This question should be fairly simple, I think. What happens when you are in a space ship orbiting around earth and you shoot a bullet out the window right towards earth, i.e. towards earth's center of gravity, assuming the mozzle velocity is much smaller then the orbital speed of the space ship. It looks like the bullet should assume a higher orbit then the ship even though you shot it right towards earth. But I'm not sure.
Perhaps, I think it would go into some sort of elliptical orbit with part near the earth and part further away, but I'm not sure either. I think a lot depends on height of the orbit and the speed of the space ship and muzzle velocity. It might even hit the earth.
There's no way to tell without more info. We would need the orbital height and the bullet velocity to find out.
I assume that the space ship is in a circular orbit. The velocity of the bullet directly after being fired is somewhat higher than the velocity of the spaceship. There are two contributions to the total velocity of the bullet directly after firing. - The velocity it already had as it was onboard the spaceship - The additional velocity from being fired As those two velocity components are perpendicular the vector sum is larger than the components. The bullet will proceed in an elliptic orbit. If the bullet does not hit the Earth then it will continue in its orbit. Question: will that eccentricly orbiting bullet spend more time at higher altitude than the spaceship it was fired from, or at lower altitude? Apogee is the point of farthest distance from Earth, perigee is the point of closest approach. As we know, in an eccentric orbit the orbiting velocity is slower at apogee, and faster at perigee. (For an extreme example compare the orbit of Halley's comet. At aphelion it just crawls, at perihelion it screams though the solar system.) The slower-at-apogee-faster-at-perigee makes the bullet spend more time outside the spaceship's orbit than inside. Also, the bullet has more orbital energy per unit of mass than the space ship has. That too is a factor that raises the orbital altitude, making it spend more time outside the space ship's orbit. So in general the answer to your question is yes, I think. For any pair of orbiting objects you can compare their total orbital energy (the space ship and the bullet have different mass, so you must compare total orbital energy per unit of mass) The bullet will have more orbital energy (per unit of mass). By the way, that is one of the reasons there have been very few spacecrafts that have visited the planet Mercury. You can't just fire rockets and make a blast to Mercury. Like in the example of the bullet if you fire the rockets at the wrong time in the wrong direction you end up increasing the total orbiting energy. It's hard to climb to an orbit at greater distance from the Sun, but to descend towards the Sun is just as hard.
I think basically you need to come up with some equations of motion, that made harder with it being a circular planet, I will see if I can come up with anything - lol (don't hold your breath).
It doesn't matter. If the bullet is fired in the direction of the center of mass of the earth, sooner or later, it will strike the earth at some point. Whatever velocity the bullet might have travelling around the earth does not negate the velocity in the direction of the ground.
Assuming the bullet is fired with speed w towards the center of the Earth from a circular orbit with radius r where the orbital velocity is v, with w < v, I get that the perigee rp, apogee ra, semi-major axis a and eccentricity e of the bullets orbit to be [tex] r_p = r \frac{v}{v+w}[/tex] [tex] r_a = r \frac{v}{v-w}[/tex] [tex]a = r \frac{v^2}{v^2-w^2}[/tex] [tex]e = \frac{w}{v}[/tex] It can be seen that as w approaches v, the apogee approaches infinity and the orbit will approach a parabola.
Correct me if I'm wrong, but if I remember correctly the escape velocity from altitude h is the same as velocity of circular orbit at altitude h. I think the parabola result that Flip Larsen mentions is related to that. It's tempting to think of escape velocity as velocity straight away from the planet. I'm taking Flip Larsen's word for it. The result shows that if a bullet is fired from an orbiting spacecraft, straight down, at escape velocity (and the resulting trajectory doesn't make the bullet impact the planet) then the bullet will escape the planet. I find it big fun to get a counterintuitive result like that.
Escape speed at a given distance is [itex] v_{esc} = \sqrt{2} v[/itex] where v is the circular orbit speed at the same distance. If the bullet is fired with that same speed v relative to the spaceship, the speed of the bullet relative to Earth will be exactly [itex]v_{esc}[/itex] and the bullet will leave Earth, unless of course, that it enters the atmosphere first. Given that the perigee in the case of w = v is half the circular orbit radius r, the spacecraft should be in a circular orbit of radius at least 2R, where R is the radius of the upper atmosphere, if it is to escape Earth without impacting first. Indeed.