Posers and Puzzles
11 Nov 09
11 Nov 09
So a car that has good seals and an air supply gets accidentally launched into space by mistake. It is a nice gas powered vehicle but of course engines don't run too well in a vacuum and there is not a whole lot of friction in it's orbit so the tires won't work even if the engine will. So no oxygen to run the engine, but enough for a couple of months of human breathing with rebreathing machines and so forth. The question is, using the headlights, assuming you have an extremely long lasting battery, storing megawatt hours of energy, you turn on the headlights which shines a bright light in front of the car. The two headlights are very powerful, each a thousand watts (he was a hunter who used his lights to freeze deer when he was on Earth) and at about 2% efficient at producing light but also about 90% efficient at producing infrared, how long would it take to get some motion backwards due to the momentum imparting property of photons? How long to travel one meter? one Km? Can he reach escape velocity?
11 Nov 09
The momentum of a photon is E/c, isn't it?
In which case, the rate of change of momentum of the car would be P/c, where P is the output power.
What I'm not sure we can assume is that the infra-red radiation is generated in any particular direction. If it is, then our output power is 1840W (if I've understood the question correctly). If it's not, it's 40W.
If we assume the mass of the car is about 1000kg, that gives us an accelaration of 1.3 x 10^-10 m/s^2 or 6.1x10^-9 m/s^2.
In the "better" scenario, that'll take about 5 hours to go one metre, and nearly a week to go a kilometre.
Any good?
11 Nov 09
1 edit
Originally posted by mtthwThe IR and visible light are both reflected in the same way since the reflectors are silver and very wide band, the reflectors are usually a parabolic job so the main bang of light would go out the front in a fair imitation of a beam, which means it is 1800+ watts and like you assumed, 1000Kg mass.
The momentum of a photon is E/c, isn't it?
In which case, the rate of change of momentum of the car would be P/c, where P is the output power.
What I'm not sure we can assume is that the infra-red radiation is generated in any particular direction. If it is, then our output power is 1840W (if I've understood the question correctly). If it's not, it's 4 ll take about 5 hours to go one metre, and nearly a week to go a kilometre.
Any good?
Your #'s are similar to mine.
BTW, did you read about the new Vasimir Ion Rocket that has been recently touted as allowing a trip to mars in about 40 days as opposed to 6 months or more with present day technology? Try to figure out the acceleration from those figures. With a constant acceleration, you accel half-way and decel the other half so you have a max velocity at the halfway point, but what is the acceleration?
They are projecting using an upgraded version of the Vasimir with a 200 megawatt nuclear power source. I assumed a distance of 160 million Km as the trip distance total, so accel for 80 Million Km halfway, etc.
23 Nov 09
Originally posted by sonhouseExcuse the notation. Not sure accuracy, as I did this at work quickly when I am supposed to be working. 😛
The IR and visible light are both reflected in the same way since the reflectors are silver and very wide band, the reflectors are usually a parabolic job so the main bang of light would go out the front in a fair imitation of a beam, which means it is 1800+ watts and like you assumed, 1000Kg mass.
Your #'s are similar to mine.
BTW, did you read ab ...[text shortened]... distance of 160 million Km as the trip distance total, so accel for 80 Million Km halfway, etc.
x = final position relative to starting point
x0 = position at starting point
t = time to reach x
v0 = initial velocity
a = accelaration
x - x0 = v0t + (1/2) at^2
x0 = 0
x = 80 million km = 8 x 10^10 m
t = 20 days = 1.728 x 10^6 s
v0 = 0
a = 2 x ((8 x 10^10) / (1.728 x 10^6)^2) = 0.0536 ms^-2
A little over 5 centimeters per second per second.
25 Nov 09
1 edit
Originally posted by lauseyPerfectly correct, about 5 milli G's. A lot of distance covered for such a tiny thrust, eh! I wonder if you would feel it if you were inside that spacecraft.
Excuse the notation. Not sure accuracy, as I did this at work quickly when I am supposed to be working. 😛
x = final position relative to starting point
x0 = position at starting point
t = time to reach x
v0 = initial velocity
a = accelaration
x - x0 = v0t + (1/2) at^2
x0 = 0
x = 80 million km = 8 x 10^10 m
t = 20 days = 1.728 x 10^6 s
v0 = 0 ...[text shortened]... 10^10) / (1.728 x 10^6)^2) = 0.0536 ms^-2
A little over 5 centimeters per second per second.
If you kept that up for 200 days accel and 200 days decel with the same acceleration, you would be 16 billion Km from earth and stopped or 10 billion miles, 2.7 times the distance of Pluto.
To get to Pluto with that acceleration, it would take 121 days to get halfway where you turn your engines around and decel for the rest of the trip, so it takes 242 days or about 8 months and you attain a maximum velocity at the halfway point of about 550 Km/sec or 350 Miles/sec. That is pretty fast indeed.
My favorite dream trip would be to the solar foci, which starts at about 55 billion miles or about 80 billion Km from the sun (where gravitational lensing of the sun starts to focus light from distant suns) and I think about a bit further out, the 1000 AU point, 1.4E11 Km, or 1.4 E14 meters, also at that 0.053 M/S^2 accel, takes about 846 days to the halfway point or about 1700 days for the whole trip (a bit over 4.5 years one way) and a maximum velocity of 3,920 Km/sec or 2450 Mi/sec. Pretty impressive velocity if you ask me!
My most favorite dream trip would be to Alpha Centauri, at .053 M/S^2, would take 26 years and at the halfway point you would be doing 22,000 Km/sec or about 14,000 miles/second, about 0.07C. Sounds almost doable for humans. Of course if you spend say, 5 years studying AC, and come back it would take 57 years round trip but still possible. I would think a journey like that would be one way however. Then you send back data at light speed so about 30 years after you leave, assuming you make it ok, Earth would start receiving close up and personal data from Alpha Centauri! What a trip that would be, and with more or less existing technology.
One thing to note: At 0.07C there would not be much in the way of relativistic time dilation so the trip times are pretty much what you see is what you get, as opposed to getting about ten times faster where the trip time for you on the craft seems to take only about half the time it looks like it took from Earth's perspective.
Anyone want to check my figures? I use S=(AT^2)/2 as starting point for distance and reworking it to T=(2S/A)^0.5 which is the halfway point so to get the total trip time, multiply time 2. And using that time, plugging it into V=AT gives maximum velocity (at the halfway point) then the engines turn around and slowly decel the rest of the way. Using meters/sec^2 and meters for distance.
25 Nov 09
3 edits
Originally posted by sonhouseI got just under 56 years total trip. Assuming Alpha Centauri is 4.37 ly away.
Perfectly correct, about 5 milli G's. A lot of distance covered for such a tiny thrust, eh! I wonder if you would feel it if you were inside that spacecraft.
If you kept that up for 200 days accel and 200 days decel with the same acceleration, you would be 16 billion Km from earth and stopped or 10 billion miles, 2.7 times the distance of Pluto.
To ge turn around and slowly decel the rest of the way. Using meters/sec^2 and meters for distance.
Also worked out that the lorentz factor for 0.07C is around 1.00000000023, so the amount of extra thrust required to travel that distance will be negligible. If it was a much greater proportion of the speed of light, noticeably more thrust will be required.
Anyone willing to do the calculations where the half way point is at 0.5C while taking into account relativistic energy required? I am guessing some calculus will be involved without looking into it further.
EDIT: Just read you put 57 years round trip there and back, but I got 56 years just one way.
4.37 ly = 4.13 x 10^16 m
Half way = 2.07 x 10^16 m
t = ((2 x 2.07 x 10^6) / 0.053)^0.5 = 8.83 x 10^8 s = 27.98 years (half way).
02 Dec 09
4 edits
Originally posted by lauseyI got the same thing, 1.0024. So if we use a 56 year number, that would be 20,440 days
Actually, incorrectly worked out the lorentz factor in my last post. It is 1.0025. Not sure if that is significant at the moment in travelling that distance.
and inverting 1.0024=0.9976 give or take, time saved is about 50 odd days over a non-relativistic version of the trip, nothing to write home about, eh!