, and the vertical motion will be unaffected by the horizontal motion.

Since the vertical force is the same in both situation, both balls will hit the ground at the same time if they start with $0$ velocity from the same height. The curvature of the Earth is unlikely to be a factor unless your bullet travels significant distances.

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  • $\begingroup$ Not sure the OP will correctly interpret your diagram but nice answer anyway. $\endgroup$ – Gert Nov 18 '17 at 22:10
  • $\begingroup$ But what about with air resistance. Then it becomes a very interesting question. $\endgroup$ – John Alexiou Nov 19 '17 at 2:11
  • $\begingroup$ @ja72 indeed it does, but if you want to model air resistance it becomes much much much more complicated. Unless one does very simple modeling - which also requires assumptions not stated in the question- various ballistic factors - muzzle velocity, rotation and shape of the bullet etc - come into play. This appears to be beyond what is OP is asking. $\endgroup$ – ZeroTheHero Nov 19 '17 at 2:23
  • $\begingroup$ It seems easy to do simulation with various values of horizontal velocity and measuring the time to hit the flat ground and see what the curve looks like. It might have a minimum or a maximum at zero speed because it has to be an even function (due to symmetry). $\endgroup$ – John Alexiou Nov 19 '17 at 14:27

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