Solution for Circular Orbit

Suppose that we have an object with mass $M$ and an object with mass $m$ that is rotating around the other in a circle with radius $R$ centered on object $M$. We would like to know a lot of things about this system.

First notice that the motion must be not only circular, but also uniform. The Conservation of Angular Momentum tells us very quickly that if object $m$ is moving in a circle, it must maintain the same speed.

The centripetal force is the same as the gravitational force. Thus, we may quickly calculate the speed: $$F_{cent}=F_{grav}$$ $$\frac{mv^2}{R}=\frac{mMG}{R^2}$$ $$v=\sqrt{\frac{MG}{R}}$$ Note that the velocity is not dependent on the mass $m$. This is fairly typical of gravity-dependent problems.

Now, we may quickly calculate the kinetic energy: $$K=\frac{1}{2}mv^2$$ $$K=\frac{mMG}{2R}$$ But wait! We already know the following potential energy formula: $$U=-\frac{mMG}{R}$$ This means that the kinetic energy is $-\frac{1}{2}$ times the potential energy! $$E_{total}=\frac{1}{2}U=-K$$ If the kinetic energy were $2$ times as much, meaning that the velocity would be $\sqrt{2}$ times higher, the two would add up to zero! What does this mean?

It turns out that the total energy equalling zero means that the object $m$ will break free of the object $M$ and fly away.

This gives us the notion of an escape velocity: $$v_{esc}=v\sqrt{2}$$ $$v_{esc}=\sqrt{\frac{2MG}{R}}$$ This is true even if the object is not already in motion; for example, one can calculate the velocity needed to escape Earth's pull from the surface, and this value is very important for all spacecraft.

One can also calculate the period and frequency: $$T=\frac{2\pi R}{v}$$ $$T=2\pi\sqrt{\frac{R^3}{MG}}$$ $$f=\frac{1}{T}$$ $$f=\frac{1}{2\pi}\sqrt{\frac{MG}{R^3}}$$ The period is generally expressed not purely, but as its square: $$T^2=\frac{4\pi^2 R^3}{MG}$$ The reason for this will become more clear with the discussion of more general elliptical orbits and Kepler's Laws.