# How Felix Baumgartner has reached the speed of sound quickly

I have watched Felix Baumgartner freefall; but I wonder how Felix has reached the speed of sound quickly, in a matter of some seconds, then we had no idea of its speed?

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I too wonder about this wikipedia quote "Baumgartner also attempted to break three other world records—the highest manned balloon flight, the highest altitude jump, and the longest time in free fall. The free fall was initially expected to last between five and six minutes; it ended after 4:22." How did they miscalculate that so badly? (Also as a side note, I think longest free fall time is not a particularly interesting thing to measure. I image people in 50 years just sitting around in empty space, far from masses, breaking records.) – Nick Kidman Oct 14 '12 at 20:19

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– Nick Kidman Oct 14 '12 at 20:31
Has anybody posted his velocity profile during the fall? I always wondered about this. – ja72 Jan 5 at 22:56

## 4 Answers

The Red Bull Stratos project involving the 43-year-old Austrian man Felix Baumgartner is to break the sound barrier. Within the first 15,000 feet of his jump he was traveling well over the cruising speed of a commercial jetliner, reaching some 625 mph. The maximum velocity reached by Felix is about some 380 km/s.

How did he do that? During a free-fall, there would be two forces acting on the object. The Air-resistance (or Drag) and Gravity ($mg$). So, Two parts come into our play...

The acceleration due to gravity $(g=\frac{GM}{(R+h)^2})$ value is almost a constant 9.8. Yes, It varies from 9.6 to 9.8 within that 39 km range. For example, at a height of 39000 m, it is some 9.684 and at a height of 10 km, it's about 9.7. At last at the sea level, it is 9.8 as you know. But, this $g$ value has no higher difference than 0.2 (even at 39 kms).

Now, the major part... The air resistance acting on a free-falling object depends on its velocity. As the velocity increases, the drag also increases and a period of time arrives when the body falls with a constant average velocity called terminal-velocity. It's probably around 50 m/s in air. At the terminal velocity, the force due to gravity equals air resistance.

But, His free-fall is from about an altitude of some 120,000 ft. (39 km) from ground. It's in the stratosphere (8-50 kms). In the stratosphere, the pressure is too low which provides the fact that the density of air is too small. This is because the weight acting upon molecules is low for higher molecules and vice-versa. This is where Drag gets weaker. 'Cause the air-resistance also depends upon the density of the medium (air). Hence, only gravity accelerates him down and he'd reach the sound barrier soon. Once he enters our troposphere (up to 8 km from ground), he slows down due to drag and he'd reach 170 m/h which is enough for parachuting, touch-down and landing at last..! On comparison to the effect caused by $g$, I'd say that air resistance plays a major role.

In addition to these, the speed of sound is also a low value at the stratosphere (a supportive factor). Because the pressure wave also depends on the density of the medium. The Laplace correction is given by $v=\sqrt{\frac{\gamma P}{\rho}}$. As the speed is inversely proportional to density $\rho$ or directly proportional to pressure, the $v$ value increases at stratosphere. (i.e.) It's not a constant 330-343 m/s. At about 30 km, it's 305 m/s. This further reduces the effect caused by $g$ difference case..!

Edit: It seems that my approximation of $g$ and $v$ has become such an issue. So, I've updated to a newer version. Some calculators I've used - Variation of $g$ with altitude and Variation of Speed of sound. If we use the Mach value 1.2, we could get the output in that calculator.

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Note that the speed of sound is not constant, but depends on the density, temperature and pressure of the fluid, hence why mission control needed to look at data to determine whether the sound barrier was actually broken during the fall. (Baumgartner himself said that he was too busy tumbling head over feet while falling to notice any sonic boom effects.) – Jerry Schirmer Oct 15 '12 at 18:00
@JerrySchirmer: Yeah, I agree with that, Jerry... It's because that Sound's just a longitudinal pressure wave of vibrations (requires medium) & not even close to EM. Hence, it depends on these factors... I went through Wiki already... – Ϛѓăʑɏ βµԂԃϔ Oct 15 '12 at 18:38
And $g$ is not constant, but varies with altitude. There is also a centrifugal component also, but it is negligible I think. – ja72 Oct 23 '12 at 20:36
@ja72: Hello ja72. I think the variation of $g$ could be approximated to be around 9.8. Ok, As the issue has become somewhat bigger, I've inserted everything... What about now guys? – Ϛѓăʑɏ βµԂԃϔ Oct 24 '12 at 4:16

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@CrazyBuddy: I assure you touchdown against the ground at 170 miles/hour moving straight down, definitely takes form of a splashdown in the sense mentioned in my prior comment. It may be a good speed for opening the parachute but not for landing. – SF. Oct 24 '12 at 20:07
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For N universal gravitational law `g=go(r^2/(r+z)^2)` where `r` is the radius of the earth and `z` is the altitude reached with respect to the earth `crest`, so the gravity is not constant during his fall and doesn't depend on the mass of the body.

As for the high increase in speed, the decrease in the effect of air resistance at z= 39 km causes a rapid increase due to gravitational strength but this increase after a minute starts to fall back till it reaches its max speed. The temperature, pressure and density of air also affect his fall.

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The value $g$ is not a constant $9.8$ $ms^{-2}$ in this case.

Given: His free fall is at a height of $39,045$ $m$, the mean radius of Earth, $6,371,000$ $m$ and the mass of Earth $5.9736\times10^{24}$ $kg$

the value $g=9.7026$ $ms^{-2}$ at the start of his dive and $g=9.8219$ $ms^{-2}$ at the end of his dive. These figures may vary based on the elevation of his landing spot, but it's clear nonetheless that $g$ is a variable in this case..!

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I think you meant for this to be a comment, rather than an answer... – Warrick Oct 15 '12 at 18:47
i think that they had made the astronaut suit heavier just to break the record quickly and to reach the max speed before reaching the dense air. – geogeek Oct 15 '12 at 19:33

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@geogeek the suit being heavier or ligther makes no difference at all. the accel g is equal for any mass, as demonstrated by Galileu centuries ago. – Helder Velez Oct 16 '12 at 1:57

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Of course, I would expect that this would cause less than a 1% error in calculating fall times. – Jerry Schirmer Oct 23 '12 at 22:31

In the differential equation it's not enough to put a constant value for the density of air. If you do so, the limit velocity turns out to be too slow, even if using reasonable values for the cross-section area and for the drag coefficient.

Instead, if you use the experimental data (included in Mathematica), the density of air changes as a function of the altitude. In this case, you see that the solution (obtained via numerical integration) shows a peak in the velocity during the first minute which is in rough agreement with the actual value.

See my post, I explain everything here: http://disipio.wordpress.com/2012/10/21/the-felix-baumgartner-equation/

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## protected by Qmechanic♦Dec 23 '12 at 22:34

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