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Space Shuttle Debris

Date: 06/24/2003 at 18:13:30
From: Scott
Subject: Space Shuttle Debris

Why are they shooting it at the wing at 550 mph to assess damage?
How do you get anywhere near 550 mph?

If the space shuttle is lifting off, how fast and with how much actual 
force f = m x v can a 1.68 lb of foam hit the shuttle? The length from 
the tip of where the fuel tank attaches to the wing is only around 50 
ft. They are "firing" this piece of foam at around 550 mph. If the 
shuttle is traveling 500 mph (I don't know the actual speed straight 
up, but I'm giving huge leeway here), the foam is travelling at 550 
mph at the time of separation. It has 50 feet to fall prior to 
striking the craft. Gravity plays a part, but we all know about 
terminal velocity and there is no way it gets there - coefficient of 
drag plays a part - but to accelerate to 550 mph seems mind boggling.


Date: 06/25/2003 at 13:57:58
From: Doctor Rick
Subject: Re: Space Shuttle Debris

Hi, Scott.

That's an intriguing question! I can't answer everything 
definitively, but I can make some rough calculations that might tell 
us something.

First, I found a transcript from the Challenger explosion, for 
comparison, indicating that at 75 seconds after liftoff, that vehicle 
had a velocity of 2900 feet/sec = 1980 miles/hour. I assume that at 81 
seconds, when the foam broke off, Columbia was traveling at least that 
fast. It's faster than you thought, isn't it?

There are at least three factors that would accelerate the foam 
relative to the wing once it broke off. One is gravity acting on the 
foam. Another is the acceleration of the shuttle due to the rocket 
engines, in which the foam no longer shared. 

The third factor is the force of the air on the foam, which was 
traveling some 2000 miles/hour relative to the air. This is equivalent 
to a stationary object being hit by a blast of wind going 200 miles 
per hour. I have no experience with such winds, but let's just see 
what acceleration would be required to bring the foam to 550 miles/
hour in 50 feet.

We have the formulas

  v = at
  s = (1/2)at^2

which we can solve to find a when we know v (550 mi/hr = 800 ft/sec) 
and s (50 ft).

  s = (1/2)a(v/a)^2
  s = v^2/(2a)
  a = v^2/(2s)
    = 800^2 / (2*50)
    = 6400 ft/sec^2

which is 200 G's, so it does sound like a lot - but then it's a strong 
wind! One thing this number makes clear is that the first two factors 
in accelerating the foam relative to the wing are insignificant. The 
acceleration of the shuttle must be on the order of 1 G, and of course 
the acceleration of gravity is 1 G.

The force required to cause this acceleration is found from the 
equation

  F = ma
    = (1.68 lb)(6400 ft/sec^2)

The English units get very confusing, but to put it in terms of 
"pounds force," we just note that we have 200 times the acceleration 
of gravity, so it is equivalent to gravity acting on a mass 200 times 
as great. The force is 200 * 1.68 lbf = 336 lbf.

One formula I found on the Web may help us see if this magnitude is 
reasonable. The wind pressure on an object is

  P = (1/2)(density of air)(v^2)(shape factor)

where the shape factor depends on the shape and orientation of the 
object but is of order of magnitude 1. The density of air (at sea 
level) is listed as 1.25 kg/m^3, which comes out to about 0.078 
lb/ft^3. With these figures, the wind pressure on the foam would be 
something like

  P = (1/2)(0.078 lb/ft^3)(2900 ft/sec)^2
    = 327990 lb-ft/sec^2/ft^2

which is about 10,000 lbf/ft^2. How big was this piece of foam? If 
it's a square foot in cross-section area, we'd have 10,000 lbf, 
considerably more than the 336 lbf we needed to account for.

Of course, I have ignored the fact that the shuttle was maybe 7 miles 
up at the time, so the air density was considerably less. But I'll 
leave that calculation to the reader, okay?

My calculations may be way off; I won't guarantee that I didn't make 
a major error somewhere.

One last point: The force of impact on the shuttle is not figured by 
the equation F = mv. As I said above, the correct equation is F = ma, 
but here the acceleration is actually the DECELERATION by which the 
wing stops the debris. We don't know enough about the details of the 
impact to make this meaningful. What I can do is compare it to more 
familiar collisions. One relevant factor is the kinetic energy of the 
foam:

  KE = (1/2)mv^2
     = 0.5(1.68 lb)(800 ft/sec)^2
     = 537600 lb-ft^2/sec^2

We can compare this with the kinetic energy of a typical car hitting 
a tree. Let's say the car has a mass of 2000 lb (I'm not sure how 
typical this is). To get the same kinetic energy as the foam, the car 
would have to be going at a speed given by

  537600 = (1/2)(2000 lb)v^2
  v = sqrt(537600*2/2000)
    = 23 mi/hr

We can at least say that the impact of the foam would be greater than 
what automobile bumpers are designed to withstand. What the shuttle 
wing should be able to withstand, I don't know!

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Physics
High School Physics/Chemistry

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