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#101 Dram

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Posted 26 April 2006 - 06:22 AM

Suppose I would be time travelling back in the past, so I exist now two times. And then I get an arrangement that one of us is going to work, and the other one is staying at home. Most certainly we would get into a fight, because both of us would want to saty at home. :) Strange, eh?


It'd be great entertainment lol :)

Not really to do with quantum physics, but I thought you guys would like to see what I've been conversing with my brother, who is a physicist, on complexities of travelling to Mars with current or forseeable tech:

This is a simplified trajectory analysis of travel between Earth and Mars. In order to keep things simple, I have ignored the orbital mechanics aspects of the solar system. So, for instance, I am ignoring the gravitational pull of the sun, initial velocity conditions due to the travelerís initial start in the Earthís reference system, any velocity needed to match Marís velocity upon arrival, any drag and/or resistance due to solar wind, light pressure from the Sun, or changes in inertia due to the ejection of propellant. Some of these issues may be important. I could in fact attempt to perform the gravitational field part of the analysis, and the answer would be somewhat different than the one Iím about to yield, but I think the analysis I am performing will get the point across, and the analysis is far less difficult to execute. The analysis I am going to perform is basically straight-line motion, between Earth and Mars.

So, letís start by summarizing the only assumptions needed for this analysis:

The distance to Mars from Earth is 100 million km, per Wikipedia
D1 = 100 ∙ 109 m
D1/2 = 50 ∙ 109 m
The desired time to get to Mars is 45 days, per PhysOrg
t1 = 3.888 ∙ 106 s
t1/2 = 1.944 ∙ 106 s
We are going to engage in constant acceleration motion, with a direction of acceleration reversal halfway there, and no initial velocity.
This means that for the first half of the trip, our rocket is firing in the direction of Earth, to speed us up, and in the second half of the trip it is firing towards Mars, to slow us back down.

Now, letís derive the equation of motion for this system.

By definition:
1. a = (d2x / dt2)

So, integrating to solve for x:
2. x = x(0) + v(0)t + 0.5at2 (valid for constant acceleration)
Now, assuming that we have no initial velocity, as we stated, and that the start position is labelled as zero, then we can solve for the acceleration:
3. a = 2x/t2

So, letís make some substitutions, and solve for the acceleration:
ē x = D1/2
ē t = t1/2

a = (2)(50∙109) / (1.944∙106)2 = 0.0265

So, weíve got an acceleration of 0.0265 m/s2. If you compare that against the standard surface-of-Earth gravitational acceleration g, which is 9.81 m/s2, then you might say that the acceleration of the spacecraft would be 0.0270 g.

So, with this low level of acceleration, the travelers in our spacecraft would not really notice too much the acceleration level at all.
As an afterthought, the moment at which the spacecraft velocity is highest is going to be the moment of turn-around. What is that velocity? Well, using the same type of analysis as above, it can be shown that the maximum velocity is given by vmax = a∙ t1/2, which can be solved to be 51440 m/s, which is roughly 185185 km/hr.


Very interesting :)

Tell your brother about Orbiter Space Simulator if he's interested. It basically simulates orbital mechanics with gravity inetia etc. It does not take into account solar wind etc however. But it's interesting to try and get to Mars :)

http://orbit.medphys.ucl.ac.uk/ - thats the website.

#102 OrbWeaver

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Posted 26 April 2006 - 08:14 AM

I just memorised the equations of constant acceleration; I could never remember how they were derived.

s = ut + 0.5at^2
v^2 = u^2 - 2as
v = u + at

#103 demagogue

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Posted 26 April 2006 - 08:45 PM

Tell your brother about Orbiter Space Simulator if he's interested.


I have to second that. I ran into this little jewel on HOTU and the fact that it refers to an official NASA spaceflight manual for its own manual I thought was pretty cool. And it has a great mod community that keeps introducing new stuff.
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#104 Dram

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Posted 26 April 2006 - 09:49 PM

I have to second that. I ran into this little jewel on HOTU and the fact that it refers to an official NASA spaceflight manual for its own manual I thought was pretty cool. And it has a great mod community that keeps introducing new stuff.


Yeah that was pretty cool :)

I tried XPlane before but was severely dissappointed by it's useless physics - realistic my ass.

#105 Maximius

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Posted 27 April 2006 - 06:58 AM

I still need a way of posting these Feynmann mp3s! Any suggestions?

#106 sparhawk

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Posted 27 April 2006 - 07:14 AM

You can upload them to my server. *HEHE*
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#107 Dram

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Posted 27 April 2006 - 08:13 AM

You can upload them to my server. *HEHE*


I knew he'd lose it someday.... :P

#108 Maximius

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Posted 27 April 2006 - 11:34 AM

I knew he'd lose it someday.... :P



Har har, now how do I distribute these dang mp3s?

#109 demagogue

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Posted 27 April 2006 - 05:58 PM

Why don't you just pick one of these services and put the link in this thread:

http://www.freewebsp...skstorage.shtml

I'm sure one of them is bound to work well.

Edited by demagogue, 27 April 2006 - 06:00 PM.

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#110 Maximius

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Posted 29 April 2006 - 01:39 PM

Why don't you just pick one of these services and put the link in this thread:

http://www.freewebsp...skstorage.shtml

I'm sure one of them is bound to work well.



I cant get any of these to upload the files. :huh: They all start and either run forever or eventually say "transmission expired" or some crap. Ill keep trying though.

#111 Maximius

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Posted 14 May 2006 - 12:51 PM

I cant get any of these to upload the files. :huh: They all start and either run forever or eventually say "transmission expired" or some crap. Ill keep trying though.



Okay, I give up on these file sharing sites, Ive tried them both from home and work and none will load the files. So here is another idea. If anyone who wants those Feynman tapes, which are pretty neat after listening to the first one, and has a file sharing account like Kazaa or something similar, I will sign up to exchange the files. This is the only alternative I can think of.

#112 obscurus

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Posted 21 May 2006 - 09:22 PM

Not really to do with quantum physics, but I thought you guys would like to see what I've been conversing with my brother, who is a physicist, on complexities of travelling to Mars with current or forseeable tech:

This is a simplified trajectory analysis of travel between Earth and Mars. In order to keep things simple, I have ignored the orbital mechanics aspects of the solar system. So, for instance, I am ignoring the gravitational pull of the sun, initial velocity conditions due to the travelerís initial start in the Earthís reference system, any velocity needed to match Marís velocity upon arrival, any drag and/or resistance due to solar wind, light pressure from the Sun, or changes in inertia due to the ejection of propellant. Some of these issues may be important. I could in fact attempt to perform the gravitational field part of the analysis, and the answer would be somewhat different than the one Iím about to yield, but I think the analysis I am performing will get the point across, and the analysis is far less difficult to execute. The analysis I am going to perform is basically straight-line motion, between Earth and Mars.

So, letís start by summarizing the only assumptions needed for this analysis:

The distance to Mars from Earth is 100 million km, per Wikipedia
D1 = 100 ∙ 109 m
D1/2 = 50 ∙ 109 m
The desired time to get to Mars is 45 days, per PhysOrg
t1 = 3.888 ∙ 106 s
t1/2 = 1.944 ∙ 106 s
We are going to engage in constant acceleration motion, with a direction of acceleration reversal halfway there, and no initial velocity.
This means that for the first half of the trip, our rocket is firing in the direction of Earth, to speed us up, and in the second half of the trip it is firing towards Mars, to slow us back down.

Now, letís derive the equation of motion for this system.

By definition:
1. a = (d2x / dt2)

So, integrating to solve for x:
2. x = x(0) + v(0)t + 0.5at2 (valid for constant acceleration)
Now, assuming that we have no initial velocity, as we stated, and that the start position is labelled as zero, then we can solve for the acceleration:
3. a = 2x/t2

So, letís make some substitutions, and solve for the acceleration:
ē x = D1/2
ē t = t1/2

a = (2)(50∙109) / (1.944∙106)2 = 0.0265

So, weíve got an acceleration of 0.0265 m/s2. If you compare that against the standard surface-of-Earth gravitational acceleration g, which is 9.81 m/s2, then you might say that the acceleration of the spacecraft would be 0.0270 g.

So, with this low level of acceleration, the travelers in our spacecraft would not really notice too much the acceleration level at all.
As an afterthought, the moment at which the spacecraft velocity is highest is going to be the moment of turn-around. What is that velocity? Well, using the same type of analysis as above, it can be shown that the maximum velocity is given by vmax = a∙ t1/2, which can be solved to be 51440 m/s, which is roughly 185185 km/hr.



When you take gravity into account, the point at which you need to flip the craft around is much further along into the trip, and the deceleration force required much less (if you need to do it at all), because you can use planet's gravity to make the trip much more efficient. This is of course how NASA does it - they use gravitational slingshots to get a free boost to acceleration, and carefully guided orbital insertion so the spacecraft uses the planet's gravity to capture it into orbit, requiring much less effort than slamming on the brakes halfway into the trip. If you make the right calculations, you don't need more than a few puffs of the steering rockets to fine tune your trajectory into orbital capture.

IIRC Cassini for example, was flung around Venus, around Earth, around Venus again and past Jupiter to get it into orbit around Saturn, with almost no extra thrust once the main rocket had broken free from Earth and few adjstments to the trajectory needed - the guys at NASA got their initial calculations so spot on it went almost flawlessly, just hitching a ride from the gravity of various planets. Pretty amazing given the complexity of the mathematics involved.

#113 Dram

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Posted 30 May 2006 - 07:40 AM

IIRC Cassini for example, was flung around Venus, around Earth, around Venus again and past Jupiter to get it into orbit around Saturn, with almost no extra thrust once the main rocket had broken free from Earth and few adjstments to the trajectory needed - the guys at NASA got their initial calculations so spot on it went almost flawlessly, just hitching a ride from the gravity of various planets. Pretty amazing given the complexity of the mathematics involved.


Also taking into account they calculated it by hand (computers were'nt sophisticated enough at the time), it was pretty damn crazy that they were off by only about 20m.

All these slingshot maneuvers you can try out in Orbiter - very good sim.

#114 Order of the Hammer Bureaucrat

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Posted 12 November 2006 - 12:35 AM

I didn't want to start a new useless thread about this, and I'm 1% sure it might be interesting or useful to someone here. I was just googling isentropic expansion to brush up some basics, and apparently nasa presents an excellent (although unorganized) website for K-12, (meaning kindergarten to grade 12) students discussing all that. http://www.grc.nasa.gov/WWW/K-12/
the good page is http://www.grc.nasa....e/isentrop.html well describing supersonic reversible engine. What I don't get is why is the educational curriculum to cut and shortened that they don't touch basic thermodynamics in high-school. This stuff is clearly for the K-12 crowd, as indicated on the site.

#115 Ishtvan

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Posted 12 November 2006 - 07:23 PM

Hmm, while we're bumping this thread, is anyone well versed in the physics of optical absorption?

I was reviewing the derivation of gain/absorption in Coldren & Corzine, and my question is:
Is there an intuitive explanation for why the transmission matrix element divided by photon density is proportional to 1/frequency, or h/<photon Energy>?

I.e., in the expression: absorption is propritional to 1/E * <reduced density of states = Sqrt(E) for bulk, parabolic bands>, can anyone explain why it makes physical sense for the 1/E to be there?

I went thru the math, and yes, you can see algebraicly why that appears, because the ratio |matrix element|/<photon density> is proprtional to <frequency>/<frequency^2>, but it seems like there should also be an intuitive explanation for that 1/Energy term that makes all absorption go to zero for large energies. Maybe something about energy density.

I guess this is an academic question anyway, because in reality higher energies will just ionize the electron and I doubt the two-band model applies anymore.




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