Spiral trajectory under gravity? I know that the general equation of motion for a two body problem under gravity is usually an ellipse.
$$r(\theta) = \frac{R}{1-e\cos(\theta-\theta_0)}$$ where $R$ and $e$ depend on initial conditions.
But surely there is some way of changing the initial conditions so that an object spirals into the source of gravity? If so, what is the equation of this spiral in polar coordinates
 A: An orbit must simultaneously conserve angular momentum and total energy (KE + PE). These two requirement restrict orbits in a $1/r$ potential to be conic sections and there is no way an orbit can spiral towards $r=0$ regardless of the initial conditions.
A: The trajectory can also be a parabola or hyperbola, which are not closed paths. In general, any conic section, but not a spiral.
Perhaps you are expecting that if the bodies do not have enough angular momentum then they will spiral into each other. This does not happen. Ideally they are point particles. If they have no angular momentum then they will fall directly towards and pass through each other and oscillate in a straight line, because they have no size so they cannot collide. But if they have even a small amount of angular momentum they will follow very flat elliptical orbits and miss each other because they are so small. Real objects have a finite size, so practically their surfaces may touch as they pass close by.
If they lose energy as they orbit they will spiral into each other. This could happen if one object orbits in the atmosphere of the other, so that it is slowed by air resistance. It would not be easy to calculate the equation of this spiral. The equation would depend on such things as how the density of the atmosphere varies with position, and whether the drag force is proportional to speed or square of speed or has some other form.
A: Let's think about the total energy of a particle moving in a plane under a $1/r$ potential.  Its potential energy is $-GMm/r$;  and its kinetic energy is
$$
T = \frac{1}{2} m \left[ \left( \frac{dr}{dt} \right)^2 + r^2 \left( \frac{d \theta}{dt} \right)^2 \right].
$$
Thus, the total energy is 
$$
E = \frac{1}{2} m \left[ \left( \frac{dr}{dt} \right)^2 + r^2 \left( \frac{d \theta}{dt} \right)^2 \right] - \frac{GMm}{r}.
$$
But we also know that the angular momentum of a particle orbiting a central force is conserved;  and in terms of the particle's motion, it works out to be equal to
$$
L = m r^2 \left( \frac{d \theta}{dt} \right).
$$
This allows us to get rid of the $d \theta/dt$ in our expression for $E$ above:
$$
E = \frac{1}{2} m \left( \frac{dr}{dt} \right)^2 + \frac{L^2}{2 m r^2} - \frac{GMm}{r}.
$$
Now, it's obvious that the first term on the right hand side can't be negative, so we therefore have
$$
E \geq \frac{L^2}{2 m r^2} - \frac{GMm}{r} \equiv U_\text{eff}(r).
$$
This function on the right-hand side is called the effective potential.  Importantly, it has the property that $U_\text{eff}(r) \to + \infty$ when $r \to 0$;  this is because the first (positive) term is proportional to $1/r^2$, while the second (negative) term is proportional to $1/r$, and for sufficiently small $r$ you always have $a/r^2 > b/r$ for any positive $a$ and $b$.  
This means that any particle moving under the force of gravity will have some minimum distance it can get to the origin;  at smaller values of $r$, we will have $U_\text{eff}(r) > E$.  Thus, the particle can never actually get to $r = 0$.  The only loophole is if $L = 0$, in which case this positive term vanishes;  but if that's the case, then you're just dropping the particle straight into the attracting object, which wouldn't qualify as an inward spiral in my book.
You can generalize this argument, by the way, to prove that a particle moving in a potential of the form $r^{\alpha}$ will never hit the central object so long as $\alpha > -2$.  For potentials like $U \propto r^{-2}$ or $U \propto r^{-3}$, though, there do exist trajectories that "spiral in" towards the central object in the way you want them to.
A: When kinetic energy is conserved in this manner, orbits would not follow a spiral pattern. If energy was lost somehow, say for example electromagnetic radiation from a Rutherford-model electron, or gravity waves from an asymmetrical mass distribution orbit, then the orbit could exhibit spiral patterns. But the equation would necessarily be changed to incorporate these effects.
