Here's part of a solution. I didn't get to finish it in time. I'll try again later.

If I understand correctly, you have a planet's position & velocity, as well as a ship's position and speed. You want to get the ship's movement direction. I'm assuming the ship's and planet's speeds are constant. I also assume, without loss of generality, that the ship is at \$(0,0)\$; to do this, subtract the ship's position from the planet's, and add the ship's position back onto the result of the operation described below.

Let:

 * \$s_s\$ = the ship's speed (\$s_s.x, s_s.y\$, likewise)
 * \$s_a\$ = the ship's bearing (angle of movement, __what we want to calculate__)
 * \$p_p\$ = the planet's initial position, global coords
 * \$p_r\$ = the planet's distance (radius) from the center of orbit, derivable from \$p_p\$
 * \$p_a\$ = the planet's initial angle in radians, relative to the center of orbit
 * \$p_s\$ = the planet's angular velocity (rad/sec)
 * \$t\$ = the time to collision (this turns out to be something we must calculate as well)

Here's the equations for the position of the two bodies, broken down into components:
$$ ship.x = s_s.x * t * cos(s_a) $$
$$ ship.y = s_s.y * t * sin(s_a) $$
$$ planet.x = p_r * cos(p_a + p_s * t) + p_p.x $$
$$ planet.y = p_r * sin(p_a + p_s * t) + p_p.y $$

Since we want \$ship.x = planet.x\$ and \$ship.y = planet.y\$ at some instant \$t\$, we obtain this equation (the \$y\$ case is nearly symmetrical):

$$ s_s.x * t * cos(s_a) = p_r * cos(p_a + p_s * t) + p_p.x $$
$$ s_s.y * t * sin(s_a) = p_r * sin(p_a + p_s * t) + p_p.y $$

Solving the top equation for \$s_a\$:

$$ s_s.x * t * cos(s_a) = p_r * cos(p_a + p_s * t) + p_p.x \\
\implies s_a = \frac{arccos((p_r * cos(p_a + p_s * t) + p_p.x)}{(s_s.x * t))}$$

Substituting this into the second equation results in a fairly terrifying equation that [Wolfram alpha won't solve for me](https://www.wolframalpha.com/input/?i=solve+b+*+t+*+sin%28arccos%28%28r+*+cos%28h+%2B+s+*+t%29+%2B+c%29+%2F+%28a+*+t%29%29%29+%3D+r+*+sin%28h+%2B+s+*+t%29+%2B+d+for+t). There may be a better way to do this not involving polar coordinates. If anyone wants to give this method a shot, you're welcome to it; I've made this a wiki. Otherwise, you may want to take this to the [Math StackExchange](http://math.stackexchange.com).