Atmospheric drag
Part I: Sub-sonic, Low Altitude
This page originally compiled Feb 2007 by TFR
It is well known that bodies falling through the Earth's atmosphere do not, in fact, continue to accelerate at the rate g suggests. Such bodies reach a terminal velocity when atmospheric drag increases to the point where it equals -g, and at that point acceleration drops to zero and velocity remains constant.
The drag equation is Fd = 1/2ρv2ACd, where:
Fd is the force of the drag on the object
ρ is the density of the medium through which the object is moving, about 1.2 kg/m3 for low-altitude air
v is the velocity of the object through the medium
A is the cross-sectional area presented to the medium
Cd is the drag coefficient, a number describing the smoothness or degree of streamlining of the object
The variable A is the area facing forward into the flow. For instance, a 3 cm cube which presents one of its faces forward would have an A of 3 × 3 = 9 cm2; a sphere 3 cm in diameter would present an area of π × 1.52 = 7.07 cm2; and so on. The coefficient Cd nominally describes how air flows around the object, a flat plate being, in theory, the reference of 1.0. A real flat plate spills around the edges as it passes through the medium, and then leaves a turbulent wake, so its Cd will be different. Our 3 cm cube roughly approximates a flat plate, so we could guess its Cd at approximately 1.0. Lists of Cd for various shapes are available on the web.
Newton's second law is F = ma, where:
F is the force on the object
m is the mass of the object
a is the acceleration caused by the force
Solving this for a = F/m, we can substitute to form an expression for the acceleration of the atmospheric drag:
ad = [ ρv2ACd ] / 2m
Equating this to g and solving for v, g = [ ρv2ACd ] / 2m, 2mg = ρv2ACd, v2 = [ 2mg ] / [ ρACd ],
vterminal =
√[ 2mg ] / [ ρACd ],
giving us an expression to directly calculate the terminal velocity.
We can now calculate the effect of drag on the velocity of a falling object. We must plug our information into Newton's law, summing together the forces acting on the object, in this case its weight (mg) and the force of the drag (1/2ρv2ACd), to eventually give us the combined acceleration:
ma = -mg + 1/2ρv2ACd,
Solving for a, simplifying, and combining with the equation for vterminal we get:
a = -g(1 - v2 / vt2)
This gives us a useful expression for acceleration of a body subject to gravity and air drag. To solve the problem of velocity and position, we need to resort to calculus which we won't go into here. Integrating the acceleration equation, we get velocity:
v = -vt tanh(gt / vt), and then integrating the velocity equation, we get position:
s = (-vt2 / g) ln[cosh(gt / vt)]
In practice, these formulae are of limited use due to the assumptions made: constant ρ, which actually varies with altitude, and constant Cd, which varies with velocity as one approaches the speed of sound. At low altitudes and airspeeds these should produce a good estimate. For our purposes, where we will be simulating the entirety of the atmosphere from bottom to top, and speeds from zero to hypersonic, we will have to use a numerical method to account for all of these variables. More on that in "Atmosphere Part II" and "Part III".
Examples:
A baseball is dropped from a height.
m = 0.142 kg, diameter = 7.3 cm, we guess Cd = 0.3
What terminal velocity does it eventually achieve?
vterminal =
√[ 2 × 0.142 × 9.80 ] / [ 1.2 × 0.00419 × 0.3 ] = 43.0 m/s
What is its velocity after 5 seconds, accounting for atmospheric drag?
v = -43.0 × tanh[(9.80 × 5) / 43.0] = -35.0 m/s (compare to 49.0 m/s if we neglect air drag)
How far has it fallen by then?
s = (-43.02 / 9.80) × ln{cosh[(9.80 × 5) / 43.0]} = -103 m (compare to 123 m if we neglect air drag)
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Last Revised: Mar 2007