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This replaces a previous similar question that I could have formulated better.


I'm in orbit and I fire a bullet in an arbitrary direction.

I could shoot

(1) Along the direction of the orbit forwards (maybe achieving escape velocity)

(2) Along the direction of the orbit backwards (maybe cancelling out forward velocity entirely causing the bullet to drop vertically)

(3) Directly toward the planet

(4) Directly away from the planet

(5) 'Sideways' in the direction of the poles

(6) All the in-between directions

I'm looking for a general expression for the path of the bullet. Is there such an expression?

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  • $\begingroup$ grc.nasa.gov/www/k-12/airplane/newton.html ? $\endgroup$ – Organic Marble Nov 21 '18 at 21:06
  • $\begingroup$ This is a two-body mechanics problem reducible to a one-body mechanics problem with Keplerian orbits as solutions. Initial direction doesn't matter until you consider the existence of the atmosphere, at which point there's no general expression, and of course if the orbit intercepts the planet itself there's no general expression. So really, as asked, no there's no such general expression for the trajectory that covers all of these cases. $\endgroup$ – uhoh Nov 21 '18 at 21:10
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I'll let someone else handle the mathematical approach and just describe the general results.

I'm in orbit and I fire a bullet...

I'm going to assume you're in a circular low Earth orbit at 200km altitude to start. In each case, the bullet fired continues in an elliptical orbit that intersects the point at which it was fired.

(1) Along the direction of the orbit forwards (maybe achieving escape velocity)

The bullet enters an elliptical orbit, with perigee at the altitude of your initial orbit and a higher apogee dependent on the muzzle velocity of the bullet. A pistol bullet (muzzle velocity around 360 m/s) would get to about 1600 km; a rifle bullet might reach a 4500 km altitude. The fastest bullets fired from hand weapons have a muzzle velocity around 1200-1400 m/s, which is not sufficient to escape Earth orbit, or even reach the moon.

(2) Along the direction of the orbit backwards (maybe cancelling out forward velocity entirely causing the bullet to drop vertically)

Muzzle velocity a fraction of orbital velocity (~7800 m/s), so the bullet won't come to a dead stop, but it will slow enough to drop perigee to below zero altitude, so the bullet will re-enter the atmosphere and burn up in less than half an orbit.

(3) Directly toward the planet (4) Directly away from the planet

In either case the bullet goes into an elliptical orbit with a lower perigee and higher apogee than the original. I don't know off the top of my head how to calculate the altitudes, but the perigee is probably in the atmosphere, so the orbit will rapidly decay.

(5) 'Sideways' in the direction of the poles

This changes the inclination of the orbit by a degree or so while leaving it nearly circular.

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  • $\begingroup$ Is it interesting to see what these trajectories look like from the (orbitting) perspective of the firer. $\endgroup$ – Steve Linton Nov 21 '18 at 21:58
  • $\begingroup$ @SteveLinton In the instant, no different than bullets fired on Earth - they zip off at high speed. The tricky one is #5 playing boomerang and revisiting your general vicinity as it crosses your orbit repeatedly. #1 also returns, but after covering a larger orbit - so you aren't there anymore. $\endgroup$ – Saiboogu Nov 21 '18 at 22:12
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    $\begingroup$ Beyond that, if you could clearly see the entire path the bullet took: #1 would appear to loop up, over, and behind you; #5 would appear to wiggle sinusoidally back and forth across your path, intersecting it every half-orbit. $\endgroup$ – Russell Borogove Nov 21 '18 at 22:18
  • $\begingroup$ It's a yes or no question: "I'm looking for a general expression for the path of the bullet. Is there such an expression?" I think the OP is asking about a mathematical expression, not a literary expression. ;-) In this case I think the answer is no. what do you think? $\endgroup$ – uhoh Nov 21 '18 at 22:23
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    $\begingroup$ Of course there's an expression -- the elliptical orbit for velocity v at point p with Earth considered as a point mass, so that Earth's surface and atmosphere isn't relevant. That the expression is valid only for a limited distance in some cases doesn't matter; we use limited-application approximations constantly. I'm not going there because I hate math. $\endgroup$ – Russell Borogove Nov 21 '18 at 22:42
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I'll supplement the other answer by answering the OP"s question.

Title: General expression describing the trajectory of a bullet in orbit - fired in an arbitrary direction (version 2)

First/Last Sentences: I'm in orbit and I fire a bullet in an arbitrary direction. I'm looking for a general expression for the path of the bullet. Is there such an expression?

Everything in between is a list of some sample directions in case we can't imagine what "arbitrary direction" means.

If we ignore Earth's oblate and lumpy gravity and call it spherically symmetric, then abstract it to a single point of gravity at its center using Newton's Shell theorem then yes, sort-of. All trajectories would be Kepler orbits. However there are no solutions that give you the position ($r, \theta$) as a function of time in a nice way. However they can be nicely solved with Newton's Method.

But the sample list of arbitrary directions includes the word "atmosphere" so there will be no abstracting today.

Trajectories sampling the atmosphere do not have anything even close to general solutions, aerodynamics is too complex and unpredictable, even before you consider tumbling and breakup.

So is there a general solution?

No.

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