I am seriously confused. We are working on a supplement mineral product. This product is about the surface of the particles, but is sold in grams. So, when you make the particles smaller, the consumer needs to ingest less mass. So my question is, how much does decreasing the radius of the particles (assuming balls) contribute to increasing the total surface area in a given grammage?
For example: Decreasing the particle size from 9 to 6 microns would about double the surface pro gram is this right?
Assume that all balls have the same radius $r$ and density $\rho$. Then the number of balls ($M$ is the total mass) is: $$ N=\frac{M}{\frac{4\pi}3\rho r^3} $$ and the overall area: $$ A=N 4\pi r^2=\frac{3M}{\rho r}. $$ so that the overall surface is proportional to $\frac1r$.
In this derivation I neglected the fact that $N$ has to be integer. For large $N$ it should be an acceptable approximation.