- #1
Anama Skout
- 53
- 13
Straight from my physics textbook:
I want to know how can one - formally - derive ##\sf eq.(6.7)##? (they didn't show a derivation for that formula) And what's the intuition behind it? For instance why is ##\sf R\propto v^2## and not ##\sf R\propto v##? Where does that ##\sf \frac12## coefficient comes from?
For objects moving at high speeds through air, such as airplanes, skydivers, cars, and baseballs, the resistive force is reasonably well modeled as proportional to the square of the speed. In these situations, the magnitude of the resistive force can be expressed as $$\sf R=\tfrac12D\rho Av^2\tag{6.7}$$ where ##\sf D## is a dimensionless empirical quantity called the drag coefficient, ##\sf r## is the density of air, and ##\sf A## is the cross-sectional area of the moving object measured in a plane perpendicular to its velocity. The drag coefficient has a value of about ##\sf 0.5## for spherical objects but can have a value as great as ##\sf 2## for irregularly shaped objects.
I want to know how can one - formally - derive ##\sf eq.(6.7)##? (they didn't show a derivation for that formula) And what's the intuition behind it? For instance why is ##\sf R\propto v^2## and not ##\sf R\propto v##? Where does that ##\sf \frac12## coefficient comes from?