$\gamma$ in Newton's Second Law of Motion in Differential Form I am teaching myself Differential Equations from a website. In the website I am up to Direction Fields and an example of a differential equation is Newton's Second Law of Motion. It is written on the website like this:
$$m\frac{dv}{dt}=mg-\gamma v$$
I know that $m$ is the mass, $g$ is the gravitational acceleration, $v$ is the velocity, and $t$ is the time, but what does $\gamma$ stand for?
 A: The general form of Newton's Law for constant mass is
$$
m \frac{dv}{dt} = F
$$
so in your case, $F = mg - \gamma v$ is the provided force law. In your case your force happens to depend on the velocity; the greater the velocity, the more negative the force, so it is a kind of friction or drag. $\gamma$ is just the proportionality constant between the friction force $F_{fric} = -\gamma v$ and the velocity, just like $k$ is the proportionality constant between the position and the force $F_{spring} = -kx$ for a spring.
A: The $\gamma$ here is some coefficient/constant (which has the units $\mathrm{kg\,s^{-1}}$). This formula is saying that the resultant force $F = \mathrm{d}v/\mathrm{d}t$ is equal to the force on the mass from gravity minus some force $\gamma v$ which is dependent on velocity. 
To me, you are right, this does not make too much sense from a Newtonian standpoint (IMO) as this is representing a drag linearly dependent on velocity which is never the case in my experience. It would make more sense if the resistive force was written as $\lambda v^{2}$ (and the units of $\lambda$ amended accordingly for some arbitrary example case), this could then represent an air/fluid resistance where $\lambda$ is some type of 'drag coefficient'.
I hope this helps.
