## Rentry Physics

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thailazer
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### Rentry Physics

A member of a Radio Controlled Aircraft flying forum (link below) asked an interesting question: If you throw a light balsa glider toward earth from the orbiting space shuttle, will it have enough energy to re-enter the atmosphere? If it does, will it burn up or safely glide to earth? Guesses are all over the map so we're looking for a steely-eyed missile man to set us straight.

(http://www.rcuniverse.com/forum/m_10320 ... m#10321478)

BMAONE23
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### Re: Rentry Physics

The Space Shuttle achieves a speed of 17,500 mph upon reaching low earth orbit (LEO). Once in LEO, it, and all that are with it are traveling at 17,500 mph. When you "Throw" the balsa plane from orbital altitude it will continue traveling at 17,500 +/- throwing speed until it reaches atmosphere (it is likely to have a slight increase in speed due to gravity and no atmospheric drag. Once it makes contact with the atmosphere, the drag will increase causing friction which will heat the Balsa wood until it catches fire and burns up.

rstevenson
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### Re: Rentry Physics

If you can throw really hard in the backwards direction -- say, at about 17,500 mph -- so that the glider comes to a dead stop then just drops straight down from LEO, what will its terminal velocity be? (Yeah, I know, zero. But I mean before it hits.)

Rob

neufer
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### Re: Rentry Physics

rstevenson wrote:
If you can throw really hard in the backwards direction -- say, at about 17,500 mph -- so that the glider comes to a dead stop then just drops straight down from LEO, what will its terminal velocity be? (Yeah, I know, zero. But I mean before it hits.)
Such things would reach a maximum velocity in the upper atmosphere of that of a bullet
but I'm guessing that a nylon shuttlecock would probably survive a straight drop from 100 km.
Art Neuendorffer

thailazer
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### Re: Rentry Physics

Forget about throwing something backwards.. That's not the question. It is tossed toward earth at a slow rate (make an assumption like 20 km per hour) from the space shuttle at its orbital velocity. Assume the glider weighs 1 kg on earth and has a wing span of 2 meters, wing chord of 20 cm, and a total tail plane area of 600 square cm. Can a re-entry trajectory occur where it does not burn up with normal mylar and balsa construction? And remember what Lord Kelvin said...."If you can not express the worth of your work with numbers, it is of a meager and unsatisfactory kind."

bystander
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### Re: Rentry Physics

I think you are missing the point. Any object ejected from the shuttle will have the velocity of the shuttle, ~ 17,500 mph. The rest of your specifications become moot. Even if your glider survived the turbulence at these speeds as it entered the atmosphere, it would surely burn up.
Know the quiet place within your heart and touch the rainbow of possibility; be
alive to the gentle breeze of communication, and please stop being such a jerk.
— Garrison Keillor

neufer
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### Re: Rentry Physics

thailazer wrote:
Forget about throwing something backwards.. That's not the question. It is tossed toward earth at a slow rate (make an assumption like 20 km per hour) from the space shuttle at its orbital velocity. Assume the glider weighs 1 kg on earth and has a wing span of 2 meters, wing chord of 20 cm, and a total tail plane area of 600 square cm. Can a re-entry trajectory occur where it does not burn up with normal mylar and balsa construction? And remember what Lord Kelvin said...."If you can not express the worth of your work with numbers, it is of a meager and unsatisfactory kind."
No successful reentry from earth orbit has ever taken place without some sort of heat shield.

Kinetic energy of a 1 kg glider at 7800 m/s orbital velocity = 30,420,000 Joules.

Specific heat capacity of balsa = 2900 Joules/(kg.ºC)

Temperature of reentering balsa glider ~10,500º C (= 30,420,000/2900).
Art Neuendorffer

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### Re: Rentry Physics

So then, for us simple folks, you are saying that it will burn up so fast that we won't even be able to see it go POOF!
To find the Truth, you must go Beyond.

neufer
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### Re: Rentry Physics

Beyond wrote:
So then, for us simple folks, you are saying that it will burn up so fast that we won't even be able to see it go POOF!
Click to play embedded YouTube video.
Art Neuendorffer

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### Re: Rentry Physics

I take it that is a {Q-Q} version of a yes. ((it would have been a better song if it was a no))
To find the Truth, you must go Beyond.

thailazer
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### Re: Rentry Physics

neufer wrote:
thailazer wrote:

Kinetic energy of a 1 kg glider at 7800 m/s orbital velocity = 30,420,000 Joules.

Specific heat capacity of balsa = 2900 Joules/(kg.ºC)

Temperature of reentering balsa glider ~10,500º C (= 30,420,000/2900).
Art... Now we are getting somewhere! Agreed that the energy has to be dissipated before it can touch earth, but that dissipation can occur over a very long time. So it does not have to reach the 10,000 degree temperature. The question still remains on how much deceleration can occur with reasonable heating.

neufer
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### Re: Rentry Physics

thailazer wrote:
neufer wrote:
Kinetic energy of a 1 kg glider at 7800 m/s orbital velocity = 30,420,000 Joules.

Specific heat capacity of balsa = 2900 Joules/(kg.ºC)

Temperature of reentering balsa glider ~10,500º C (= 30,420,000/2900).
Art... Now we are getting somewhere! Agreed that the energy has to be dissipated before it can touch earth, but that dissipation can occur over a very long time. So it does not have to reach the 10,000 degree temperature. The question still remains on how much deceleration can occur with reasonable heating.
Balsa doesn't have to reach ~10,500º C; ~451º F will do quite nicely.
Art Neuendorffer

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### Re: Rentry Physics

451 degrees huh? That means you gotta keep your balsa stuff out of the oven while making PIZZA
To find the Truth, you must go Beyond.

C_Roundy
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### Re: Rentry Physics

Hi, guys, I was just compelled to register here at SA* for starters specifically to see what the brain trust here is doing to vanquish this burning question. I read through the whole thread where it initially came up over on "RC Universe" forum, and obviously the guys over there are um, fairly far removed from being physicists... But, from your posts on the subject here, I feel intuitively that you guys are missing some of the fundamental principles involved too.
Can anyone here prove to me that the variables I specify in my post over there are not centrally relevant, and or framed incorrectly?

Thanks, Chuck
My original post follows,Quote:
---------------------------------------------------------------------------------------------------------------------------------------------------------------------
Just read the whole thread, awsome! Even if you work the numbers to accurately predict the results of variations on this experiment,
(like the Japanese already have before deciding to go ahead and actually fund the experiment), the main range of responses guys have posted still teach an awful lot about the prevalence of certain misconceptions in popular culture!

Example: Assuming a stable orbit at launch that is lower and therefore faster than geo-synchronous, and only slight velocity changes imparted by the launch (like we're talking about here), orbital mechanics 101, or even basic physics shows that:
>Launching inward takes you forward from your launch vehicle's trajectory (not down).
>Launching outward takes you backward.
>Launching forward takes you outward.
>Launching backward takes you inward.
Only 2 or 3 guys out of the first 48 posts seemed to understand this!

Now, here is a little tidbit that may help open the minds of the guys who jump to the "its gonna burn up / blow apart of course" conclusions instead of managing to actually think it through successfully.

>Picture said balsa (or paper etc) airplane having only the mass of one atom instead of one ounce. Our heroic astronaut throws it backward to start its historic journey back down to the rest of our artfull, brilliant tribe of balsa and paper airplane throwing monkeys. Our cutting edge, high tech test vehicle hurtles along at the obligatory bazillion mph forward, and say, a mind blowing 50 mph downward. After a relatively short period (on a scale compared to the age of universe) in which humanity watches Superbowl commercials, eats popcorn, watches paint dry, and takes a nap, eventually our hotrod collides brutally with a whopping atmospheric density of ONE atom per cubic yard, then miles deeper into our gravity well, one atom per cubic foot, miles still further inward, one per cubic inch etc.
Nothing that is not made of unobtanium could possibly survive this kind of soul crushing torture test right?
OH, Wait! I totally forgot! the first atom it hit absorbed half its kinetic energy, and cut its speed by half! (forgive me, I'm assuming it really centerpunched that evil mother'), and again with the next atom, and the next etc.

Now..., can some more of us guys glimpse the deeper understanding that there IS a directly proportional relationship between the re-entry vehicle's [[mass divided by cooling surface area] multiplied by drag coefficient], and how much kinetic heating is caused by re-entry; and consequently whether or not any heat shielding is required to survive the mission?

Really, the main thing left to do is a proper set of engineering calculations to determine exactly what numerical constant (compound ratio) we need to beat for a given re-entry trajectory, to shed heat fast enough to stay under a given material's maximum allowable temperature,
and the Japanese have already done that... They decided to fund the experiment, the principle actually has far reaching potential to lower vehicle and thusly mission costs .............

Remember boys and girls, your mind is a muscle, so always do some stretching before you think you can power lift without hurting yourself!

Best regards to all, Chuck Roundy

rstevenson
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### Re: Rentry Physics

If you're convinced you're right while disengenuously asking for proof that you're wrong, I suspect no proof will suffice.

Be that as it may, there is I think a significant difference between a single atom and a balsa (or Space Shuttle) glider. The latter will push between many of the air molecules, coming into close contact with only a relatively few of them, where the single atom might well behave as you suggest.

I'm very interested in this topic myself and would welcome a link to this Japanese effort you refer to.

Rob

bystander
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### Re: Rentry Physics

Wow, such a simple solution! How could all the aerospace engineers and rocket scientists of the world have missed this.
Congratulations, you have just earned the coveted role as pilot and crew on the first balsa wood glider to be deorbited.
Know the quiet place within your heart and touch the rainbow of possibility; be
alive to the gentle breeze of communication, and please stop being such a jerk.
— Garrison Keillor

Sam
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### Re: Re-entry Physics

rstevenson wrote:I'm very interested in this topic myself and would welcome a link to this Japanese effort you refer to.
http://www.msnbc.msn.com/id/23827045/
"No avian society ever develops space travel because it's impossible to focus on calculus when you could be outside flying." -Randall Munroe

Beyond
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### Re: Rentry Physics

Hey, THAT is quite something, sam. Now they are going to have to look extremly hard to find a 'small' crew to pilot the future ones, if the test from the space station succeeds.
To find the Truth, you must go Beyond.

neufer
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### Re: Rentry Physics

neufer wrote:
thailazer wrote:
Assume the glider weighs 1 kg on earth and has a wing span of 2 meters, wing chord of 20 cm, and a total tail plane area of 600 square cm. Can a re-entry trajectory occur where it does not burn up with normal mylar and balsa construction? And remember what Lord Kelvin said...."If you can not express the worth of your work with numbers, it is of a meager and unsatisfactory kind."
No successful reentry from earth orbit has ever taken place without some sort of heat shield.

Kinetic energy of a 1 kg glider at 7800 m/s orbital velocity = 30,420,000 Joules.

Specific heat capacity of balsa = 2900 Joules/(kg.ºC)

Temperature of reentering balsa glider ~10,500º C (= 30,420,000/2900).
To avoid a temperature of ~10,500º C:
• 1) the glider should be made of a hard elastic material such that most of the 30,420,000 Joules goes into heating the air and

2) the glider should be made of a high temperature material such that the absorbed energy can be radiated away.
Mylar and balsa clearly satisfies neither criteria.

However, a glider of reinforced carbon-carbon would work:
http://en.wikipedia.org/wiki/Reinforced_carbon-carbon wrote:
<<Carbon fibre-reinforced carbon (aka carbon-carbon, abbreviated C/C) is a composite material consisting of carbon fibre reinforcement in a matrix of graphite. It was developed for the nose cones of intercontinental ballistic missiles, and is most widely known as the material for the nose cone and wing leading edges of the Space Shuttle. The Brabham team pioneered its use in the brake systems of Formula One racing cars in 1976; carbon-carbon brake discs and pads are now a standard component of Formula One brake systems.>>
Art Neuendorffer

hstarbuck
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### Re: Rentry Physics

C_Roundy wrote:
>Picture said balsa (or paper etc) airplane having only the mass of one atom instead of one ounce.

...Best regards to all, Chuck Roundy
If it had the mass of one atom wouldn't it be an atom? Or at least a nucleus. And what element would we choose, there are over a hundred. Lets choose lead. Maybe we could use an electron microscope to remove a single atom and then set it free. Now, would the lead atom make it down to Earth? Wait a minute, what does this have to do with a balsa glider??

thailazer
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### Re: Rentry Physics

So still no definitive answer here. As Lord Kelvin once said, "If you can not express the worth of your work with numbers, it is of a meager and unsatisfactory kind." I like the post using an assumption of 1kg, but the little balsa gliders I am referring to are about .5 ounce in weight, so much less "M" but the same "V squared" term. I think it's time for a space walk at the ISS!

neufer
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### Re: Rentry Physics

Click to play embedded YouTube video.
thailazer wrote:
So still no definitive answer here. As Lord Kelvin once said, "If you can not express the worth of your work with numbers, it is of a meager and unsatisfactory kind." I like the post using an assumption of 1kg, but the little balsa gliders I am referring to are about .5 ounce in weight, so much less "M" but the same "V squared" term. I think it's time for a space walk at the ISS!
What you are basically talking about is minimal mass per unit area. That's great for an object moving slowly in the dense troposphere but it becomes less & less of an advantage the higher one goes.

More importantly: it is absolutely useless for anything moving Mach 25 at any altitude
Art Neuendorffer

BMAONE23
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### Re: Rentry Physics

Consider too that most all "Shooting Stars" are Micrometeroid Dust particles to pea sized rocks of space debris that are generally 50 µm to 2 mm in sige and are significantly lighter than a Balsa Glider to begin with. Yet surprisingly none of them survive reentry. (remember, the average pinky nail size is about 8 -10 mm wide) The Balsa craft is much larger than 2 mm or even Atom sized and will encounter a vast quantity of atomic Hydrogen nitrogen and Oxygen before the atmospheric pressure could begin to slow it down. 1 atom strike on the wing will not rob the entire vehicle of momentum, only the specific atom it struck causing the surrounding atoms to heat up