# Angular Momentum vs. Force Due to Gravity

I'm getting my feet wet with orbital mechanics and have a very basic question. Kepler's 2nd Law shows that 2 objects in an elliptical orbit sweep out equal areas in equal time, implying objects increase in velocity as they get closer to one another.

This makes sense to me re: inverse square law of gravity - as distance decreases the force of gravity increases. However, the conservation of angular momentum also has a radius component showing that objects spin around a body faster the closer they get to it.

How do I separate these two concepts cleanly in my mind re: orbital bodies?

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.

If a force $$\vec F = \hat n F(r)$$ than $$\vec F$$ is a central force, where $$\hat n$$ is the unit radial vector in spherical coordinates. $$\vec F$$ is only in the radial direction and only depends on the radial distance $$r$$ from the center of the force. For any central force, the angular momentum is constant since the torque from the force $$\vec r \times \vec F = 0$$. Also, a central force is a conservative force.

Using the conservation of energy and the conservation of angular momentum, for any central force it can be shown that the rate at which area is swept out by the radius to a particle acted on by the force is $${L \over {2m}}$$ where $$L$$ is angular momentum and $$m$$ is the mass of the particle. (For the derivation, see the textbook Symon, Mechanics.) So, Kepler's second law is a consequence of the conservation of angular momentum and energy for a central force.

The force of gravity is a special case of a central force where $$f(r) \propto -{1 \over r^2}$$.

In the case of orbital bodies, energy and angular momentum are conserved.

Maybe it is useful to compare with a situation where only energy is conserved: a simple pendulum. The speed increases as the gravitational potential decreases as it happens for orbital movement. But there is not a central force, what is required to conserve angular momentum, as mentioned by the other answers. So, only energy is conserved.