First Public Shuttle Engine Test 14
Guppy06 writes: "NASA's John C. Stennis Space Center (Mississippi, near the gulf coast) will be opening its doors to the public for the first time this Saturday. As part of its celebrations of the 20th anniversary of the first space shuttle launch (as well as flight-certifying a Pratt & Whitney fuel turbo pump), there'll be a 520-second static test of an SSME around 2000 CDT. Translating that into English for the non space geeks, that means they'll be lighting up a space shuttle main engine (attatched to a large steel frame, grounded in a big chunk of concrete so it doesn't go anywhere) for about 9 minutes around 8:00. The press release is available here. Now if only they did stuff like this more often, there might be more interest in NASA ..."
Re:Addendum (Score:1)
I didn't want to write anything long because your posts will get archived while mine will disappear (unless you quote what I say). But given the effort that you put, I think that you deserve something in return.
You're not hopeless. You're quite knowledgeble, and you can undoubtedly solve many physics problems involving mass, momentum, and the like. But that's not enough. To be a Jedi, you need the feel conservation of energy and conservation of momentum as second nature. These are the two most important concepts in mechanics, and will save your butt countless times.
You mentioned 5 other things to take into account, and accurately said that it wasn't an exhaustive list, but sometimes there's a trick. If I ask you what is (exp(sin(Pi/9))+sqrt(17+log(23))) - (exp(sin(Pi/9))+sqrt(17+log(23))), you're still busy parsing the first few calculations while anyone with a global view will immediately say this is equal to zero, without even worrying whether I meant log base-10 or base-e.
Same here. In the rocket engine problem, you put a big box around the Earth such that the boundary of the box will be space. That's a closed system. The total momentum inside of that box doesn't change, as long as nothing gets out of the box (which is the case since I chose a large box, going halfway to the moon, say). Conservation of linear momentum says that the integral of density(x,y,z)*velocity(x,y,z) over that box remains a constant, unchanged by rocket engines, volcanos, or people jumping up and down.
Now you can say "ah ha but I don't count people or the air to be part of Earth". Fine, but you still have: integral of density(x,y,z)*velocity(x,y,z) over the Earth + integral of density(x,y,z)*velocity(x,y,z) over not-Earth = constant. In other words, the average velocity of the Earth times Earth's mass + the average velocity of the non-Earth times non-Earth is a constant. But these two average velocities are equals, else the Earth and people would be spreading apart and be completely separated given enough time. It really makes no difference if you don't count people and air, as long as people and air don't fly off the Earth.
The upshot: Earth's orbit won't change at all, and you don't need to worry about the deformation of the steel frame and concrete block or anything else to get that answer. That is why you should worship conservation of momentum as your new God.
Previous post is worth archiving... (Score:2)
This is it:
Who cares about the position of the Earth right after the push? The main thing happening is a change of velocity of v=a*520. The difference in position is then a function of time, roughly x(t) = v*t. (think t = days or weeks)
I didn't want to write anything long because your posts will get archived while mine will disappear (unless you quote what I say). But given the effort that you put, I think that you deserve something in return.
You're not hopeless. You're quite knowledgeble, and you can undoubtedly solve many physics problems involving mass, momentum, and the like. But that's not enough. To be a Jedi, you need the feel conservation of energy and conservation of momentum as second nature. These are the two most important concepts in mechanics, and will save your butt countless times.
You mentioned 5 other things to take into account, and accurately said that it wasn't an exhaustive list, but sometimes there's a trick. If I ask you what is (exp(sin(Pi/9))+sqrt(17+log(23))) - (exp(sin(Pi/9))+sqrt(17+log(23))), you're still busy parsing the first few calculations while anyone with a global view will immediately say this is equal to zero, without even worrying whether I meant log base-10 or base-e.
Same here. In the rocket engine problem, you put a big box around the Earth such that the boundary of the box will be space. That's a closed system. The total momentum inside of that box doesn't change, as long as nothing gets out of the box (which is the case since I chose a large box, going halfway to the moon, say). Conservation of linear momentum says that the integral of density(x,y,z)*velocity(x,y,z) over that box remains a constant, unchanged by rocket engines, volcanos, or people jumping up and down.
Now you can say "ah ha but I don't count people or the air to be part of Earth". Fine, but you still have: integral of density(x,y,z)*velocity(x,y,z) over the Earth + integral of density(x,y,z)*velocity(x,y,z) over not-Earth = constant. In other words, the average velocity of the Earth times Earth's mass + the average velocity of the non-Earth times non-Earth is a constant. But these two average velocities are equals, else the Earth and people would be spreading apart and be completely separated given enough time. It really makes no difference if you don't count people and air, as long as people and air don't fly off the Earth.
The upshot: Earth's orbit won't change at all, and you don't need to worry about the deformation of the steel frame and concrete block or anything else to get that answer.
That is why you should worship conservation of momentum as your new God.
Good for future generations (Score:1)
marshmallows on a stick? (Score:3)
Guppy06 - please spell check.
Sounds like fun... (Score:3)
FWOOSH!
Tour guide: Er... we have, or rather had, the souvenir stand. Thanks you all for coming and, uh, here's a free pin for y'all, and please let's keep this a secret betwen ourselves, OK?
Tongue-tied and twisted, just an earth-bound misfit, I
Re:Addendum (Score:1)
Important Questions (Score:2)
Will having all the people in India stamp their feet for those 9 minutes cancel the effect?
Re:marshmallows on a stick? (Score:2)
Actually, from what I've heard, there'll be a good deal of water spray. Water makes a good coolant and shock absorber, you know...
Guppy06 - please spell check.
Where's the fun in that? And what's the point in spell-checking when /. is never going to post it, anyway? Oh, wait a sec...
Re:Important Questions (Score:2)
Hrmm... an SSME puts out 1862 kilonewtons of thrust at sea level. The earth is about 5.92E24 kilograms. Newton tells us that the earth will experience an acceleration of 3.15E-19 m/s^2, or 0.000000315 picometers per second per second. Newton also tells us that the 520 second firing time means that the earth will move 0.000425 picometers, or 1.67 trillionths of an inch. Not accounting for the angle the engine will sweep as the earth turns, of course...
"Will having all the people in India stamp their feet for those 9 minutes cancel the effect?"
We'll use conservation of momentum. 1862 kilonewtons for 520 s is 9.68E8 kg*m/s.
Let's say the average person masses in at 75 kilograms and can jump 30 cm up in the air. Newton tells us that they will be leaving the ground at 2.43 m/s. 75 kg moving at 2.43 m/s is 181.9 kg*m/s. We'll assume for simplicity's sake that they can jump once per second. We'll also assume that it's enough time to gently absorb the shock of their landing with their knees, making the momentum change of the landing negligible compared to the jumping.
At that rate, it will take about 5,322,000 people to counteract the rocket. According to the CIA World Factbook, the nearest match is Denmark's 5,336,394.
Unfortunately for the Danes, in order to work, they'll have to be jumping on the other side of the world as the engine. Stennis Space Center is located at 30.369 degrees north latitude, 89.613 degrees west longitude. The opposite side of the earth would be 30.369 degrees south latitude, 90.387 degrees east longitude, or about half-way between Bangladesh and Antarctica, in the middle of the Southern Indian Ocean. Only about a 1400 mile swim due east to Perth, Australia, though...
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Ask a silly question, inspire somebody to give you a serious answer...
Addendum (Score:2)
Re:Addendum (Score:2)
Re:Addendum (Score:2)
It's called "normal differential equations."
Acceleration = a
Velocity = v
Displacement = x
Time = t
v = dx/dt
a = dv/dt
v(0) = 0
x(0) = 0
dv = a*dt
v = integral of a * dt
v = at + C (something Galileo figured out w/o calculus)
v(0) = 0
0 = a*0 + C1
C1 = 0
v = at
dx/dt = at
dx = at*dt
x = integral of at*dt
x = 0.5*at^2 + C2 (something Galileo figured out w/o calculus)
x(0) = 0
0 = 0.5*a*0^2 + C2
C2 = 0
x = 0.5*at^2
a(t) = 3.15E-19 m/s
t = 520 s
x = 0.5*(3.15E-19 m/s)*(520 s)^2
x = 4.25E-14 m
Is that a rigorous enough explaination for you, or do I need to explain what "derivative" and "integral" mean? How about "limit?"
The only problem I see is that I stated the wrong metric prefix (0.0425 pm), but you failed to mention that. The inch figure still stands (4.25E-14 m = 1.67E-12 in), which suggests that the raw numbers I was using work fine.
The only real fault you've caught me on so far is momentarily confusing "momentum" with "kinetic energy" during some on-the-fly calculations. If I really wanted to spend time and effort on this problem, I would have taken into account:
And this is still not an exhaustive list.However, if you take such infinite delight in my brainfart of confusing mv with 0.5mv^2, then good for you. I only wish I were so easily amused.
Update (Score:3)
The scheduled firing was scheduled for 8:00. I got to the gate around 7:00, and spent the next hour stuck in traffic along the five-mile stretch to the designated parking area. In that traffic jam, most of the lisence plates were from Mississippi and Louisiana. There were plenty from Alabama (70+ miles away), too many for them all to be military. There was even one from Ontario.
Upon reaching the parking area (where the people trying to organize parking ran out of designated space), I joined the crowd of 400-500 people waiting on about 4-5 busses (and I only saw one bus owned and operated by NASA). 8:00 came and went with no firing, and around 8:45 (still in line for a bus) somebody started moving through the crowd with a bullhorn saying something about "indefinately postponed" because of technical difficulties.
I can't really blame them for not expecting the turnout they got. I don't know how well the press release was covered in the Biloxi area, but I know that the bit barely got three paragraphs in some sidebar in New Orlean's Times-Picayune. I guess everybody wanted to get an up-close view of part of that thing that sets off everybody's car alarm whenever it comes in for a landing.
There's been no official word that I can see yet when they'll try again, and whether it will be public again or not.
All in all, anybody who think Americans don't care about their space program deserves to get smacked hard.
Re:Sounds like fun... (Score:1)