I just watched a very good YouTube video on why nuclear engines might be useful, and it also goes into why Hall Effect thrusters are super good at squeezing obscene ISP out of things.

During the video, however, the author talks about using hydrogen as remass (more specifically about how hydrogen is a pain in the ass to store), because it gives better ISP than, say, Helium.

There is an attempt at explanation (here:

), which doesn't connect for me, here's my readback of it so that someone can help me understand what's actually going on:

  1. ISP (gas mileage in space) is maximized by optimizing for the exhaust velocity of your remass.
  2. Because hydrogen has a lower atomic mass, a given quantity of power dumped into it results in it moving faster than, say, helium. (For the thermal nuclear rocket design being discussed, hydrogen was giving an ISP ~880s while helium was giving ~650s.)

I can understand why a given kick would make the heavier object move more slowly, that's literally how power and mass work.

What I don't understand is why that results in less force on the rocket? If I'm pouring a megawatt of power into the propellant to chuck it out the back, regardless of how fast I end up accelerating it, I should have the same equal, opposite reaction on me, no?


Consider two vehicles, A and B, which have identical, 1MW motive systems but use different remass, A has hydrogen, B has helium.

Each expends 1 helium atom's worth of fuel (identical mass).

A exerts 1MW of power on four atoms of hydrogen.

B exerts 1MW of power on one atom of helium.

The same force was applied to the same mass, but supposedly A gets 25% more velocity from the burn? Why does the hydrogen have a higher exhaust velocity when it's having to share the force the vehicle exerts on it among four times as many particles?

I would expect a given kg of fuel to have the same 'push' to it given identical propulsion schemes.

I'm sure the physics make sense here, but I'm clearly missing something happening so why do I care how fast my propellant is going, relative to me? If I shove on two hydrogen atoms with 1MN of force, and my buddy next to me shoves on the helium atom with 1MN of force, why do we not end up going the same speed?

  • 1
    $\begingroup$ I believe OP is referencing this Youtube video from Real Engineering: youtu.be/MMLgJlJX0Rk $\endgroup$ Apr 20, 2021 at 18:56
  • 1
    $\begingroup$ Would it be fair to say your question is really about thrust, not Isp? Based on this part of your question "What I don't understand is why that results in less force on the rocket?" (emphasis mine) $\endgroup$ Apr 20, 2021 at 18:58
  • 2
    $\begingroup$ Thrust is mass flow x exhaust velocity. If you throw away the same mass slower, you get less thrust. $\endgroup$ Apr 20, 2021 at 19:00
  • 2
    $\begingroup$ OK, it sounds like you want the derivation of the thrust equation. I can work with that. Thanks for the clarification. $\endgroup$ Apr 20, 2021 at 19:04
  • 1
    $\begingroup$ Happy to clarify in the original copy of the question, too. I'm at "I know enough to be confused, but not enough to know what to ask." ;) $\endgroup$ Apr 20, 2021 at 19:05

5 Answers 5


Why lighter atoms work better as fuel for a rocket:

The simple explanation, with concept only no numbers.

A rocket takes an amount of energy, puts that energy into matter, and that causes the matter to be shoved out the rear of the rocket, producing thrust.

(The energy usually comes from chemical reactions, thus heat. But it could also be pure thermal, or electrostatic, or whatever... It does not matter for this discussion.)

So you have an amount of energy
being put into a mass of matter by accelerating that mass to velocity

If you manage to squeeze the same amount of energy into less mass, that mass is moving faster. This produces more thrust for the same mass, thus better fuel efficiency.
Higher exhaust velocity = higher ISP = more thrust from the same fuel. (but much more energy needed)

Why does a lighter atom go faster? Whether from thermal heat, or an applied electric field, or whatever.. Your engine is applying a certain force on the propellant.
The force being applied depends on the engine.
Applying the same force on a heavy object, imparts a slow speed on the object.
Applying the same force on a light object, imparts a lot of speed to the object.
There is no atom lighter than a monatomic Hydrogen atom!

Simple explanation, with a few numbers. (But no fancy units, constants etc.)

One thing to remember:
A rocket engine needs Energy to accelerate its Propellant
But it's not the energy that drives the rocket, it is the Momentum.

Let's give your rocket motor an energy of 100 thingies per second.

If this rocket is accelerating 1 Hydrogen (mass 1), it gets it up to a speed of sqrt(100/1) = 10
This imparts momentum of Speed * Mass = 10 * 1 = 10 to the rocket

If this rocket is accelerating 1 Helium (mass 4), it gets it up to a speed of sqrt(100/4) = 5
This imparts momentum of Speed * Mass = 5 * 4 = 20 to the rocket

NOTE that using the same energy input, you got twice the speed out of the Hydrogen, per item
But the Hydrogen only masses 1/4 as much, so for the same FUEL MASS, you get two times the thrust total. (While burning 4 times the energy.)

The hydrogen gives 2 * the ISP of the Helium.

Note, you can only realistically play with substitutions like this when the propellant is only propellant, not also your energy source. For Chemical engines, the Fuel is both energy source and propellant, and changing the composition of the propellant changes the energy from the burning hereof, etc.
But in the example the OP was looking at, the energy source is separate from the propellant, and thus allows some leeway in propellant selection.

  • $\begingroup$ "If you [apply] the same amount of energy into less mass... this produces more thrust for the same mass." This is the part that I snag on, how does this not violate conservation of energy? If I have 1kg of hydrogen atoms, and 1kg of helium atoms, I can shove on both kilograms with 1MN of force and both should move away from me at 1Mm/sec, having imparted upon me the same portion of that total systemic acceleration. The only way I can see your explanation working is if I'm applying the same force to equal numbers of atoms, in which case I get 4x the bang from the same mass, H vs. He, not 25%? $\endgroup$ Apr 20, 2021 at 20:27
  • 1
    $\begingroup$ @WilliamWalkerIII Ok, adding part 2: a few numbers. $\endgroup$ Apr 20, 2021 at 20:31
  • 1
    $\begingroup$ @WilliamWalkerIII The processes in both thermal (such as chemical and nuclear thermal) and electric engines apply on average equal energy to all atoms or molecules in a parcel (for gases it's a statistical process — see Maxwell-Boltzmann distributions). In the thermal case, a lighter molecule will have a higher velocity than a heavier one, and the de Laval nozzle converts the randomly-oriented initial velocities to a collimated flow with residual randomly-oriented components; the higher the initial random velocities (thus, the lighter the molecules), the faster the collimated flow. $\endgroup$ Apr 21, 2021 at 0:02
  • 1
    $\begingroup$ @WilliamWalkerIII For your question about relativistic reaction mass - we're nowhere close to that yet. We have trouble exceeding 5000s with wimpiest least thrust possible drives, and nuclear propulsion will operate in realm of 800-2000s. Meanwhile, the highest possible (by laws of physics) specific impulse is of order of 30mln.seconds. As the exhaust approaches speed of light, you're starting to convert energy into mass - no more specific impulse is gained, only thrust increases and with external energy source you could cheat extra "free" reaction mass out of energy. $\endgroup$
    – SF.
    Apr 21, 2021 at 7:59
  • 1
    $\begingroup$ @William This might help. In a sample of gas at a particular temperature, the gas molecules have a spectrum of kinetic energies. The temperature of the gas is directly proportional to the mean (translational) kinetic energy of the molecules. So if you have some helium atoms and some hydrogen atoms at the same temperature, the hydrogen atoms have twice the average speed of the helium atoms. $\endgroup$
    – PM 2Ring
    Apr 21, 2021 at 17:56

For an intuitive understanding of the answer, it helps to keep a few points in mind:

Momentum is not kinetic energy. Momentum is conserved, while kinetic energy is not (it converts to other forms of energy). Two colliding lumps of clay will conserve momentum, but not kinetic energy. Two pool balls colliding conserve both momentum and kinetic energy.

Momentum is proportional to velocity; kinetic energy is proportional to the square of velocity.

A rocket works by throwing momentum, not energy, out the tailpipe. That’s why the Rocket Equation has “v”, not “v2”.

Temperature is, by definition, the average kinetic energy of gas molecules. Two gasses with different molecular weights (like H2 and He), at the same temperature, will have different molecular velocities inversely proportional to the square roots of their molecular weights. He is slower than H2.

A gas cannot expand faster than the average velocity of its molecules. At the same temperature, H2 will expand into a vacuum at 20.5 times the velocity of He. So, for a given mass of H2 and He at the same temperature, the H2 will ideally provide about 40% more delta v than He.

Also, He has 20% higher specific heat than H2 so it will take more energy to raise a given mass of He to a specific temperature than for the same mass of H2. He loses again.

  • 1
    $\begingroup$ "Also, He has 20% higher specific heat than H2." According to Wikipedia, hydrogen's heat capacity is 14.304 J/(g K), while helium's is 5.193. And it makes sense: first, a gram of hydrogen contains more molecules to provide energy to than a gram of helium; second, hydrogen molecules have not just translational degrees of freedom, but also rotational ones, unlike helium molecules. $\endgroup$
    – Litho
    May 3, 2022 at 17:36

Okay, so my previous approach didn't work out. So let's try another.

Hydrogen need not have any more ISP than reinforced concrete. Indeed, there's little difference between the two if you are simply throwing hydrogen canisters out of the nozzle to get thrust. That is, if you are accelerating the reaction mass that is stationary macroscopically and microscopically, straight backwards.

But you aren't. Your reaction mass is already moving at a microscopic scale. You are merely converting this random motion into directional macroscopic movement. This is when low molecular mass helps

  • $\begingroup$ "thermal energy in a given mass of gas at a given temperature is proportional to number of gas molecules, independent of molecular mass; E = PV = NKT" This would certainly make the thing make sense. Can you elaborate on this bullet from the physics perspective? Why is this the case? It seems to me that more mass means you can have more total thermal energy (certainly true of solids and liquids)? What makes gasses ignore mass in favor of particle count? $\endgroup$ Apr 20, 2021 at 20:37
  • $\begingroup$ @WilliamWalkerIII ah, there you go. Feel free to ask if you still don't get it. $\endgroup$ Apr 21, 2021 at 7:38
  • 2
    $\begingroup$ @WilliamWalkerIII " Why is this the case? It seems to me that more mass means you can have more total thermal energy (certainly true of solids and liquids)? What makes gasses ignore mass in favor of particle count?" It's more or less the same for solids and liquids as well. To heat 1 kg of lithium by 1 Kelvin, you need about 8 times more energy than to heat 1 kg of iron by 1 Kelvin, because 1 kg of lithium has about 8 times more atoms than 1 kg of iron. $\endgroup$
    – Litho
    Apr 21, 2021 at 10:37
  • 1
    $\begingroup$ "The relevance here is that it is the collision of one hydrogen atom with too much energy with one atom of the engine that causes the engine atom to detach; AKA melting. [...] Hence temperature is dependant only on particle kinetic energy. Nowhere here is atomic mass even mentioned - because it's irrelevant." It's not that simple. How much energy an atom of propellant can transfer to an atom of engine over a collision is determined not only by how much energy it has, but also by their mass ratio. If a moving train hits a dust mote, it cannot transfer all of its KE to it, only a tiny part. $\endgroup$
    – Litho
    Apr 22, 2021 at 11:29
  • 1
    $\begingroup$ Atoms of the engine jiggle in places, so when atoms of propellant hit them, sometimes energy is transfered from propellant to engine, sometimes the other way around. The important thing is not how much energy can be transferred over a single collision, but whether on the average, over many collisions, energy goes from propellant to engine or the other way around. And it turns out that propellant will transfer energy to the engine until the avg energy of engine's atoms' oscillations is equal to the avg KE of propellant's atoms, independently of their mass ratio. But it's not easy to prove. $\endgroup$
    – Litho
    Apr 22, 2021 at 11:35

If hydrogen and helium are being used as reaction mass, they will be molecular, not atomic, at non-ionizing exhaust temperatures. The atomic mass of hydrogen is 1 but the molecular mass of H2 is 2. The atomic and molecular mass of He are both the same =4. At the same temperature, the H2 will have 1.4 times the velocity of the He. ISP will be 40% higher for H2 assuming the same mass of reaction fuel and the same exhaust temperature.

  • $\begingroup$ I was surprised the bond dissociation energy for H2 is so high; ~4.5 eV corresponding to about 50,000 K. Durable stuff! $\endgroup$
    – uhoh
    Jan 5 at 0:35

Because it's cheaper..... Liquid hydrogen can be produced from natural gas or water. liquid hydrogen costs about 9-15 dollars per kilogram Liquid helium costs 40 dollars per kilogram.

  • $\begingroup$ rats! I voted to delete then did it again by accident which irrnversably retracted it. $\endgroup$
    – uhoh
    Jan 5 at 0:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.