Very interesting article requires math I don't have, guru? If not, interesting NTLby merovingian Tags: article, guru, interesting, math, requires 

#1
Jan2507, 04:28 PM

P: 8

These are the problems...
1. At 1G of acceleration, how long does it take to reach the speed of light (299 792 458 m / s) relative to the object accelerating? Obviously it never reaches the speed of light relative to the earth but I'm also curious what this speed is relative to us. 2. This is a little more difficult. How long does it take at this rate of acceleration does for the object to... Hmm, okay, let's say a star is 5 light years away, and at the rate of 1G of acceleration, it takes 5 years relative to the the object to get there relative to the object. I'm looking for this number. Does this make sense? I should be able to do this but, apparently I'm a little retarded. I'm done with school so this isn't homework or anything but I would appreciate it you can show your work. Very curious, thank you for your time. How this whole discussion started Basically, using 8 probes travelling 30K km/s, each with 8 sub probes, it would take 10Billion years to explore 4% of the milkyway. Basically, ET's haven't had enough time to find us according to the article. Article has many flaws but great for discussion about interstellar life. Anyway, can you shed some light? 



#2
Jan2507, 04:51 PM

Sci Advisor
P: 8,470

Why did the mods move this from the relativity section to the homework section? The post said "I'm done with school so this isn't homework or anything", and the questions are clearly being asked in a relativistic context.
Anyway, I'm not sure what you mean when you ask how long it would take to reach the speed of light "relative to the object accelerating". When we say "relative to the Earth" we really mean "in the Earth's inertial rest frame", so do you mean something like "in the noninertial rest frame of the accelerating object" here? If so, then since this is a rest frame, the object is always at rest. And in a more physical sense, the object will never be moving fast enough that it can catch up with a light beam, in fact local measurements on the accelerating craft will always show light moving at c relative to the craft. As for the question about time lengths, you'd probably find this page useful: http://math.ucr.edu/home/baez/physic...SR/rocket.html As the page mentions, if an object is accelerating at a constant rate a (meaning the Gforce felt onboard is always a, which implies that from the point of view of an inertial frame the rate of acceleration actually decreases as its speed increases), then for it to travel to a destination at a distance d (as measured in the frame where it started accelerating from rest, like the frame of the Earth if it took off from there), the time T to reach the destination as measured by clocks onboard the craft is: T = (c/a) cosh^1 [ad/c^2 + 1] (cosh^1 is the inverse hyperbolic cosine function, sometimes written as arcosh or acosh...if you type acosh(x) into the calculator here it'll compute the inverse hyperbolic cosine of x, just make sure you have it set to radians rather than degrees) Likewise, the time t to reach the destination as measured by clocks in the frame where the ship first started accelerating from rest is: t = sqrt[(d/c)^2 + 2d/a] (if you're interested in how long it would take to explore the galaxy at 1G acceleration this is the equation you want, since it would give the time from the galaxy's perspective rather than the much shorter time experienced onboard the ships) For convenience, you might want to use units of light years and years for distance and time, so c=1, and the page also mentions that 1G acceleration works out to a = 1.03 lyr/yr^2 in these units. As an example, if you want to know how long it'd take in Earth's frame for a ship accelerating away from Earth at 1G to reach a destination 100 light years away, the answer would be: t = sqrt[(100/1)^2 + 2*100/1.03] = sqrt[10000 + 194.2] = 100.97 years 



#3
Jan2507, 09:12 PM

P: 8

When a proton leaves the sun and smashes into the earth, how long did it take it? No time at all from it's perspective, right? It's travel is instantanious. But from our perspective it took however long it takes a proton to travel... 1.496 x 10^11 m. Let's say that the clock on board an object traveling 1.496 x 10^11 m towards the earth at a constant speed were to match the time it took it to get here at "rest" according to our clocks. What would that (earth) time be? Am I making any sense? 



#4
Jan2507, 09:55 PM

P: 312

Very interesting article requires math I don't have, guru? If not, interesting NTL 



#5
Jan2607, 01:06 AM

P: 8





#6
Jan2607, 01:22 AM

P: 8

There is a star 5 light years from earth. At what speed does the ship need to be travelling to get there in five years relative to the ship? In otherwords, you're going to be in a ship and you're going to go a constant speed and cover 5 light years, when you arrive not including acceleration or deceleration, you will be 5 years older. If the ship is moving at c, it will arrive instantaniously at the location 5 LY away, and five years relative to earth, right? So, at some point as the ship moves slower and slower, it acutally takes the ship 5 yrs to cover 5 LY's relative to the ship. I'm curious how fast that ship appears to be moving relative to us... and I guess relative to it. That make sense this time? I appoligize I can't seem to communicate this in a more clear fashion. 



#7
Jan2607, 04:00 AM

Sci Advisor
P: 1,883

v= dx/dt dtau (proper Time) = 5 y dx (distance in Earth frame) = 5 ly dt (time in Earth frame) = ? Now dtau² = dt²  dx²/c² > dt = sqrt(dtau²+dx²/c²) = sqrt(25 y² + 25 y²) = 7.07 y v=dx/dt = 5 ly / 7.07 y = 0.707 c. 



#8
Jan2607, 04:25 AM

P: 8

You're a pimp! Thanks dude! =)
.707c Wow, that's pretty fast. Interesting, so this is probably pretty hard to figure out but... Let's say your 20 years old and you expect to live till your 70 in space, how far would you get at a constant rate of acceleration of 1G over the course of 50 years? I'm curious how many light years of space we can cover in a lifetime. Again, this isn't homework or anything and it really is just curiosity so if you don't have time to mess with what seems like would be a nasty math problem, I would understand. Something for you... An interesting way to explore the milkyway, self replicating probes We were making fun of the author of the article in the initial post for lack of imagination based upon his assumption that over 10Billion years, an alien race would only launch 200 probes each with 8 sub probes, plus the author does not take into consideration that there maybe more than one space faring alien race which may exist. These criticisms and the link posted above got this response... "I'd hold off on criticizing others for a lack of imagination. Don't you realize that self replicating probes will doom us? We will be galactic spammers, the aliens will wipe us out as a nuisance. Or our probes will harvest the planet they pray towards, the aliens will wipe us out as heretics and blasphemers. At a very minimum the probes will be crossing the border without proper documentation, the fines and impound fees could leave us in "debtors prison" for millennia." HAHAHAHA! 



#9
Jan2607, 11:02 AM

Sci Advisor
P: 8,470

T = (c/a) cosh^1 [ad/c^2 + 1] with units of distance in lightyears and units of time in years, so c=1, and 1G acceleration means a = 1.03 ly/y^2. In this case, for d = 1000000000 ly, we have: T = (1/1.03) * cosh^1 [1030000001] and plugging acosh(1030000001) into the calculator here we get 21.445971820718775, so if you divide by 1.03 you get around 20.8 years of proper time. Beyond that, as they say, you need to worry about the effects of the universe's expansion so the math would be more complicated. But just for the sake of curiosity, if we lived in a nonexpanding universe, how far would you get at 1G for 50 years? In this case we can use this formula from the relativistic rocket page: d = (c^2/a) [ch(aT/c)  1] Where ch is the hyperbolic cosine function, represented as cosh(x) on that calculator. aT/c = 1.03*50 = 51.5, and the calculator says cosh(51.5) = 1.1618119050195011e+22, so d = (1/1.03)*(1.1618119050195011*10^22  1), which works out to around 1.128 * 10^22 lightyears, or just over 10 billion trillion ly. 



#10
Jan2607, 01:38 PM

P: 8

OMG! We can go anywhere! This is sweet! Thank you.
Did you read about the self replicating probes? 



#11
Jan2907, 05:56 AM

Sci Advisor
P: 1,883





#12
Jan3007, 08:26 AM

P: 8

Yeah, the fuel part is very interesting but it assumes that the propulsion device is on board the ship. Not that we can really go ANYWHERE anyway.




#13
Feb707, 07:22 AM

P: 8

Not to mention the whole expansion of the universe issue, difficult to calculate you think?
If anyone is puzzled by my last post. Wiki solar sail, for example is don't dependent on an on board propulsion system. I also hear that it maybe possible to sling shot a ship off a really big black hole. 


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