# Law of Universal Gravitation

1. Nov 23, 2009

### Wildcatfan

Update: Thanks for the help! I'll work on it....

Last edited: Nov 23, 2009
2. Nov 23, 2009

### Identity

I'm not sure what your equation is, but this is how I would derive Newton's Law

$$T^2 = \frac{4\pi^2}{GM}R^3$$

In circular motion,

$$F = \frac{mv^2}{R}$$

The period in circular motion is

$$T = \frac{2\pi R}{v} \Rightarrow v = \frac{2\pi R}{T}$$

So, by substitution,

$$F = \frac{m\left(\frac{2\pi R}{T}\right)^2}{R}=\frac{4\pi^2 mR}{T^2}$$

$$\Rightarrow T^2 = \frac{4\pi^2 mR}{F}$$

Substituting into Kepler's 3rd,

$$\frac{4 \pi ^2 mR}{F} = \frac{4\pi^2}{GM}R^3$$

$$\therefore F = \frac{GMm}{R^2}$$