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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.

Curious
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Tariq
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1 Answers1

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$T = (R + H)(2H/(GM))^{0.5}$ 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? =)

McGarnagle
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timberhill
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