Some physics/math help

[IKE]

Originally Posted by francistexas

The page says, "There is a change of 1 mile for every 2 feet per second (fps) change in velocity when you are below a 500-mile altitude above the Earth." Sorry if I wasn't clear. I didn't mean they actually gave me two ratios.

I think it's much easier than it is made out to be, and others have mentioned that using that exact equation isn't even necessary. At the moment, I'm just stumped on the velocities.

EDIT: Delta-V, change in velocity, is defined as (Vf-Vo), correct? If so, would Delta-V or the 'a' in the above mentioned formula be the given ratio?

a = delta V/t.


Say initial altitude is 500 miles and final altitude is 400 miles, the change in altitude would be 100 miles therefore the change in velocity would be:
(100 miles)/(1 mile/2 fps)=200 fps change in velocity.