Monday, October 15, 2012

The Physics Behind the Red Bull Stratos Project


The Physics Behind the Red Bull Stratos Project
by Chelsea Gottschalk


            As mentioned in class and outlined in a NY Times article, Felix Baumgartner recently became the first man to break the sound barrier by gravity alone. Felix was taken to a vertical height of 128,100 ft (24 miles) by a hot air balloon, and then jumped down to the ground, using a special space-like suit and a parachute. Part of this mission, called the Red Bull Stratos project, was to test new spacesuits as well as escape tactics for extreme altitudes. Although the figures are not exact yet, it was estimated that Felix dropped when he was 128,100 ft above the ground. He fell for approximately 4 minutes (240 seconds), reaching a maximum speed of 833.9 mph (372.787 m/s) before deploying a parachute about 1 mile (1609.3 m) above a New Mexico desert.

         Let’s take a look at the physics behind Felix’s super stunt.

Let’s assume that Felix has a mass of approximately 75 Kg and his space suit weighs around 70lbs (31.75Kg) (around what some models weigh), therefore: Total mass= 106.75 Kg

Also, since Felix will be constantly accelerating, the maximum speed will be equal to the point just before he opens his chute

There are two reference frames that we should assess:

1) The descent from the highest point to right when Felix deploys his parachute (this represents the total time under complete free fall) (37,435.5 m)
2) The descent from the point Felix opens his parachute to the time he touches the ground (1609.3 m)

Let’s look at the free fall first:

We know that Wnet=Wc+Wnc, where the Wnc represents the air resistance encountered from the fall

From the law of conservation of energy:

KEi + PEi = KEf + PEf – Wnc

1/2mv^2 + mgh = 1/2mv^2 + mgh –Wnc

Vi= 0 m/s                hi= 37,435.5 m
Vf= 372.787 m/s     hf = 0 m

0 + (106.75)(9.8)(37,435.5) = (1/2)(106.75)(372.787)^2 – Wnc

Wnc = -3.17 x 10^7 J of air resistance

F x d = -3.17 x 10^7 J   d=37,435.5 m

Fair= 846.79 = 847 N of air resistance opposing the direction of motion

Now, let’s look at what Felix’s max speed would be if there were no nonconserved forces:

Vf ^2 = Vi^2 + 2ay

Vf= sqrt (9.8 x 37435.5)

Vf= 856.58 = 857 m/s, over 2x as fast!

The speed of sound is 340.29 m/s…..

Now, let’s look at the second reference frame, from the moment Felix opens his parachute until he reaches the ground, covering a height of 1 mile, or 1609.3 m

We can use the same equation as before to find the force of the drag produced by the parachute:

KEi + PEi = KEf + PEf – Wnc

1/2mv^2 + mgh = 1/2mv^2 + mgh –Wnc

Vi = 372.787 m/s    hi= 1609.3 m
Vf= 0 m/s                hf= 0 m

(106.75)g(1609.3) + (1/2)(106.75)(372.787)^2 = 0 + 0 – Wnc

Wnc= -9.1 x 10^6 J

F x d = -9.1 x 10^6 J   d=1609.3 m

Fparachute= 5654.63 = 5650 N of force in the direction opposing motion- that is a lot!

Let’s look at the magnitude of the deceleration to see how it compares to the limits of the human body (5g’s before loss of consciousness):

Vf ^2 = Vi^2 + 2ay

0 = 372.787^2 + 2(1609.3)a

a= -43.1772 = -43.2 m/s^2 / 9.8 = 4.4g’s
It looks like Felix was very close to approaching the limit of the human body, but he survived, no doubt due to the careful physical calculations of this awesome stunt!

References: http://www.nytimes.com/2012/10/15/us/felix-baumgartner-skydiving.html?_r=0

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