>I understand that, in the space station, there is no gravity.
Not true. There's always gravity ... it's just that you feel weightless.
>Huh? How is that possible. If there's gravity, then how...?
You and the space ship travel in the very same orbit. So you don't fall to the 'floor'.
It's like falling inside an elevator that's dropping with the acceleration of gravity. You'd float inside that elevator.
Look at it this way:
 If I throw a space ship into the air it will travel along a certain path.
 If I throw you into the air (with the same initial velocity), you'd travel the very same path.
 Now I throw the space ship with you inside it. Same path for you and the ship. See? You just float inside the ship.
>Oh, I remember seeing people floating inside a plane. The plane was traveling along a parabolic arc, right?
Yes ... a parabolic arc is the path an object would take when thrown into the air ... almost.
>Almost?
Well, that'd be true if there weren't any frictional drag due to the air. With drag, it's not quite parabolic.
>Then what kind of path is it?
Do you really want to know? There's a bit of math involved, so have a nap.
>zzzZZZ
We consider an object with velocity V acted upon by forces of gravity mg (acting down) and air resistance aV (opposing the motion).
Imagine the object tossed off the edge of a tall cliff with air resistance proportional to the object's speed.
[1]
In vector form:
m dV/dt = mg  aV
In component form, with V = [u,v]
[1a] du/dt =  k u
[1b] dv/dt =  g  k v
where k = a/m




We can solve each differential equation:
[2a] u(t) = u(0) e^{k t}
[2b] v(t) = g/k + (v(0)+g/k) e^{k t}

Since u = dx/dt and v = dy/dt we get (with x(0) = y(0) = 0):
[3a] x(t) = (1/k) u(0) { 1  e^{k t} }
[3b] y(t) = (g/k) t + (1/k) {v(0) + g/k}{ 1  e^{k t} }


trajectories, using units of feet and seconds.

Note:
For large values of time t:
x(t) ∼ u(0)/k
y(t) ∼ (g/k)t.

so, eventually, the object is going straight down (at x = u(0)/k) with constant speed (namely g/k).
>zzzz... huh? That's what happens when somebody jumps out of a plane, with a parachute, right?
Yes. Lots of drag and, soon, a constant downward speed of (g/k).
When 'k' is larger (with a parachute), the speed is smaller.
For u(0) = 10 ft/sec, V(0) = 100 ft/sec and k = 0.1 sec^{1}, we'd get
where the eventual downward speed is 322 ft/sec (or 215 mph).
>zzzz
In case you're wondering, the path described by equations [3a,b] has the form:
y = A x + B log(1  x / d) where d is the eventual horizontal distance travelled.
>zzzz


