I'm curious to know this. Neglect air friction and imagine a bullet that were shot normal to the Earth's surface, from the Equator. I will have to consider the Coriolis effect and so I expect the path of the bullet will follow a spiral rather that a straight line (relative to Earth's centre). The gravity will reduce with altitude as well and so it would be difficult to apply basic laws of motion, but I really need to know how this would look like and how long it will take for the bullet to reach around 36000 km above the Earth's surface. Will it come back, stay in that orbit, or escape (assuming that normal velocity reached 0 @ that orbit)? I expect if it comes back, then it will follow a similar path it traveled along during the shot. This is just a curiosity and thanks for help in advance.

  • $\begingroup$ For a radial free fall, see e.g. Wikipedia and e.g. this Phys.SE link. $\endgroup$
    – Qmechanic
    Nov 10 '12 at 13:21
  • $\begingroup$ I think I can't understand your question. Your bullet travels about 300-1500 m/s. The orbital velocity of earth is roughly some 8000 m/s and escape velocity is some 11.2 km/s. So, could you explain me what you require? (Do you really mean 36,000 km which is past the moon..?) $\endgroup$ Nov 10 '12 at 14:00
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    $\begingroup$ @CrazyBuddy Moon is at 380,000 km from earth. $\endgroup$
    – mythealias
    Nov 10 '12 at 15:44
  • $\begingroup$ @mythealias: Oh, Yeah... Forgot the last zero. But, that's just to point out his GEO assumption :-) $\endgroup$ Nov 10 '12 at 15:53
  • $\begingroup$ Thanks for reply. This is just an imaginary question and so, you could consider a bullet that can theoretically travel @ any initial velocity we supply. To make it more practical, assume the reverse case when a bullet with 0 orbital velocity located on the geostationary orbit (i.e. apparently moving in the reverse direction of a geostationary satellite) free falls towards the Earth. $\endgroup$
    – Tariq
    Nov 10 '12 at 19:27

$T = (R + H)(2H/(GM))^{0.5}$ for simple case without atmosphere. For GEO it's about 5 hours with zero velocity in GEO point and enormous shooting velocity. But air resistance is proportional to ~$V^2$, so it would be hard for bullet to leave atmosphere ;)

Where does the question came from? Maybe you've just read "From the Earth to the Moon" Jules Verne? =)

  • $\begingroup$ Although even without an atmosphere a bullet doing 1000m/s will only have enough ke to reach = 1000^2/2g = 50km $\endgroup$ Nov 13 '12 at 2:47
  • $\begingroup$ That's why I used word enormous ;) $\endgroup$
    – timberhill
    Nov 13 '12 at 6:28
  • $\begingroup$ @pink.fascist thanks, but I think you used the direct Newton's law of gravitation and motion formula which don't apply directly here (g is varying with altitude which causes the laws more complex than uniform acceleration). If I use the formula from Wikipedia, or even Kepler's approximation, I would get around 3 hours + 20 minutes. The problem is that, it made no difference between free fall with or without Coriolis effect (Earth's spin). $\endgroup$
    – Tariq
    Nov 13 '12 at 9:58
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    $\begingroup$ You offended me with that =) Of course, I know that g is function of distance. Same is for air resistance, which makes precise formula a bit complicated. $\endgroup$
    – timberhill
    Nov 13 '12 at 15:54
  • $\begingroup$ g variations are minimal at those distances. $\endgroup$
    – harogaston
    May 24 '14 at 17:21

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