Atmosphere

The atmosphere of a celestial body slows the movement of any object passing through it, a force known as atmospheric drag (or simply drag). An atmosphere also allows for aerodynamic lift. The celestial bodies with atmospheres are the planets Eve, Kerbin, Duna and Jool, as well as Laythe, a moon of Jool. Only Kerbin and Laythe have atmospheres that contain oxygen.

Atmospheric pressure diminishes exponentially with increasing altitude. An atmosphere's scale height is the distance over which atmospheric pressure changes as a factor of e, or 2.718. For example, Kerbin's atmosphere has a scale height of 5000 m, meaning the atmospheric pressure at altitude n is 2.718 times greater than the pressure at altitude n + 5000.

Atmospheres vary in temperature, though this has no bearing on gameplay.

Atmosphere allows aerobraking and easier landing. When atmosphere contains oxygen, it allows jet engines to work. However, atmosphere makes taking off from a planet more difficult and increases a stable orbit altitude.

Drag

File:Ml16-XL parachute.JPG

A Mk1-2 pod with a Mk16-XL parachute being slowed by drag in Kerbin's atmosphere.

In the game, the force of atmospheric drag (F_D) is modeled as follows:

$F_{D}=0.5\,\rho \,v^{2}\,m\,d\,A$

where ρ is the atmospheric density (kg/m³), v is the ship's velocity (m/s), m is the ship's mass (kg), d is the coefficient of drag (dimensionless), and A is the cross-sectional area (m²).

Note that the m term is not present in the real-word drag equation. In the game, this causes acceleration due to drag (a = F_D / m) to be unaffected by a ship's mass. (It also causes the units of the drag equation to have an extra "kg" term.)

ρ can be derived from atmospheric pressure (p), which is a function of the atmosphere's pressure at altitude 0 (p₀) and scale height (H):

$p=p_{0}\cdot e^{-altitude/H}$

$\rho =p\cdot 1.2230948554874\cdot 0.008$

The coefficient of drag (d) is calculated as the mass-weighted average of the max_drag values of all parts on the ship. For most ships without deployed parachutes, d will be very near 0.2, since this is the max_drag value of the vast majority of parts.

As an example, the coefficient of drag for a craft consisting simply of a Command Pod Mk1-2 (mass 4, drag 0.2) and a deployed Mk16-XL Parachute (mass 0.3, drag 500) is:

${\frac {4\cdot 0.2+0.3\cdot 500}{4+0.3}}=35.07$

Note that the game does not calculate the ship's cross-sectional area (A) and assumes a value of 1 instead. This means the net force of drag on a spacecraft is unaffected by the width or orientation of a spacecraft, even if the spacecraft has very large wings. (Wings do affect the force of lift, however.)

Terminal velocity

The terminal velocity of an object falling through an atmosphere is the velocity at which the force of gravity is equal to the force of drag. Terminal velocity changes as a function of altitude. Given enough time, an object falling into the atmosphere will slow to terminal velocity and then remain at terminal velocity for the rest of its fall.

Terminal velocity is important because:

1. It describes the amount of velocity which a spacecraft must burn away when it is close to the ground.
2. It represents the speed at which a ship should be traveling upward during a fuel-optimal ascent.

The force of gravity (F_G) is:

$F_{G}=m\,a=m\,{\frac {GM}{r^{2}}}$

where m is still the ship's mass, G is the gravitational constant, M is the mass of the planet, and r is the distance from the center of the planet to the falling object.

To find terminal velocity, we set F_G equal to F_D:

$m\,{\frac {GM}{r^{2}}}=0.5\,\rho \,v^{2}\,m\,d\,A$

${\frac {GM}{r^{2}}}=0.5\,\rho \,v^{2}\,d\,A$

$v=v_{T}={\sqrt {\frac {2\,GM}{r^{2}\,\rho \,d\,A}}}$

Assuming d is 0.2 (which is a good approximation, provided parachutes are not in use) and given that A is 1, this simplifies to:

$v_{T}={\sqrt {\frac {10\,GM}{r^{2}\,\rho }}}$

For the Mk16 pod and parachute example pictured above, the drag coefficient is 35.07, so its terminal velocity at sea level on Kerbin (which is 600 km from Kerbin's center) is:

$v_{T}={\sqrt {\frac {2\,GM}{r^{2}\,\rho \cdot 35.07}}}$

$\rho =1\cdot e^{-0/5000}\cdot 1.2230948554874\cdot 0.008$

$v_{T}={\sqrt {\frac {2\cdot 6.674e\cdot 10^{-11}\cdot 5.2915793\cdot 10^{22}}{600000^{2}\cdot 1.2230948554874\cdot 0.008\cdot 35.07}}}=7.56\operatorname {m/s}$

Examples

Altitude (m)	vT (m/s)
	Eve	Kerbin	Duna	Jool	Laythe
0	58.385	100.13	212.41	23.124	115.62
100	58.783	101.01	214.21	23.162	116.32
1000	62.494	109.30	231.16	23.508	122.83
10000	115.27	240.52	495.18	27.272	211.77