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How will a bullet having very high velocity and having little mass behave when fired from the Earth perpendicular to the surface? It will be a great help if somwone explained it? Will it come back? Or will it escape into the outer space?

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  • $\begingroup$ Depends on how fast you fire the bullet! If you fire it faster than escape velocity $v_{\text{esc}}$, it will never come back. If you fire it slower than $v_{\text{esc}}/\sqrt{2}$ it will hit the ground. Otherwise it'll come back. $\endgroup$ – knzhou May 20 '18 at 11:28
  • $\begingroup$ @knzhou don't forget the rotation of earth=))) $\endgroup$ – OON May 20 '18 at 11:36
  • $\begingroup$ @OON You're right, of course! But also the air drag, the effect of the Sun and moon, ... $\endgroup$ – knzhou May 20 '18 at 11:36
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    $\begingroup$ @knzhou The highest speed bullets 1.2-1.7 km/s (en.wikipedia.org/wiki/Muzzle_velocity) are still much slower than the escape velocity of 11 km/s (en.wikipedia.org/wiki/Escape_velocity). Fortunately they still fall near those who fire them. $\endgroup$ – my2cts May 20 '18 at 13:12
  • $\begingroup$ @knzhou Well, obviously there are more important effects than others. Unless you fire close to the moon it will give quite small contribution. The biggest variable is the air drag that may be huge for the light bullet but small if you consider heavier shell with appropriate aerodynamics. If I have more time later I may write answer addressing this thing. $\endgroup$ – OON May 20 '18 at 16:39
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Yes, it will come back with a speed high enough to kill people. Firing shots in the air is a stupid as urinating against the wind but far more dangerous and damaging.

See https://en.wikipedia.org/wiki/Celebratory_gunfire

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  • $\begingroup$ There is actually a Mythbusters episode exactly about this, where they found that depending on the angle the bullet is fired, the velocity at which it falls back down could or could not be lethal. $\endgroup$ – neradha May 20 '18 at 13:33
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If you fired the bullet greater than the escape velocity (more acurately greater the escape-velocity/√2), the bullet will not return in the earth. For earth, the vallue of escape velocity is 11.2 km/sec. Because at a higher velocity, the bullet will get a huge kinetic energy and so that the bullet will easily overcome gravitational force of the earth. Note: Normally a bullet can not achieve this high velocity. So, it will hit on the ground.

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  • $\begingroup$ escape-velocity/√2 gives the speed of a circular orbit, but that's not relevant to this question, since the bullet is fired vertically, not horizontally. And of course, to get an orbit at ground level requires a perfectly smooth planet with no atmosphere. $\endgroup$ – PM 2Ring May 22 '18 at 2:58

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