Atmospheric Pressure $\rho$ = $$ \rho = p_0 \cdot (1 - \frac{L \cdot h}{T_0} )^\frac{g \cdot M}{R \cdot L} $$ where

  • $\rho$ = pressure at altitude
  • $p_0$ = standard pressure at sea level (101325 Pa)
  • $L$ = Temperature lapse rate (0.0065 K/m)
  • $h$ = The altitude in question
  • $T_0$ = standard temperature at sea level (288.15 K)
  • $g$ = Earth surface gravitational acceleration (9.80665 $m/s2$)
  • $M$ = Molar mass of dry air (0.0289644 kg/mol)
  • $R$ = Universal gas constant (8.31447 J/(mol•K))

Terminal Velocity $v$ = $$ v = \sqrt{\frac{2\cdot m \cdot g}{\rho \cdot A \cdot C_d}}$$ where

  • $v$ = velocity at the specified altitude
  • $m$ = Mass of falling object
  • $g$ = gravitational acceleration
  • $A$ = projected area of the object
  • $C_d$ = Coefficient of drag of the object
  • $\rho$ = Density of the fluid through which the object is falling

Density of Air $\rho$ = $$ \rho = \frac{p}{R_{specific} T} $$ where

  • $p$ = Absolute pressure
  • $R_{specific}$ = Specific gas constant for dry air (287.058 J/(kg·K))
  • $T$ = Absolute temperature

Gravity at Altitude $g$ = $$ g_h = g_0 \cdot (\frac{R_e}{R_e + h})^2 $$ where

  • $g_h$ = Gravity at elevation
  • $g_0$ = Gravity at surface
  • $R_e$ = Mean radius of the Earth
  • $h$ = Elevation in question

Will need to combine these four functions over time and altitude (third order diff-eq?) to get the result.

Sources:

eli5/redbull.txt · Last modified: 2017/02/05 22:43 (external edit)
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