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Some Soviet space stations reportedly had anti-aircraft cannons installed. Could such a cannon hit the firing space station accidentally on a subsequent orbit? The muzzle velocity of the cannon is under 700 m/s, significantly slower than orbital velocity so the projectiles should have similar orbits to the station.

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The space station could shoot itself, but it's extremely unlikely to happen by accident.

Assuming your space station is in a circular orbit you can calculate it's position in polar co-ordinates as a function of time $(r(t), \theta(t))$ very easily since $r$ is constant and $\theta = 2\pi t/\tau$ where $\tau$ is the orbital period. When you fire the cannon the shell is in a different orbit, and specifically it's in an elliptical orbit $(r'(t), \theta'(t))$. For the station to shoot itself the two orbits must intersect, i.e. at some time $t$ you have simultaneously:

$$r(t) = r'(t)$$ $$\theta(t) = \theta'(t)$$

The problem is that for a generic elliptical orbit the expressions for $r'(t)$ and $\theta'(t)$ are not at all simple so there is no easy way to solve the above simultaneous equations and work out at what time, if ever, they intersect. There's certainly no obvious reason to suppose they should intersect.

You can see that it is possible for the spaceship to shoot itself. If you fire the cannon radially outwards the shell will fall behind the space station. If you fire in the direction of motion the shell will move ahead of the space station. So there must be some angle in between where the shell hits the space station. However this will be the exception rather than the rule.

I'm aware this isn't a great answer since I can't solve the equations of motion and give you a rigorous answer. If anyone else can do this I'd be very interested to see the calculation.

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Actually, this is a great answer as it does mention why there is doubt. As to the idea that shooting behind leads to a lower orbit and shooting forward leads to a higher orbit, the "in between" that you are looking for is to release the projectile with quasistatic separation speed from the space station. They will then orbit together. – dotancohen Jul 8 '12 at 10:42
@dotancohen If the station is in a circular orbit then you could fire the cannon horizontally (with respect to the Earth below) at a velocity such that the shell is in a different circular orbit. Then in theory the orbits should intersect once they've gone halfway around the Earth. But space is very very big compared to the size of a space station, so scoring a direct hit would probably be quite an engineering challenge. If the shell ends up in orbit it will become space junk, posing a small but ever-present danger to all spacecraft with nearby orbits. – Nathaniel Jul 8 '12 at 11:08
In certain respect the cannon would "hit" the station always: "actio and reactio" Would the recoil action on the stations orbit be tolerable? – Georg Jul 8 '12 at 13:37
@Nathaniel, that's not correct. If an object is in a circular orbit, it takes two velocity changes to enter a new circular orbit. Only one velocity change puts the projectile in a new elliptical orbit that must be circularized on the "other side". – Alfred Centauri Jul 9 '12 at 0:05
@Alfred: Not quite! You are thinking if the projectile is shot in the same direction as the space station. However, if the new velocity vector has an equal tangential component to the original and a zero radial component, but a different inclination, then it will still be in a circular orbit. This is known as an inclination change. It is how we have geostationary satellites launched from ~30 north of the equator! – dotancohen Jul 9 '12 at 7:43

First, let's establish the context of Newtonian mechanics, an ideal vacuum, and the mass of the Earth concentrated at a point to eliminate uncertainties due to varying drag, "mascons", etc. Let's also stipulate that there are no other sources of orbital perturbation including the gravitational attraction between the station and the projectile.

Now, it is the case that the combination of position and velocity at a particular time uniquely determines the orbit. Assume the ISS is in a perfectly circular orbit and fires a projectile (assuming the mass of the ISS is so much greater than the projectile that the ISS orbit is essentially unchanged), the projectile, having a different velocity at the "instant" it is fired, necessarily is in a different, elliptical orbit.

However, it is also true that this orbit is closed and so both the ISS and the projectile will periodically return to that point in space (with the same velocities as at the time of firing). The question is, will they arrive there at the same time?

I think it is the case that the answer is yes after some whole number of revolutions.

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Thank you, you perfectly describe the mechanical system as I perceived it. – dotancohen Jul 9 '12 at 7:45

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