Breaking the Sound Barrier in Free Fall

Find out where an object or person needs to drop from to exceed the local speed of sound in free fall

On October 14th, 2012 in Roswell, New Mexico, Felix Baumgartner jumped from 39,045 meters, or 24.26 miles, above the Earth. At this altitude, in a layer of the atmosphere called the stratosphere, the air pressure is about 0.33% of the air pressure at sea level. This was important to Felix's goal to break the sound barrier in free fall because the rate of drag is directly related to air pressure. The fact that the speed of sound varies with altitude also helped. At high altitudes, because of the cold and the low density of the atmosphere, sound travels more slowly.

Obtain the air pressure of the atmosphere at Felix's starting altitude. Note that in the stratosphere, the air pressure is only 3.3 millibars:

In[1]:=

StandardAtmosphereData[Quantity[24.26,"Miles"],"Pressure"]

Out[1]=

3.30214

mbar

Use

StandardAtmosphereData

to access and visualize the speed of sound:

In[2]:=

ListLinePlotTable[{Quantity[x,"Miles"],StandardAtmosphereData[Quantity[x,"Miles"],"SoundSpeed"]},{x,0,30,0.5}],



Out[2]=

Next, let's predict the speed that Felix reached in his descent:

Use

FormulaData

to examine the equations of motion. Air resistance is dominated by the quadratic term for high velocities:

In[3]:=

formulas=FormulaData[{"WindPressure","AirDensity"},"Association"]

Out[3]=







Obtain a table of the relevant variables and their physical quantities are:

In[4]:=

FormulaData[{"WindPressure","AirDensity"},"QuantityVariableTable"]

Out[4]=

symbol	description	physical quantity	dimensions
P	pressure	Pressure	{{LengthUnit,-1},{MassUnit,1},{TimeUnit,-2}}
V	wind velocity	Speed	{{LengthUnit,1},{TimeUnit,-1}}
ρ	air density	MassDensity	{{LengthUnit,-3},{MassUnit,1}}
F	wind force	Force	{{LengthUnit,1},{MassUnit,1},{TimeUnit,-2}}
A	area	Area	{LengthUnit,2}
C d	drag coefficient	MultiplicativeConstants	Dimensionless

Substitute the equation for pressure into the second expression to get the force that the air exerts on Felix as he falls.:

In[5]:=

formulas〚2〛/.Normal[formulas]

Out[5]=

The acceleration due to gravity doesn't vary very much over the range the jump: 9.69 meters per second when he jumps to 9.831 meters per second when he lands. This isn't surprising, since relative to the radius of the Earth (3,956.6 miles), 24.26 miles is a short distance.

Visualize how air density varies:

In[6]:=

ListLinePlotTable[{StandardAtmosphereData[Quantity[x,"Miles"],"Density"],Quantity[x,"Miles"]},{x,0,30,0.5}],



Out[6]=

The full equation of motion is:

v

t

=0.5A

ρ(z)

-mg(1)

where m is the mass of Felix and his gear, ρ(z) is the air density which varies with altitude, and g is the acceleration due to gravity, assumed to be constant and equal to its value at sea-level. Rearrange this equation to look at the velocity as a function of altitude:

z

t

v

z

=mv

v

z

=0.5A

ρ(z)

-mg

We will solve the equation numerically. We'll make simplifying assumptions to estimate approximate parameters of the fall, such as mass, cross-sectional area and drag coefficient.

Obtain the air density, ρ(z), and sound speed from StandardAtmosphereData:

In[7]:=

airdensity[z_?NumericQ]:=QuantityMagnitude[StandardAtmosphereData[Quantity[z,"Meters"],"Density"],"Kilograms"/"Meters"^3];(*airdensityinkg/m^3*)

In[8]:=

soundspeed[z_?NumericQ]:=QuantityMagnitude[StandardAtmosphereData[Quantity[z,"Meters"],"SoundSpeed"],"Meters"/"Seconds"]

Estimate approximate values for mass, cross-sectional area and drag coefficient, and taking g to be constant:

In[9]:=

m=90(*massinkilogramsassumingalargeamountofgear*);Cd=1.3(*highendofdragcoefficientforuprighthuman*);A=0.4(*cross-sectionalaream^2assumingheissomewhattumbling*);g=QuantityMagnitude[GeogravityModelData[{33.3734,-104.529,Quantity[0,"Meters"]},"Magnitude"],"Meters"/"Seconds"^2](*takegtobealwaysequaltoitsvalueatsea-level*);

Before we look at how his speed changed through his descent, let's also consider what the maximum free fall velocity as a function of altitude would be. In that case, substituting

v

z

into equation (1) gives:

0=0.5A

ρ(z)

-mg,

since velocity is unchanging. We can rearrange the terms to solve for v:

mg/0.5Aρ(z)

Let's put this all together and see how his fall progressed:

In[10]:=

sol=NDSolve[{mv[z]v'[z]0.5*Cd*A*airdensity[z]*v[z]^2-m*g,v[39045]0.0001},v,{z,0,39045},Method"StiffnessSwitching"];

Visualize the solution for Felix's velocity together with the terminal velocity and the speed of sound. Note that Felix's speed rapidly increases until it meets or exceeds the speed of sound around 32,000 meters. It also briefly exceeds terminal velocity, increasing his instability as the atmosphere decelerates him:

In[11]:=

Plot{v[z]/.sol,soundspeed[z],Sqrt[m*g/(0.5*Cd*A*airdensity[z])]},{z,0,39045},



Out[11]=

	Felix's velocity
	terminal velocity
	speed of sound

Find where the maximum occurs. Note that compared to official reports of about 30,000 meters, the result is off by a couple kilometers, likely due to an inaccurate accounting for the drag coefficient at supersonic speeds:

In[12]:=

maximumz=Maximize[{v[z]/soundspeed[z]/.sol〚1〛,0<z<39045},z]

Out[12]=

{1.2215,{z27838.7}}

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(2)

Solve the differential equation for different initial heights and visualize the results. Note that Felix needed to jump from higher than 33,000 meters in order to achieve supersonic speeds:

In[1]:=

values=Tablesol=NDSolve[{mv[z]v'[z]0.5*Cd*A*airdensity[z]*v[z]^2-m*g,v[height]0.0001},v,{z,0,height},Method"StiffnessSwitching"];height,First@Maximizev[z]soundspeed[z]/.sol1,0<z<height,z,{height,10000,50000,5000}

Out[1]=

{{10000,0.274828},{15000,0.384794},{20000,0.521973},{25000,0.686756},{30000,0.868526},{35000,1.06136},{40000,1.25945},{45000,1.45767},{50000,1.65218}}

In[2]:=

ListLinePlotvalues,



Out[2]=

Or seen another way, check how his speed would have changed over time depending on his initial height:

In[3]:=

sols=Table[NDSolve[{mv[z]v'[z]0.5*Cd*A*airdensity[z]*v[z]^2-m*g,v[x1000+45]0.0001},v,{z,0,x1000+45},Method"StiffnessSwitching"],{x,33,39}];PlotEvaluate[Max[v[z],0]/soundspeed[z]/.sols],{z,0,39045},



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Breaking the Sound Barrier in Free Fall

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Breaking the Sound Barrier in Free Fall

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