The force of gravity (let's say it's one object orbiting a much more massive object - the sun) is an inward pointing force. That's why angular momentum is conserved - because the force is always parallel to the object's position (relative to the sun) so the torque is zero. And so it's a property of any "central force" potential (a potential that's only a function of $r$, or a force that's only inward/outward). The other day I used it for the potential $e^{-r/a}/r$.
In fact, Kepler's 2nd law is just the conservation of angular momentum. Angular momentum is $mv_\theta *r$ (the theta component of velocity in coordinates centered on the sun times the radial distance to the sun). The area swept in a differential time $dt$ is $v_\theta *r*dt$.
The potential $V\propto-1/r$, $F\propto 1/r^2$ then tells you how the radial position changes over time. Notice that kepler's second law doesn't say anything about $r(t)$. It's just once you have $r$, and lets say you can get the angular momentum from the initial conditions, you can quickly get $v_\theta$. So then $dv_r/dt=F_g/m+v^2/r$. That is, the acceleration $dv_r/dt$ is the gravitational force plus the centrifugal force.
Many central force questions can be answered with just two concepts - the conservation of energy and the conservation of angular momentum (kepler's second law). For example, if the question is "how close does a comet from interstellar space come to the sun given blah blah blah" this can be answered right away by calculating the initial energy and angular momentum of the comet from the initial conditions. Then set $E=U_g(r_{\text{min}})+mv_\theta(r_{\text{min}})^2/2$. Where $U_g$ is the gravitational potential energy, and $v_\theta(r)=v_{\theta\text{ initial}}*r_\text{initial}/r$
I hope this helps a bit. Not sure if I specifically addressed exactly what you wanted me to.
