B Energy production that converts Hydrogen to Iron?

[PeterDonis]

would the craft get more delta v if it used all the energy to accelerate all the iron

What is the exhaust if it isn't the iron?

 

[Devin-M]

What is the exhaust if it isn't the iron?

The exhaust is Iron.

Suppose you have the energy in some kind of battery from converting Hydrogen into 20 iron atoms, do you use all the energy in the battery to accelerate 2 of those iron atoms or all 20? Which gives the craft more delta v?

Im imagining something like the LHC but instead of collisions you have a beam dump on both sides (beam dumps added by myself for illustration):

 

[PeterDonis]

The exhaust is Iron.

Then you can't possibly accelerate all of the iron, since you have to use at least some of it as exhaust. So I don't understand what you mean by "accelerate all the iron".

 

[Devin-M]

By "accelerate all the iron" I mean "exhaust all the iron" interchangeably.

 

[PeterDonis]

By "accelerate all the iron" I mean "exhaust all the iron" interchangeably.

You can't exhaust all the iron either since that leaves no payload. So again I don't understand what you mean. I think you need to be more precise in specifying exactly what process is going on and exactly what variable you are varying.

 

[Devin-M]

Lets talk about a falcon 9 size vehicle with a dragon capsule sized payload, but instead of kerosene and oxygen it's filled up with deuterium, which will be converted to iron with the onboard reactor. The iron will be accelerated/"exhausted" with an onboard particle accelerator.

 

[PeterDonis]

Lets talk about a falcon 9 size vehicle with a dragon capsule sized payload, but instead of kerosene and oxygen it's filled up with deuterium, which will be converted to iron with the onboard reactor.

Ok. Then there is no variable at all; you run the nuclear reaction to completion, that gives you a certain amount of energy, and how much thrust that gives you is determined by the exhaust velocity of the iron, which is determined by the mass per particle of the iron.

The iron will be accelerated/"exhausted" with an onboard particle accelerator.

Why? The nuclear reaction already gives the iron exhaust velocity. Trying to give it extra velocity with an accelerator is pointless because the energy has to come from somewhere and the only energy source is the nuclear reaction. You might as well just let the reaction heat up the iron directly.

 

[Devin-M]

Converting deuterium to (20) iron-56 atoms gives us a certain amount of energy in the battery, 8400MeV of total energy if my math is correct.

So does the Falcon 9 sized craft get more delta V using 8400MeV of energy to accelerate 2 or 20 iron atoms “out the back?”

With the option with 2 iron atoms, they would be exiting the back of the craft faster because more energy per atom. With this option the other 18 iron atoms are released non-energetically and allowed to float away from the craft.

 

[PeterDonis]

the Falcon 9 sized craft

Is way, way too big for 8400 MeV of energy to do anything useful or for the comparison you are making to be useful. You need to pick a payload that is of comparable size to the fuel.

Try this: you have a payload of 1 uranium-238 atom and your fuel is 280 deuterium atoms, which will react to form 20 iron-56 atoms giving 8400 MeV of energy. We want to exhaust N iron atoms. Any iron atoms that aren't exhausted become part of the payload. (That last item is something you left out of your previous analysis, and it makes a big difference.)

 

[Devin-M]

Try this: you have a payload of 1 uranium-238 atom and your fuel is 280 deuterium atoms, which will react to form 20 iron-56 atoms giving 8400 MeV of energy. We want to exhaust N iron atoms. Any iron atoms that aren't exhausted become part of the payload. (That last item is something you left out of your previous analysis, and it makes a big difference.)

Yes, I agree with the scenario.

So from the example on post #50, I believe the U-238 gets more delta v is N=2 than if N=20.

 

[Devin-M]

Wait, suppose the payload is 1000 U-238 atoms, not 1.

 

[Devin-M]

So we have 1000 U-238 atoms payload, we have 20 iron atoms, 8400MeV of energy and N=2 or N=20. I think the Delta V on the 1000 U-238 atoms is higher with N=2

 

[PeterDonis]

from the example on post #50

That calculation is wrong because you don't include iron atoms not in the exhaust as part of the payload.

 

[PeterDonis]

For a good discussion of the relativistic rocket equation and how it is derived, see here:

https://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html

Your scenarios are somewhat different because your exhaust is not light and you don't convert all of the mass of the fuel into energy, only a small fraction of it; but the analysis in that article can easily be adapted to your case.

 

[Devin-M]

That calculation is wrong because you don't include iron atoms not in the exhaust as part of the payload.

If I’m a skater, and I have a 1kg ball with a 0.1kg core, I could either throw the entire 1kg ball with 100J or I could take out the 0.1kg core, drop the rest of the ball before throwing the 0.1kg core with 100J.

 

[PeterDonis]

drop the rest of the ball

If you drop the rest of the ball, that changes the momentum and energy conservation equations. You can't just ignore what you drop; you have to include it in your analysis. You didn't.

 

[PeterDonis]

If I’m a skater, and I have a 1kg ball with a 0.1kg core, I could either throw the entire 1kg ball with 100J or I could take out the 0.1kg core, drop the rest of the ball before throwing the 0.1kg core with 100J.

Where is the 100J coming from? You have to include the energy source in your analysis as well.

 

[Devin-M]

Ok lets say a 100kg astronaut at rest has in their left hand a 0.9kg ball and their right hand 0.1kg ball… they throw the 0.1kg ball with their right hand with 100J, while opening their fingers on their left hand. The 0.9kg ball remains at rest, the astronaut goes 0.427m/s and the 0.1kg ball goes 42.6m/s in the opposite direction.

 

[PeterDonis]

Ok lets say a 100kg astronaut at rest has in their left hand a 0.9kg ball and their right hand 0.1kg ball… they throw the 0.1kg ball with their right hand with 100J, while opening their fingers on their left hand. The 0.9kg ball remains at rest, the astronaut goes 0.427m/s and the 0.1kg ball goes 42.6m/s in the opposite direction.

Again, where is the 100 J coming from? You're ignoring that. Perhaps you are relying on the non-relativistic approximation, but if so, it makes no sense to talk, as you have, about mass-energy equivalence.

Also you keep changing scenarios. But ok, fine, let's analyze this one. Here are what your numbers say:

Energy before: 100J

Energy after: (1/2) 100 (0.427)^2 + (1/2) 0.1 (42.6)^2 = 99.85 J, not quite the same but I'll assume that's rounding error.

However:

Momentum before: zero

Momentum after: 100 * 0.427 - 0.1 * 42.6 = 42.7 - 4.26 = obviously not zero

So your numbers violate momentum conservation. So however you are getting your numbers, you're doing it wrong.

 

[PeterDonis]

while opening their fingers on their left hand

The fact that an astronaut can obviously do this with an item in their hand does not mean you can obviously do this with iron that is a product of a nuclear reaction.

 

[Devin-M]

I think I found the error in my spreadsheet. After fixing it, now I get opposite results where the astronaut is better off throwing the full 1kg with 100J rather than dropping 0.9kg and then throwing 0.1kg with 100J.

 

[Devin-M]

1kg Ball, 100J, 100kg Astronaut:
Astronaut: 0.14m/s
Ball: 14.08m/s

0.1kg Ball, 100J, 100kg Astronaut:
Astronaut: 0.044m/s
Ball: 44.7m/s

 

[PeterDonis]

1kg Ball, 100J, 100kg Astronaut:
Astronaut: 0.14m/s
Ball: 14.08m/s

0.1kg Ball, 100J, 100kg Astronaut:
Astronaut: 0.044m/s
Ball: 44.7m/s

How are you determining the final speeds for these cases?

 

[Devin-M]

How are you determining the final speeds for these cases?

I used:

C=(D/B)*A

where:

A=Astronaut_Final_V_m/s
B=Ball_Mass_kg
C=Ball_Final_V_m/s
D=Astronaut_Mass_KG

By slowly increasing the value of A until:

((1/2)*B*C^2)+((1/2)*D*A^2)=100J

 

[Devin-M]

Fully solving:

A=(sqrt(2)*sqrt(B)*sqrt(E))/sqrt(D*(B+D))

C=(D/B)*A

A=Astronaut_Final_V_m/s
B=Ball_Mass_kg
C=Ball_Final_V_m/s
D=Astronaut_Mass_KG
E=Total_Energy_J=100J

 

[PeterDonis]

I used:

C=(D/B)*A

where:

A=Astronaut_Final_V_m/s
B=Ball_Mass_kg
C=Ball_Final_V_m/s
D=Astronaut_Mass_KG

By slowly increasing the value of A until:

((1/2)*B*C^2)+((1/2)*D*A^2)=100J

Ok, good, so basically you are solving for the combined constraints of momentum and energy conservation, with the total energy being pre-set at 100 J. (You're using the non-relativistic approximation, but as I think I've already noted, that should be fine for all of the scenarios you've posed.)

The same method should work for the other scenarios you have posed, since all of the relevant quantities are known.

 

[PeterDonis]

The same method should work for the other scenarios you have posed, since all of the relevant quantities are known.

Just to expand on this some and address the question of whether it makes sense to "drop" any propellant, the relevant equations for energy and momentum conservation, with a more intuitive choice of notation, are as follows:

$$
e = \frac{1}{2} M V^2 + \frac{1}{2} \left( k - s \right) v^2
$$

$$
M V = \left( k - s \right) v
$$

where $e$ is the energy available from whatever source is being used (for example the nuclear reaction burning deuterium to iron), $M$ is the final payload mass, $V$ is the final payload velocity (in the frame where everything is initially at rest), $k$ is the total mass of propellant available, $s$ is the mass of propellant that is "dropped", i.e., not used as exhaust (so $k - s$ is the mass of propellant that is used as exhaust), and $v$ is the final velocity of the exhaust. We can use the second equation to eliminate $v$ from the first, and then rearrange to find:

$$
V^2 = \frac{2 e}{M \left( 1 + \frac{M}{k - s} \right)}
$$

All of the quantities on the RHS are fixed except $s$, so the question is what choice of $s$ will maximize $V$ (which is equivalent to maximizing $V^2$, and that means maximizing the RHS of the above). It should be obvious that any value of $s$ greater than zero will increase the factor in the denominator of the RHS above and hence will decrease $V^2$, so if we want to maximize $V^2$, we want $s = 0$, i.e., we want to use all of the propellant as exhaust and not "drop" any.

 

[Devin-M]

So from what I find online:
Falcon 9
Dragon Capsule (payload) 12000kg
Booster 1st+2nd Stage Combined Dry 26200kg
Booster 1st+2nd Stage Combined Wet 544600kg
Total Fuel Mass 518400kg

For simplicity we can call the fusion craft a single stage combined booster and payload with:
Dry Mass: 38200kg (of which 12000kg is useful payload)
Wet Mass: 556600kg
Propellant Mass: 518400kg
Propellant Energy: 719TJ/kg

 

[PeterDonis]

For simplicity we can call the fusion craft a single stage combined booster and payload with:
Dry Mass: 38200kg (of which 12000kg is useful payload)
Wet Mass: 556600kg
Propellant Mass: 518400kg
Propellant Energy: 719TJ/kg

The Falcon 9 propellant mass is optimized for chemical reactions as the energy source, not fusion reactions. If you were designing an actual fusion rocket you would probably want quite a bit less propellant.

Also, with that large a propellant mass relative to the payload mass, and fusion as the energy source, the non-relativistic approximation is no longer very accurate.

 

[Devin-M]

On a relativistic kinetic energy calculator I found 1 kg has an exhaust velocity of 0.125c with 719TJ/kg.