Green propellants — LMP-103S, AF-M315E (vs hydrazine): Worked examples

The scenario matrix

Before working anything, let us list every kind of input these three equations can be handed. Each row is a "cell" — a distinct behaviour we must show at least once.

Cell	What varies / edge condition	Which equation	Example
A	Normal forward conversion $I_{s p} \to v_{e}$	$v_{e} = I_{s p} g_{0}$	Ex 1
B	Reverse: given $v_{e}$ , recover $I_{s p}$	$I_{s p} = v_{e} / g_{0}$	Ex 2
C	Tank-metric comparison (density enters)	$ρ I_{s p}$	Ex 3
D	$T_{c} / M$ — hotter but heavier (which wins?)	scaling	Ex 4
E	Degenerate: $M$ or $T_{c}$ shrinks toward a limit	scaling limit	Ex 5
F	Zero-ish fuel: tiny burn, $m_{0} \approx m_{f}$	Tsiolkovsky	Ex 6
G	Large fuel fraction: $m_{f} \to$ small, log blows up	Tsiolkovsky	Ex 7
H	Real-world word problem (mission planning)	all three	Ex 8
I	Exam twist: a hidden unit / $g_{0}$ trap	$I_{s p} \leftrightarrow v_{e}$	Ex 9

The figures below anchor cells C, D/E, and F/G where the shape of the answer matters.

Example 1 — Cell A: forward conversion

Example 2 — Cell B: reverse conversion

Example 3 — Cell C: the tank metric (density enters)

Worked example Statement

A cubesat tank is volume-limited at $V = 2.0$ L. Compare total impulse capacity of AF-M315E ( $ρ = 1.47$ kg/L, $I_{s p} = 266$ s) vs hydrazine ( $ρ = 1.01$ kg/L, $I_{s p} = 230$ s). Total impulse $= m_{fuel} I_{s p} g_{0}$ .

Forecast: green is both denser and higher $I_{s p}$ — predict a big win, well over 50% more.

Fuel mass carried $= ρ V$ : AF-M315E $= 1.47 \times 2.0 = 2.94$ kg; hydrazine $= 1.01 \times 2.0 = 2.02$ kg. Why this step? A fixed volume holds different masses depending on density — that's the whole point of density.
Total impulse $J = m I_{s p} g_{0}$ : green $= 2.94 \times 266 \times 9.81 = 7672 N\cdots$ ; hydrazine $= 2.02 \times 230 \times 9.81 = 4558 N\cdots$ . Why this step? Each kg of fuel delivers $I_{s p} g_{0}$ of momentum ( $= v_{e}$ ), so multiply mass by $v_{e}$ .
Ratio $= 7672/4558 = 1.68$ . Why this step? The $g_{0}$ cancels, so the ratio equals $ρ_{g} I_{s p, g} / (ρ_{h} I_{s p, h})$ — the density-impulse product from the parent note.

Verify: the parent got $ρ I_{s p}$ ratio $= 391/232 = 1.69$ ; ours is $1.68$ (rounding) ✓. Look at s01: the amber bar (green fuel) towers over the cyan bar — the visual of "same bottle, more punch." Units: $[kg] [s] [m/s^{2}] = [N\cdots]$ ✓.

Example 4 — Cell D: hotter but heavier — which wins?

Worked example Statement

Propellant X: $T_{c} = 1800$ K, $M = 20$ g/mol. Propellant Y: $T_{c} = 1000$ K, $M = 12$ g/mol. Which has the higher $I_{s p}$ (proportional to $T_{c} / M$ )?

Forecast: X is hotter (good) but heavier (bad). Guess before computing — does 1.8× temperature beat 1.67× mass?

Form each ratio $T_{c} / M$ : X $= 1800/20 = 90$ ; Y $= 1000/12 = 83.3$ . Why this step? The performance driver is the ratio, not $T_{c}$ or $M$ alone — temperature lifts energy, mass drags it down.
Take square roots: X $= 90 = 9.49$ ; Y $= 83.3 = 9.13$ . Why this step? $v_{e} \propto T_{c} / M$ because kinetic energy $\frac{1}{2} v^{2}$ scales with thermal energy $\propto T_{c} / M$ — the square root converts energy to speed.
X wins ( $9.49 > 9.13$ ). Why this step? Higher $T_{c} / M$ means higher exhaust speed and thus $I_{s p}$ .

Verify: heavier exhaust ( $20$ vs $12$ ) yet X still wins — this is the parent's "Hot & Heavy beats Cool & Light" rule in action. On s02, X's point sits on a higher curve of the $T_{c} / M$ surface. Units inside the root cancel to a pure number since we compare ✓.

Example 5 — Cell E: degenerate limit (mass shrinks toward zero)

Worked example Statement

Keep $T_{c} = 1800$ K fixed. What happens to $T_{c} / M$ as the exhaust molar mass $M$ is made lighter and lighter — $M = 18, 9, 2$ g/mol (approaching pure $H_{2}$ )?

Forecast: lighter gas is thrown faster — expect the value to grow without bound as $M \to 0$ .

$M = 18$ : $1800/18 = 100 = 10.00$ . Why this step? A clean round case to anchor the trend.
$M = 9$ : $1800/9 = 200 = 14.14$ .
$M = 2$ : $1800/2 = 900 = 30.00$ . Why these steps? Halving $M$ multiplies the ratio by $2$ and the root by $2 \approx 1.41$ — we are watching the limiting behaviour $T_{c} / M \to \infty$ as $M \to 0$ .

Verify: each halving of $M$ scales the answer by $2$ : $10.00 \times 1.414 = 14.14$ ✓; $14.14 \times 4.5$ ... check directly $900 = 30$ ✓. Physical sanity: pure hydrogen exhaust is why nuclear-thermal rockets chase $I_{s p} \sim 900$ s — light exhaust is thrown fastest. But real green propellants can't hit $M = 2$ : their exhaust ( $N_{2}, H_{2} O, C O_{2}$ ) is heavy, so they must win on $T_{c}$ instead. See Thermochemistry & enthalpy of decomposition.

Example 6 — Cell F: near-zero fuel (tiny burn)

Worked example Statement

A 50 kg cubesat performs a tiny station-keeping nudge, burning only 0.10 kg of AF-M315E ( $v_{e} = 2609$ m/s). Find $Δ v$ .

Forecast: almost no mass leaves, so $m_{0} \approx m_{f}$ — expect a tiny $Δ v$ , a few m/s.

Masses: $m_{0} = 50.00$ kg (wet), $m_{f} = 50.00 - 0.10 = 49.90$ kg (dry). Why this step? Tsiolkovsky needs the before and after masses — the difference is the fuel spat out.
Ratio and log: $\frac{m _{0}}{m _{f}} = \frac{50.00}{49.90} = 1.002004$ , $ln (1.002004) = 0.002002$ . Why this step? The rocket equation uses the logarithm of the mass ratio because each little slug of fuel is thrown by an already-lighter rocket — the effect compounds.
$Δ v = 2609 \times 0.002002 = 5.22 m/s$ . Why this step? Multiply exhaust speed by the log-ratio, per Tsiolkovsky.

Verify: tiny burn → tiny $Δ v$ ✓. Sanity check with the near-linear approximation $Δ v \approx v_{e} \cdot (m_{fuel} / m_{0}) = 2609 \times (0.10/50) = 5.22$ m/s — matches because for small burns $ln (1 + x) \approx x$ . On s03 this is the flat left end of the curve.

Example 7 — Cell G: large fuel fraction (log blows up)

Worked example Statement

A 100 kg tanker burns down to just 20 kg dry mass using AF-M315E ( $v_{e} = 2609$ m/s). Find $Δ v$ . Then push further: what if it burned to 10 kg?

Forecast: 80% of the mass leaves — expect a large $Δ v$ , and the 10 kg case should be much bigger still (log grows fast near an empty tank).

Case 1: $m_{0} / m_{f} = 100/20 = 5$ , $ln 5 = 1.6094$ . Why this step? Large mass ratio → large logarithm; this is where fuel efficiency really pays off.
$Δ v = 2609 \times 1.6094 = 4200 m/s$ . Why this step? Tsiolkovsky again — same formula, but now the log is large.
Case 2: $m_{0} / m_{f} = 100/10 = 10$ , $ln 10 = 2.3026$ , $Δ v = 2609 \times 2.3026 = 6008 m/s$ . Why this step? Halving the dry mass doesn't halve or double $Δ v$ — it adds $v_{e} ln 2 = 2609 \times 0.693 = 1808$ m/s, showing the diminishing (logarithmic) return.

Verify: $4200 + 1808 = 6008$ m/s ✓ — the extra fuel adds a fixed $v_{e} ln 2$ each time you halve dry mass, the signature of the log. On s03 this is the steep right side of the curve rising toward the $m_{f} \to 0$ wall (where $Δ v \to \infty$ , the limiting case).

Example 8 — Cell H: real-world mission problem

Worked example Statement

A 120 kg smallsat must achieve $Δ v = 300$ m/s using AF-M315E ( $v_{e} = 2609$ m/s, $ρ = 1.47$ kg/L). (a) How much fuel mass is needed? (b) What tank volume?

Forecast: 300 m/s is modest vs $v_{e} = 2609$ , so expect the fuel to be a small fraction of the satellite — maybe ~13 kg.

Invert Tsiolkovsky for the mass ratio: $\frac{m _{0}}{m _{f}} = e^{Δ v / v_{e}} = e^{300/2609} = e^{0.11499} = 1.1219$ . Why this step? We know $Δ v$ and want mass; the exponential undoes the logarithm in the rocket equation.
Dry mass is the fixed satellite: $m_{f} = 120$ kg (the structure that must survive). Then $m_{0} = 1.1219 \times 120 = 134.6$ kg. Why this step? $m_{f}$ is what's left after the burn — the dry satellite; $m_{0}$ includes the fuel we're solving for.
Fuel mass $= m_{0} - m_{f} = 134.6 - 120 = 14.6$ kg. Why this step? The difference between wet and dry mass is the propellant.
Tank volume $= m_{fuel} / ρ = 14.6/1.47 = 9.93$ L. Why this step? Density converts the required mass into the physical bottle size — the true design constraint on a cubesat.

Verify: plug back: $Δ v = 2609 ln (134.6/120) = 2609 ln (1.1217) = 2609 \times 0.11487 = 300$ m/s ✓ (rounding). Fuel is $14.6/134.6 = 10.8%$ of wet mass — a small fraction, matching the forecast. Units: $[kg] / [kg/L] = [L]$ ✓.

Example 9 — Cell I: the exam twist (the $g_{0}$ / unit trap)

Worked example Statement

Trap question: "A propellant has $I_{s p} = 252$ . Its $Δ v$ for a mass ratio of 2 is $252 ln 2 = 175$ ." Spot and fix the error.

Forecast: something is missing — $I_{s p}$ is in seconds, but $Δ v$ needs m/s.

Identify the units: Tsiolkovsky is $Δ v = v_{e} ln (m_{0} / m_{f})$ , and $v_{e}$ is in m/s, not seconds. Why this step? Mixing seconds ( $I_{s p}$ ) directly into a velocity equation is the classic error the parent note warns about.
Convert first: $v_{e} = I_{s p} g_{0} = 252 \times 9.81 = 2472$ m/s. Why this step? You must turn "seconds" into a real exhaust speed before using the rocket equation.
Now compute correctly: $Δ v = 2472 \times ln 2 = 2472 \times 0.6931 = 1714 m/s$ . Why this step? Same formula, right units — the answer is ~ $9.8 \times$ bigger than the wrong "175" (exactly the missing factor $g_{0}$ ).

Verify: the wrong answer $175$ times $g_{0} = 9.81$ gives $1717 \approx 1714$ m/s ✓ — confirming the forgotten factor was precisely $g_{0}$ . Always multiply $I_{s p}$ by $g_{0}$ before it meets $ln (m_{0} / m_{f})$ .

Recall Which cell was hardest? Self-test

In Example 7, why does halving the dry mass add a constant $v_{e} ln 2$ rather than a constant $Δ v$ percentage? ::: Because $Δ v = v_{e} ln (m_{0} / m_{f})$ and $ln (m_{0} / (m_{f} /2)) = ln (m_{0} / m_{f}) + ln 2$ — the log turns the multiply-by-2 into an add- $ln 2$ , so each halving contributes the same $v_{e} ln 2 \approx 1808$ m/s.

In Example 5, what stops a real green propellant reaching $M \to 0$ (huge $I_{s p}$ )? ::: Its exhaust is $N_{2}, H_{2} O, C O_{2}$ — inherently heavy. It can't get light exhaust, so it wins on high $T_{c}$ instead (see Oxidisers — nitrate & dinitramide chemistry).

Linked topics: Green Chemistry & Sustainability · Ionic liquids · Hydrazine · Catalysis · Rocket equation (Tsiolkovsky).