Energy production that converts Hydrogen to Iron?

[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.

 

[Devin-M]

On rocket equation calculator, 37,474,057m/s exhaust velocity, initial mass 556600kg, final mass 38200kg, change in velocity 100,393,423m/s or 0.33c.

Proxima Centauri: 4.24ly
Travel Time: 12.8yrs @ 0.33c

 

[Devin-M]

It's amazing what you'd be able to do with just a tiny amount of fuel.

I tried the same dry mass with only 100kg of fuel. Initial mass 38300kg, final mass 38200kg, exhaust velocity 37474057m/s...

97km/s delta v for getting around Earth or the Solar System with only 100kg of fuel in a 38200kg craft!

 

[PeterDonis]

It's amazing what you'd be able to do with just a tiny amount of fuel.

If it's fusion fuel, sure, since the energy per unit mass is about 5 orders of magnitude larger than for chemical fuel. The hard part is actually getting the fusion reaction to go.

 

[Devin-M]

It seems possible to do even better still:

We start off the same:
Dry Mass: 38200kg (of which 12000kg is useful payload)
Wet Mass: 556600kg
Propellant Mass: 518400kg
Propellant Energy: 719TJ/kg

But, before we do any thrusting, we use all the energy in the Deuterium to make matter-antimatter:
518400kg deuterium * 0.008kg/kg = 4147kg matter-antimatter
518400kg-4147kg = 514253kg iron

Now we dump overboard non-energetically 510106kg of iron leaving us with:

4147kg matter-antimatter
4147kg iron

The propellant mass is now:

4147kg + 4147kg = 8294kg

This changes the pre-thrust mass of the rocket to:
Dry Mass: 38200kg (of which 12000kg is useful payload)
Wet Mass: 46494kg
Propellant Mass: 8294kg (1/2 matter-antimatter, 1/2 iron)
Propellant Total Energy: 372EJ

But this becomes difficult to put into the rocket equation because pre-thrust the propellant mass is 8294kg, but only half of that mass actually comes out of the engine, because the other half turns to pure energy.

So we're back to the scenario in which an astronaut has a 0.5kg ball in his left hand, a 0.5kg ball in his right hand, and he's letting go of one ball while throwing the other...

 

[PeterDonis]

It seems possible to do even better still

The best possible scenario in terms of propellant usage is a photon rocket, which is the scenario discussed in the article on the relativistic rocket equation that I referenced. That scenario assumes that all of the propellant mass is converted to photons that are ejected as exhaust with perfect collimation.

The downside of this method is that you can't just magically convert matter to photons. You have to find actual reactions that do it.

 

[Devin-M]

Actually, I’ll have to retract my last post. It isn’t quite clear that converting all the energy to antimatter and then dumping much of the iron would actually improve the delta-v.

 

[PeterDonis]

converting all the energy to antimatter

Doesn't make sense anyway. The energy is what it is. You can't make it more energy by converting it to antimatter. You might as well just use it directly.

 

[Devin-M]

I thought by reducing total mass before thrust begins (with the same energy on board) it would be beneficial, but isn’t clear whether that’s truly the case.

 

[PeterDonis]

I thought by reducing total mass before thrust begins (with the same energy on board) it would be beneficial

Which would work the same even if you didn't turn the energy into antimatter. And I've already shown you that it doesn't make things any better.

 

[Devin-M]

I guess it makes sense. It's not like a car so decreasing the propellant mass (and by extension total mass) with the same total energy won't make it faster.

If we increase the mass of the ball the astronaut throws for the same throw energy, the astronaut always goes faster.

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

2kg Ball, 100J, 100kg Astronaut:
Astronaut: 0.198m/s
Ball: 9.9m/s

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

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]

If we increase the mass of the ball the astronaut throws for the same throw energy, the astronaut always goes faster.

I already gave a general argument for why that is the case.