[chris (OP)]

A couple of days ago we were discussing the fate of a bullet fired from a gun on the International Space Station (or any orbiting body, come to that).

Specifically, if you fired the bullet in the direction of travel, its velocity would be 17,000 mph + muzzle velocity?

So would the consequence be that the bullet would effectively ascend to a "higher" orbit than the ISS, which would be going a bit more slowly?

[Halc]

The bullet (presumably not fired at escape velocity) would assume an elliptical orbit which would intercept the old ISS orbit every lap.  It would orbit slower than the ISS, so would take longer than 90 minutes to come back.

The ISS meanwhile, due to the reaction force, would also assume a slightly elliptical orbit and reduce its orbit period a bit.

This is similar to the old question of if you have two space ships in the same orbit, one behind the other, say a couple km apart.  The guy in back wants to toss a ball to the other ship (with a gentle toss).  Which way does he throw it?

[Janus]

Assuming a muzzle velocity of 1200 m/sec ( about mid range for small arms), The new bullet orbit would climb to a new apogee of ~400 km higher than ISS orbit ( perigee would remain at ISS orbit distance) and would take ~ 5 min longer to complete its orbit.

[Petrochemicals]

As

Assuming a muzzle velocity of 1200 m/sec ( about mid range for small arms), The new bullet orbit would climb to a new apogee of ~400 km higher than ISS orbit ( perigee would remain at ISS orbit distance) and would take ~ 5 min longer to complete its orbit.

I think that depends if the end of the barrel or start or middle are inline with the direction of travel. If the end of the barrel  is parrallel to the ground the bullet would be exerting force on he earth side of the barrel, dropping once it had exited into a lower faster orbit ?