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I heard about parachute jumping at an initial altitude of 30km. I want to simulate this flight numerically. How could I simulate the air drag (I mean, Which equation gives the air drag)? Normally I use,

$$\frac{\text{d}v}{\text{d}t}=a -bv^2$$

where $a$ is the acceleration and $b$ is the drag coefficient.

In this case, how does air drag coefficient variate with altitude?

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you might want to take And in place of the density you put it in the – miceterminator Sep 1 '12 at 20:41
Actually the drag coefficient does not change, the density does. See answer below that contains $\rho$ as density. – ja72 Sep 2 '12 at 15:15
up vote 1 down vote accepted

Drag force $F = C_d \frac{1}{2} A \,\rho_{\rm AIR} v^2$ and equations of motion $ m g - F = m \dot{v}$. Here $C_d$ is the coefficient of drag, $\rho_{\rm AIR}$ is the density of air (as a function of temperature and altitude), $v$ is velocity, $\dot{v}$ is acceleration, $m$ is mass, $A$ is the swept area (changes with orientation) and $g$ is gravity.

To solve $\dot{v} = g - k \frac{v^2}{2}$ with $k=\frac{\rho\,A\,C_d}{m}$ you calculate

$$ \Delta t = \int \frac{1}{\dot v}\,{\rm d} v + K_1 = K_1 - \frac{1}{\sqrt{2 g k}} \ln\left( \frac{\sqrt{\frac{k v^2}{2 g}}-1}{\sqrt{\frac{k v^2}{2 g}}+1} \right) $$ $$ \Delta x = \int \frac{v}{\dot v}\,{\rm d} v + K_2 = K_2 - \frac{1}{k}\ln \left(k \, v^2-2 g\right)$$

With appropriate initial conditions you solve for $K_1$ and $K_2$.

Note that if the final speed is $v_f = \sqrt{ \frac{2 g}{k} } = \sqrt{ \frac{2 m g}{A C_d \rho_{\rm AIR}}}$ then the velocity profile from standing still is

$$ v(t) = v_f \frac{ e^{2 g t/v_f}-1 }{ e^{2 g t/v_f}+1 } $$

with initial slope $\dot{v}(0) = g $.

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Thank you. But in fact I want to know how the fluid density change with altitude. – R. M. Sep 1 '12 at 23:06
Maybe you should edit the question to reflect that. – ja72 Sep 2 '12 at 0:54

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