# Mach 3

1. Dec 8, 2009

### delve

Here is an answer from a book of mine: From the force diagram we have $$N-mg=(\frac{mv^2}{R})e_r$$. The acceleration that the pilot feels is $$\frac{N}{m}=g+(\frac{mv^2}{R})e_r$$. I'm confused though; what happened to the g being divided by the last term as well?

2. Dec 8, 2009

### tiny-tim

Hi delve!
I'm not following you …

the equation is equivalent to N/m = g + (v2/R)er

what do you mean by g being divided by something?

3. Dec 8, 2009

### delve

Actually, you just helped me realize what I was confused about. Thank you!