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Energy required to accelerate a ship.. WITH ANTIMATTER

by cadsmack
Tags: accelerate, antimatter, energy, required, ship
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cadsmack
#1
Oct2-04, 05:22 AM
P: 4
Hey, I was reading a science fiction book that had lots of fast pitched space battles, ships manoeuvring around at 8 gees and stuff, powered by antimatter :P . So I was wondering, lets say you had a ship with a mass of 1000 tonnes.. how much power would be needed to accelerate the ship at 8 gees, assuming it used some advanced propulasion system which would accelerate a really tiny masses to absolutely fantastic velocities (so mass change of the ship would be negligible)? And if this energy was coming from a matter/anti-matter reaction which was able to harness 100 percent of the energy released to do accelerate atoms of hydrogen or something, at what rate would you use up your anti-matter?

I get stuck at the first part of the problem, I can't formulate the problem in a way that gives any kind of meaningful answer... I figured that the energy required per second would be the same energy the ship would need to accelerate a one kilogram mass to
8*9.8*1,000,000 = 78,400,000 m/s, which would be like
1/2*m(v^2) = 3^15 J.
But hey, that doesn't seem right... in the same way that I can't quite explain why I'd think that approach would solve the problem, I can't explain what is wrong with it... I've tried other approaches that "felt right" and they all gave different results that didn't feel right.

So, yeah, I need some help, I'm going nuts.. highschool physics is just a half formed memory now and I think I'm hurting my brain trying to conjure a solution out of thin air.. as for the antimatter thing, I hope it's as simple as looking up how much energy is released when a gram of antimatter is annihilated with a gram of normal matter.

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Mk
#2
Oct2-04, 05:28 AM
P: 2,056
Quote Quote by cadsmack
I hope it's as simple as looking up how much energy is released when a gram of antimatter is annihilated with a gram of normal matter.
yes, it is that easy.
Mk
#3
Oct2-04, 05:33 AM
P: 2,056
The reaction of 1 kg of antimatter with 1 kg of matter would produce 1.810^17 J of energy (by the equation E=mc2). In contrast, burning a kilogram of petrol produces 4.210^7 J, and nuclear fusion of a kilogram of hydrogen would produce 2.610^15 J.
So that would be 1800000000000000000 J/kg = 1800000000000000 J/g = 1.8x10^15 J/g

Mk
#4
Oct2-04, 05:42 AM
P: 2,056
Energy required to accelerate a ship.. WITH ANTIMATTER

Anti-matter is a friggen perfect source of energy, no waste products or radiation, around 100 times more powerful than fusion, and much much more! Who knows what we can do with 100 times the power of the thermonuclear fusion of hydrogen
cadsmack
#5
Oct2-04, 06:35 AM
P: 4
If you were using a small amount of antimatter and shooting it at a larger mass of matter and using the resulting explosion of particles for thrust, the the whole rocket would get radioactive as hell though.
Mk
#6
Oct2-04, 07:18 AM
P: 2,056
There is very little radiation. And with all that energy that would make a pretty badass bomb, the Air Force has a Positronics Research Labratory, AFRL that is working on a brick sized material capable of storing 10 micrograms of antimatter... equivelent to 900 lbs. (408 kg.) of high explosive. And since it is encased in matter we can handle it.
Mk
#7
Oct2-04, 07:30 AM
P: 2,056
Welcome to physicsforums! As I should say as well.
pervect
#8
Oct2-04, 11:43 AM
Emeritus
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P: 7,662
Quote Quote by cadsmack

I get stuck at the first part of the problem, I can't formulate the problem in a way that gives any kind of meaningful answer...
The energy requirements are going to depend on a parameter of the ship called the specific impulse. Another equivalent idea is the effective exhaust velocity.

The thrust of the rocket is going to be (dm/dt)*v_e, where dm/dt is the rate of fuel consumption, and v_e is the exhuast velocity. This is the rate of change of momentum (dm*v_e) with time (dt).

The power required to generate this thrust will be .5*(dm/dt)*v_e^2. This is the energy in the exhaust (.5*dm*ve^2) per unit time (dt). Energy/ timpe = power. This is an ideal figure with a 100% efficient rocket

Combining these two equations, we can see that the power per unit thrust will be .5*(dm/dt)*v_e^2 / (dm/dt)*v_e = .5*v_e

So the higher your exhaust velocity, the more power you will need per unit thrust. High ISP is good for long-term rocket performance, because it minimizes the mass of fuel used. It is very demanding on the amount of power required, though.

As far as antimatter thrusters go, there is some good information at

http://www.islandone.org/APC/Antimatter/02.html

The highest ISP design is the "beam core" design. The idea here is that most of the energy is going to be released in the proton/antiproton reactions. This will generate pions, about 2/3 of which will be charged. You use a very strong magnetic field to deflect these charged pions, so that they exhaust out the back. The uncharged pions have to be absorbed or otherwise dealt with.

Because only 2/3 of the pions will be charged, effiiency will be less than 66%. Because of the very high power in the exhaust with the very high ISP, you probably won't see rockets accelerating at 8g with this sort of drive.

Other sorts of anti-matter rocket will have lower ISP's, but use a lot less energy. They won't be as capable in the amount of time that they can accelerate, but they probably will be able to pull more G's without melting themselves.
cadsmack
#9
Oct2-04, 07:22 PM
P: 4
Ok, if I'm burning 1 g/s of antimatter and using totally unrealistic technology to get 1.8x10^15 J/s of harnessable energy put into moving x kg/s of hydrogen out the back then the v_e is
1.8x10^15 = 0.5*(x)*v_e^2 , v_e = sqrt(2.0*1.8x10^15/x). Then, the thrust would be
x*sqrt(2.0*1.8x10^15/x) = sqrt(2.0*1.8x10^15*x) = 6.0x10^7*sqrt(x) N.
To get an acceleration of 72 m/s^2 I'd do
78.4 = 6.0x10^7*sqrt(x) N/10^6 Kg => x = 1.7 kg/s

But, 1 gram of antimatter/second is an aweful lot of antimatter, considering today 1 gram of antimmater is estimated to be worth 25 billion dollars.
pervect
#10
Oct2-04, 07:39 PM
Emeritus
Sci Advisor
P: 7,662
Quote Quote by cadsmack
Ok, if I'm burning 1 g/s of antimatter and using totally unrealistic technology to get 1.8x10^15 J/s of harnessable energy put into moving x kg/s of hydrogen out the back then the v_e is
1.8x10^15 = 0.5*(x)*v_e^2 , v_e = sqrt(2.0*1.8x10^15/x). Then, the thrust would be
x*sqrt(2.0*1.8x10^15/x) = sqrt(2.0*1.8x10^15*x) = 6.0x10^7*sqrt(x) N.
Hmmm - I just checked, I get 1gm of matter + 1 gm of antimater = 1.8*10^14 joules.

(I used google calculator, evaluating 2gm * c^2).

Other than this, your formulas all look good. 1 joule/second is commonly called 1 watt.

To get an acceleration of 72 m/s^2 I'd do
78.4 = 6.0x10^7*sqrt(x) N/10^6 Kg => x = 1.7 kg/s
The formula is correct assuming your ship has a mass of 10^6 kg, or 1000 metric tons (I didn't double check the numbers, you might have to change the figure for the energy of 1gm of matter+1gtm of antimatter anyway).

But, 1 gram of antimatter/second is an aweful lot of antimatter, considering today 1 gram of antimmater is estimated to be worth 25 billion dollars.
Yep. Though if you're willing to use 100x as much reaction mass, you can lower the energy cost by a factor of 10.
cadsmack
#11
Oct3-04, 12:47 AM
P: 4
Ok, the ship masses 1000 tonnes (M_s = 10^6 Kg) and carriess 500 tonnes (M_r = 5*10^5 Kg) hydrogen reaction mass mass. If the ship's mission requires an acceleration of 8 gees (a = 78.6 ms^-2) for 100 hours (t_1 = 3.6*10^5 s), what is the minimum amount of antimatter needed? I think the condition when least antimatter is used is when the inneficiency of the rocket is maximized; you'd want the power per unit thrust, and therefore the exhaust velocity, to be at a minimum constant value throughout the whole trip, so that at the end of the trip you would have exhausted all your reaction mass.

dm/dt = rate of reaction mass exhaust, dm = reaction mass spent.
Acceleration is a constant, so we can use

A = (dm/dt)*V_e/(M_s - dm).That makes a differential equation like

dm(t) = M_s - (dm'(t) )*V_e/A, dm(t) = 0. Its solution is

dm = M_s - e^(-A*t/(V_e) )*M_s. Setting dm(t) = M_r and t = t_1 and solving for V_e we get

V_e = -A*t_1/ln( (M_s - M_r)/M_s ) ) = 2.0*10^7 m/s .. nice .. substituting back in to dm we get

dm = M_s(1 - (1 - M_r/M_s)^(t/t_1) ) (neat formula :) differentiating,

dm/dt = M_s*Ln(1 - M_r/M_s)*(1 - M_r/M_s)^(t/t_1)

Power usage is 0.5*(dm/dt)*v_e^2 J/s and to find the total energy used we'll integrate that over the trip time..

Integral( 1/2*(M_s*Ln(1 - M_r/M_s)*(1 - M_r/M_s)^(t/t_1) )*(-A*t_1/ln( (M_s - M_r)/M_s ) )^2 dt (0 -> t_1))

=> -A^2*M_s*t_1^2*(1 - M_r/M_s)^(t/t_1)/(2*Ln(1 - M_r/M_s)^2) [0 --> t_1]

= A^2*M_r*t_1^2/(2*Ln(1 - M_r/M_s)^2)

= 4.1*10^20 J !!!!

That's 2302.8 Kg of antimatter :P hey, at least you're now going 10% the speed of light :>

... man my head hurts now.





"ONE POINT TWENTY-ONE GIGAWATTS!?!"
pervect
#12
Oct3-04, 11:58 AM
Emeritus
Sci Advisor
P: 7,662
I'm going to assume you did that all correctly :-)

BTW, you could have saved yourself a little bit of work by using the well-known "rocket equation" instead of deriving it yourself.

http://scienceworld.wolfram.com/phys...tEquation.html

I should probably point out that ignoring relativity, you'd have to accelerate for 1 year at 1 g to reach the speed of light (i.e. 1 g = 9.8 m/s^2 is very close to 1 light year/ year^2) - so a mere 10 hours at 8g isn't going to get you up to 10% of lightspeed. You'd need 36 days at 1g or 4.5 days at 8g to get up to .1 c.
Manchot
#13
Oct8-04, 07:43 PM
P: 728
On a side note, I've been thinking about a way to create antimatter at a low energy cost. The feasibility isn't very good, mind you, but it could work in theory. Somehow, if you were to create a black hole, one with a mass sufficiently "small" (as black holes go), you could get it to an equilibrium point where the matter you're putting in equals the total energy of the Hawking radiation. Technically, Hawking radiation produces all types of particles, and about half of those should be antiparticles. Before they begin to annihilate one another, they could be separated into different containers by the different charge signs, thereby preventing any annihilation from occurring (using E&M fields). So, for example, the negative particle box would contain a bunch of anti-protons and electrons, along with all the other negative particles. The neutral particles would all be allowed to annihilate, though, amd this energy would be put into the "main" power grid.

If this were ever possible, you could just give someone equal amounts of the two types of particle soup. When they need that energy, they just have to combine the two soups to get lots and lots of instant energy.
Crauven
#14
Oct14-04, 11:16 AM
P: 7
That still does not remove the ongoing problem that we need some way to fuel it. Antimatter wont propel without a equal amount of matter correct? Matter has to then be fed into it, and there is only a limited amount of matter that one container to hold. So, what I am trying to get at, is though we may be able to go faster, we won't get any farther in space because the matter that gets used up will only last so long, and seeing as most space shuttles have limited capacity for fuel holding, we would have to assume it wouldn't get very far. Just my thoughts.

Daniel
Mk
#15
Jan26-05, 02:59 AM
P: 2,056
Faster=Farther, because of inertia, there's not that much stuf in the space part of space to slow you down.


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