Nuclear thermal propulsion — NTR Isp ~900 s concept: Worked examples

Everything below rides on one equation, so let us pin it once and never re-derive it:

The scenario matrix

Every NTR exhaust-velocity problem lives in one of these cells. The examples that follow are tagged with the cell they cover.

Cell	What varies	The physics stress it tests	Example
A	Light propellant, moderate $T_{c}$	the baseline ~900 s result	Ex 1
B	Heavy propellant, same reactor	the $1/ M$ penalty	Ex 2
C	Crank $T_{c}$ up	the $T$ (weak) gain	Ex 3
D	$T_{c} \to 0$ (degenerate: cold reactor)	limiting behaviour, sanity floor	Ex 4
E	Finite exhaust temp $T_{e} \neq = 0$	the non-ideal full formula, no shortcut	Ex 5
F	Real word problem: $Δ v$ for a Mars trip	ties $I_{s p}$ to Tsiolkovsky Rocket Equation	Ex 6
G	Exam twist: they give you $v_{e}$ , ask for $M$	inverting the formula, sign/root care	Ex 7
H	Mixed propellant (H₂ + seeded gas)	effective molar mass, weighted average	Ex 8

We deliberately have no negative-number or quadrant cases here — unlike an angle problem, every physical input ( $T$ , $M$ , $γ$ ) is strictly positive, and $v_{e}$ is a magnitude (a square root, always $\geq 0$ ). Cell D is where we prove the formula still behaves at the boundary.

Cell A — the baseline (light gas, moderate heat)

Worked example Ex 1 — Recover the famous 900 s

Given: pure hydrogen, $M = 2 \times 1 0^{- 3} kg/mol$ , $T_{c} = 2700 K$ , $γ = 1.4$ . Find: $v_{e}$ and $I_{s p}$ .

Forecast: guess before computing — will $I_{s p}$ be nearer 400, 900, or 2000 s? (The whole topic screams 900, so predict that and see if the numbers cooperate.)

Prefactor $\frac{2 γ}{γ - 1} = \frac{2 ( 1.4 )}{0.4} = 7.0$ . Why this step? It bundles the thermodynamic efficiency of converting random heat into one-directional motion; it depends only on the gas type, not on temperature.
Thermal energy per kg $\frac{R T _{c}}{M} = \frac{8.314 \times 2700}{2 \times 1 0 ^{- 3}} = 1.122 \times 1 0^{7} J/kg$ . Why this step? This is the raw energy budget locked inside each kilogram of hot hydrogen; the light $M$ in the denominator is what makes it huge.
Combine $v_{e} = 7.0 \times 1.122 \times 1 0^{7} = 7.86 \times 1 0^{7} \approx 8862 m/s$ .
Convert $I_{s p} = \frac{8862}{9.81} \approx 903 s$ .

Verify: units inside the root are $J/kg = m^{2} / s^{2}$ , so the root gives m/s ✔. Magnitude ~9 km/s is about 26× the speed of sound in cold air — hydrogen really does scream out. Matches the forecast of ~900 s.

Cell B — the $1/ M$ penalty (heavy gas, same reactor)

Worked example Ex 2 — Same 2700 K, but push steam

Given: water vapour, $M = 18 \times 1 0^{- 3} kg/mol$ , $T_{c} = 2700 K$ , $γ \approx 1.3$ . Find: $v_{e}$ , $I_{s p}$ , and the ratio to Ex 1.

Forecast: $M$ went up 9×. Since $v_{e} \propto 1/ M$ , guess $v_{e}$ drops by about $9 = 3$ . Predict ~3000 m/s.

Prefactor $\frac{2 ( 1.3 )}{0.3} = 8.667$ . Why this step? Steam has more internal degrees of freedom (lower $γ$ ), which raises the prefactor slightly — but that gain is tiny next to the $M$ loss.
Energy per kg $\frac{8.314 \times 2700}{18 \times 1 0 ^{- 3}} = 1.247 \times 1 0^{6} J/kg$ — exactly $1/9$ of Ex 1's value.
Combine $v_{e} = 8.667 \times 1.247 \times 1 0^{6} = 1.081 \times 1 0^{7} \approx 3288 m/s$ .
$I_{s p} = 3288/9.81 \approx 335 s$ .

Verify: ratio $v_{e}^{H_{2}} / v_{e}^{steam} = 8862/3288 \approx 2.70$ . The pure- $M$ prediction was $9 = 3$ ; we got a touch less because steam's higher prefactor claws back a little — sanity confirmed. Lesson: the heat source didn't change, yet $I_{s p}$ collapsed from 903 to 335. The propellant's molar mass rules.

Cell C — the weak $T$ gain (crank the reactor)

Worked example Ex 3 — What if the reactor could hit 5000 K?

Given: hydrogen as in Ex 1, but raise $T_{c}$ from 2700 K to 5000 K. Find: the new $I_{s p}$ .

Forecast: temperature nearly doubled. But it lives under a square root — so guess the gain is less than double. Predict maybe +35%.

Use the scaling shortcut $\frac{v _{e}^{new}}{v _{e}^{old}} = \frac{T _{c}^{new}}{T _{c}^{old}} = \frac{5000}{2700} = 1.852 = 1.361$ . Why this step? Everything except $T_{c}$ is identical, so all constants cancel — only the temperature ratio survives, and it enters as a square root.
Scale up $I_{s p} = 903 \times 1.361 \approx 1229 s$ .

Verify: temperature rose by 85%, but $I_{s p}$ rose only 36% — the square root damped it exactly as forecast. Reality check: solid reactor fuel elements melt near 2700–3000 K, so 5000 K is fiction for a solid core. This is why ~900 s is the material ceiling, not a physics one.

Cell D — the degenerate limit (cold reactor)

Worked example Ex 4 — Let the reactor go cold:

$T_{c} \to 0$ Given: the master formula with $M, γ$ fixed, but $T_{c}$ shrinking toward 0 K. Find: what happens to $v_{e}$ and $I_{s p}$ , and does the formula stay sane?

Forecast: no heat in → no energy to convert → nothing should come out. Predict $v_{e} \to 0$ .

Read the dependence $v_{e} = a fixed positive number \frac{2 γ}{γ - 1} \frac{R}{M} \times T_{c}$ . Why this step? Isolating $T_{c}$ shows $v_{e}$ is just a constant times $T_{c}$ .
Take the limit as $T_{c} \to 0^{+}$ , $T_{c} \to 0$ , so $v_{e} \to 0$ and $I_{s p} \to 0$ . Why this step? It confirms the formula degrades gracefully — no division blow-up, no negative root, no imaginary number.
Check the sign for all valid inputs: $T_{c} > 0$ , $M > 0$ , and $γ > 1$ (so $γ - 1 > 0$ ). Every factor under the root is positive, so $v_{e}$ is always a real, non-negative number. There is no quadrant/sign case to worry about — unlike an angle, exhaust speed is a magnitude.

Verify: at a tiny $T_{c} = 1 K$ for hydrogen: $v_{e} = 7.0 \times \frac{8.314 \times 1}{2 \times 1 0 ^{- 3}} = 2.91 \times 1 0^{4} \approx 171 m/s$ — small and positive, trending to 0 as promised. Physically sound: a cold gas barely dribbles out.

Cell E — the non-ideal full formula ( $T_{e} \neq = 0$ )

The boxed formula assumed the nozzle expands so far that exhaust temperature $T_{e} \to 0$ . Real nozzles stop at a finite $T_{e}$ . Then you must keep the full enthalpy drop, with no clean shortcut.

Worked example Ex 5 — Hydrogen with a realistic finite exhaust temperature

Given: hydrogen, $M = 2 \times 1 0^{- 3}$ , $γ = 1.4$ , $T_{c} = 2700 K$ , but the nozzle only cools the gas to $T_{e} = 300 K$ (it doesn't reach absolute zero). Find: $v_{e}$ , and compare to the idealized Ex 1.

Forecast: we're throwing away the last $300 K$ of enthalpy, which is $\frac{300}{2700} \approx 11%$ of the drop. Under a square root, guess $v_{e}$ falls by about half of that, ~5%.

Effective temperature drop $T_{c} - T_{e} = 2700 - 300 = 2400 K$ . Why this step? Only the drop in enthalpy becomes kinetic energy; whatever heat is still in the gas at the nozzle exit is energy you failed to extract.
Plug in $v_{e} = 7.0 \times \frac{8.314 \times 2400}{2 \times 1 0 ^{- 3}} = 7.0 \times 9.977 \times 1 0^{6} = 6.984 \times 1 0^{7} \approx 8357 m/s$ .
Compare $\frac{8357}{8862} = 0.943$ → a 5.7% loss.

Verify: the fraction lost obeys $1 - \frac{300}{2700} = 0.889 = 0.943$ ✔ exactly. Forecast of ~5% confirmed. Lesson: the boxed $v_{e} \approx 2 γ R T_{c} / [(γ - 1) M]$ is an upper bound; real exhaust is a few percent slower.

Cell F — the real mission (word problem)

Worked example Ex 6 — How much can 900 s buy you on a Mars burn?

Given: an NTR upper stage with $I_{s p} = 903 s$ (from Ex 1). It starts fully fuelled at mass $m_{0} = 100 tonnes$ and burns down to $m_{f} = 40 tonnes$ dry. Find: the achievable $Δ v$ using the Tsiolkovsky Rocket Equation $Δ v = v_{e} ln \frac{m _{0}}{m _{f}}$ . Then compare to a chemical stage at $I_{s p} = 450 s$ with the same mass ratio.

Forecast: $I_{s p}$ doubled vs chemical, and $Δ v$ scales linearly with $v_{e}$ , so guess the NTR gives about double the $Δ v$ .

Exhaust speed $v_{e} = I_{s p} g_{0} = 903 \times 9.81 = 8858 m/s$ . Why this step? Tsiolkovsky wants $v_{e}$ in m/s, not seconds; multiply by $g_{0}$ to undo the $I_{s p}$ definition.
Mass-ratio log $ln \frac{100}{40} = ln 2.5 = 0.9163$ . Why this step? The rocket equation says $Δ v$ depends only on this dimensionless ratio and $v_{e}$ — the logarithm is why extra fuel gives diminishing returns.
NTR $Δ v = 8858 \times 0.9163 \approx 8117 m/s$ .
Chemical, same ratio $v_{e} = 450 \times 9.81 = 4414.5$ , $Δ v = 4414.5 \times 0.9163 \approx 4045 m/s$ .

Verify: ratio $8117/4045 = 2.007 \approx 2$ ✔ — the $Δ v$ doubled, matching the $I_{s p}$ doubling because $m_{0} / m_{f}$ was identical. A trans-Mars injection needs roughly $3.6 km/s$ ; the NTR's $8.1 km/s$ clears it with margin to spare, while the same-mass chemical stage barely has more than one such burn.

Cell G — the exam twist (invert the formula)

Worked example Ex 7 — Given the exhaust speed, back out the molar mass

Given: a mystery NTR propellant produces $v_{e} = 6000 m/s$ at $T_{c} = 2700 K$ with $γ = 1.4$ . Find: its molar mass $M$ . Which real gas is it?

Forecast: $6000$ m/s is slower than pure hydrogen's $8862$ , so $M$ must be heavier than 2 g/mol. Guess somewhere around 4 g/mol (helium?).

Square both sides $v_{e}^{2} = \frac{2 γ}{γ - 1} \frac{R T _{c}}{M}$ . Why this step? The unknown $M$ is trapped inside a square root; squaring frees it. We keep the positive root only — $v_{e}$ is a speed, so no $\pm$ ambiguity.
Solve for $M$ $M = \frac{2 γ}{γ - 1} \frac{R T _{c}}{v _{e}^{2}} = 7.0 \times \frac{8.314 \times 2700}{600 0 ^{2}} = 7.0 \times \frac{2.245 \times 1 0 ^{4}}{3.6 \times 1 0 ^{7}}$ .
Compute $M = 7.0 \times 6.235 \times 1 0^{- 4} = 4.36 \times 1 0^{- 3} kg/mol = 4.36 g/mol$ .

Verify: plug $M = 4.36 \times 1 0^{- 3}$ back into the forward formula: $v_{e} = 7.0 \times \frac{8.314 \times 2700}{4.36 \times 1 0 ^{- 3}} = 3.6 \times 1 0^{7} = 6000$ m/s ✔. The value ~4 g/mol is helium ( $M \approx 4$ ) — a real (if wasteful, since He is inert and expensive) NTR candidate. Forecast confirmed: heavier than H₂, hence slower.

Cell H — mixed propellant (effective molar mass)

Sometimes hydrogen is "seeded" with a bit of a heavier gas (for opacity, so the reactor radiation heats it better). The mixture behaves like a single gas with an effective molar mass — the mole-weighted average.

Worked example Ex 8 — 95% hydrogen + 5% xenon by moles

Given: $95%$ H₂ ( $M = 2 g/mol$ ) and $5%$ xenon ( $M = 131 g/mol$ ) by mole fraction. Reactor $T_{c} = 2700 K$ , take $γ = 1.4$ . Find: $M_{eff}$ and the $I_{s p}$ , compared to pure hydrogen's 903 s.

Forecast: only 5% xenon — sounds negligible. But xenon is 65× heavier than H₂. Guess $M_{eff}$ is dragged up noticeably, maybe to ~8 g/mol, and $I_{s p}$ drops well below 903.

Weighted average $M_{eff} = 0.95 \times 2 + 0.05 \times 131 = 1.90 + 6.55 = 8.45 g/mol = 8.45 \times 1 0^{- 3} kg/mol$ . Why this step? A few very heavy molecules carry disproportionate mass, so they dominate the average — this is exactly the "small heavy contamination" trap.
Exhaust speed $v_{e} = 7.0 \times \frac{8.314 \times 2700}{8.45 \times 1 0 ^{- 3}} = 7.0 \times 2.657 \times 1 0^{6} = 1.860 \times 1 0^{7} \approx 4313 m/s$ .
$I_{s p} = 4313/9.81 \approx 440 s$ .

Verify: cross-check via the ratio $I_{s p} \propto 1/ M_{eff}$ : $903 \times 2/8.45 = 903 \times 0.4866 = 439$ s ✔. Lesson: a mere 5% of a heavy seed slashed $I_{s p}$ from 903 to ~440 s — right back to chemical-rocket territory. Purity of the light propellant is everything; this is why real NTRs guard against contamination.

Coverage recap

Figure — Nuclear thermal propulsion — NTR Isp ~900 s concept

The bar chart above places all eight examples on one $I_{s p}$ axis so you can see the story at a glance: pure hot hydrogen sits near 900 s, the $T$ crank barely lifts it, and any mass contamination (steam, helium, xenon seed) drags it down hard. Molar mass, not temperature, is the lever.

Recall Which single variable moved

$I_{s p}$ the most across all examples? Molar mass $M$ ::: It enters as $1/ M$ and spanned 2→131 g/mol across the cases, swinging $I_{s p}$ from 903 s down to 335 s. Temperature only entered as $T$ and is capped by melting, so it moves $I_{s p}$ far less.

Recall Why is there no "negative case" or "quadrant case" in this whole topic?

Because every physical input ( $T_{c} > 0$ , $M > 0$ , $γ > 1$ ) is strictly positive and $v_{e}$ is a square root — a magnitude. ::: The formula can never produce a negative or imaginary exhaust speed for valid inputs; the only boundary case is $T_{c} \to 0$ giving $v_{e} \to 0$ (Cell D).

Connections

Specific Impulse — the $I_{s p}$ every example computes.
Tsiolkovsky Rocket Equation — used explicitly in Ex 6 to cash in $I_{s p}$ as $Δ v$ .
De Laval Nozzle — the finite- $T_{e}$ expander behind Cell E.
Adiabatic Flow & Enthalpy — the energy balance $\frac{1}{2} v_{e}^{2} = c_{p} (T_{c} - T_{e})$ underneath every formula here.
Nuclear Fission — the heat source setting $T_{c}$ .
Chemical Rocket Propulsion — the ~450 s baseline that Ex 6 and Ex 8 land back on.
Nuclear Electric Propulsion — the cousin that trades thrust for even higher $I_{s p}$ .