Typical Isp values — solid (~260s), LOX - RP1 (~311s), LOX - LH2 (~450s), ion engines (~3000s): Exercises

First, the symbols this page uses — read these before the formula block below:

Definition Symbols used on this page

$g_{0} = 9.81 m/s^{2}$ — a fixed unit-conversion constant (NOT local gravity). It exists only so that $I_{s p}$ comes out in seconds.
$I_{s p}$ — specific impulse, in seconds: an efficiency score for the engine + propellant.
$v_{e}$ — exhaust velocity, in m/s: the physical speed the burnt gas leaves at.
$\overset{m}{˙}$ — propellant mass flow rate, in kg/s: how much mass the engine throws out each second.
$F$ — thrust, in newtons (N): the push force the engine makes.
$m_{0}$ / $m_{f}$ — start (wet) mass and final (dry) mass in kg; their ratio $m_{0} / m_{f}$ is the mass ratio.
$T_{c}$ — combustion-chamber temperature in kelvin (K): how hot the burning gas is before it leaves the nozzle. Hotter gas moves faster.
$M$ — mean molar mass of the exhaust gas in g/mol (or kg/mol): how heavy, on average, each exhaust molecule is. Lighter molecules move faster for the same heat.

Now the formulas, every symbol in them already defined above:

The figure below is the "wiring diagram" for exactly that idea: it shows the master relation $v_{e} = I_{s p} g_{0}$ in the centre and three arrows out to the thrust, mass-flow, and rocket-equation forms. When you get stuck on a problem, find which of the three boxes you need, then follow the arrow back to the centre to see what to substitute.

Figure — Typical Isp values — solid (~260s), LOX - RP1 (~311s), LOX - LH2 (~450s), ion engines (~3000s)

Reference numbers (memorise these — L1 leans on them):

Engine	Propellant	$I_{s p}$ (s)	$v_{e} = I_{s p} g_{0}$ (m/s)
Solid	APCP	~260	~2550
Kerolox	LOX/RP-1	~311	~3050
Hydrolox	LOX/LH2	~450	~4410
Ion	Xe/Kr	~3000	~29 400

Two problems (Q7, Q12) also use the chemical scaling law below. Here is why it holds, sketched in place:

L1 — Recognition

Recall Solution Q1

What we do: just read the ladder of benchmark values. LOX/LH2 = hydrolox ≈ 450 s. Among the four listed families (solid 260, kerolox 311, hydrolox 450, ion 3000) it is the highest of the chemical engines, but far below the ion engine. Why: exhaust velocity for chemical rockets scales as $v_{e} \propto T_{c} / M$ (the scaling law sketched above, with $T_{c}$ the chamber temperature and $M$ the exhaust molar mass). Hydrolox exhaust is mostly water and leftover H $_{2}$ — the lowest molar mass $M$ of any chemical propellant — so it gets the highest chemical $v_{e}$ , hence highest chemical $I_{s p}$ . Answer: $\approx 450$ s, highest of the chemicals.

Recall Solution Q2

What we do: $I_{s p}$ in seconds is exhaust velocity divided by $g_{0}$ , so multiply back. Why this tool: the definition $I_{s p} = v_{e} / g_{0}$ is literally "exhaust speed measured in units of $g_{0}$ ." To undo that and recover the physical speed we multiply by $g_{0}$ . $v_{e} = I_{s p} g_{0} = 3000 \times 9.81 = 29, 430 m/s \approx 29.4 km/s$ Answer: $\approx 29, 400$ m/s.

L2 — Application

Recall Solution Q3

What we do: thrust is momentum thrown per second, $F = \overset{m}{˙} v_{e}$ , and $v_{e} = I_{s p} g_{0}$ . Why this tool: by Newton's 3rd law each kilogram leaving at speed $v_{e}$ pushes back with momentum $\overset{m}{˙} v_{e}$ ; that is the force. See Thrust and Mass Flow Rate. $F = \overset{m}{˙} I_{s p} g_{0} = 500 \times 311 \times 9.81 = 1, 525, 455 N \approx 1.53 MN$ Answer: $\approx 1.53 \times 1 0^{6}$ N.

Recall Solution Q4

What we do: rearrange $F = \overset{m}{˙} I_{s p} g_{0}$ to solve for $\overset{m}{˙}$ . Why: we know force and $I_{s p}$ ; the only unknown is how much mass per second must leave. $\overset{m}{˙} = \frac{F}{I _{s p} g _{0}} = \frac{12 \times 1 0 ^{6}}{260 \times 9.81} = \frac{12 \times 1 0 ^{6}}{2550.6} \approx 4705 kg/s$ Answer: $\approx 4705$ kg/s — nearly 5 tonnes of propellant every second. That's why boosters are gone in ~2 minutes.

Recall Solution Q5

What we do: same $F = \overset{m}{˙} I_{s p} g_{0}$ , now with a minuscule $\overset{m}{˙}$ . $F = 5 \times 1 0^{- 6} \times 3000 \times 9.81 = 0.147 N$ Answer: $\approx 0.15$ N — about the weight of a few coins. Huge $I_{s p}$ , negligible thrust. See Ion and Electric Propulsion.

L3 — Analysis

Recall Solution Q6

What we do: invert the Tsiolkovsky Rocket Equation $Δ v = I_{s p} g_{0} ln (m_{0} / m_{f})$ for the mass ratio. Why the logarithm / exponential: the rocket equation says velocity gained is proportional to the logarithm of the mass ratio (each kg burned matters less than the last, because you're also carrying the still-unburned fuel). To get the ratio out of a log we apply its inverse, the exponential. $\frac{m _{0}}{m _{f}} = e^{Δ v / (I_{s p} g_{0})}$ Solid: $\frac{3000}{260 \times 9.81} = \frac{3000}{2550.6} = 1.1762 \Rightarrow e^{1.1762} = 3.242$ . Propellant fraction $= 1 - \frac{1}{3.242} = 0.691 = 69.1%$ . Ion: $\frac{3000}{3000 \times 9.81} = \frac{3000}{29430} = 0.10194 \Rightarrow e^{0.10194} = 1.1073$ . Propellant fraction $= 1 - \frac{1}{1.1073} = 0.0969 = 9.69%$ . Answer: solid needs mass ratio $3.24$ (69% propellant); ion needs $1.11$ (only 9.7% propellant) — the ion does the same job with ~7× less propellant mass.

Recall Solution Q7

What we do: with $T_{c}$ equal, the temperature cancels and only the molar-mass ratio under the square root survives. Why the square root: the hot gas's thermal energy per molecule scales with $T_{c}$ , but that energy becomes kinetic energy $\frac{1}{2} M v^{2}$ , so $v \propto T_{c} / M$ — heavier molecules move slower for the same energy. $\frac{v _{e, A}}{v _{e, B}} = \frac{T _{c} / M _{A}}{T _{c} / M _{B}} = \frac{M _{B}}{M _{A}} = \frac{40}{18} = 2.222 = 1.491$ Answer: $\approx 1.49$ — the light-exhaust engine is ~49% faster. This is exactly why hydrolox (light H $_{2}$ O/H $_{2}$ exhaust) beats heavier-exhaust chemistries.

L4 — Synthesis

Recall Solution Q8

What we do: solve $m_{0} = m_{f} e^{Δ v / (I_{s p} g_{0})}$ for each engine. Why: given a fixed payload and a target $Δ v$ , the rocket equation fixes exactly how much propellant (hence total mass) you must carry. (a) Kerolox: exponent $= \frac{9000}{311 \times 9.81} = \frac{9000}{3050.9} = 2.9500$ . $m_{0} = 20000 \times e^{2.9500} = 20000 \times 19.106 = 382, 120 kg$ . (b) Hydrolox: exponent $= \frac{9000}{450 \times 9.81} = \frac{9000}{4414.5} = 2.0388$ . $m_{0} = 20000 \times e^{2.0388} = 20000 \times 7.6813 = 153, 626 kg$ . Comment: the hydrolox vehicle is $\approx 154$ t vs $\approx 382$ t — under half the liftoff mass for the identical payload and $Δ v$ . Because $m_{0}$ depends exponentially on $1/ I_{s p}$ , a modest 45% jump in $I_{s p}$ slashes the mass more than twofold. This exponential leverage is why engineers fight for every second of $I_{s p}$ . Compare with Staging and Mass Ratio. Answer: (a) $\approx 3.82 \times 1 0^{5}$ kg; (b) $\approx 1.54 \times 1 0^{5}$ kg.

Recall Solution Q9

What we do: for each engine find $m_{0} = m_{f} e^{Δ v / (I_{s p} g_{0})}$ , then subtract $m_{f}$ . Hydrolox: exponent $= \frac{6000}{450 \times 9.81} = \frac{6000}{4414.5} = 1.3592$ . $m_{0} = 1000 \times e^{1.3592} = 1000 \times 3.8931 = 3893.1$ kg. Propellant $= 3893.1 - 1000 = 2893.1$ kg. Ion: exponent $= \frac{6000}{3000 \times 9.81} = \frac{6000}{29430} = 0.20388$ . $m_{0} = 1000 \times e^{0.20388} = 1000 \times 1.2262 = 1226.2$ kg. Propellant $= 1226.2 - 1000 = 226.2$ kg. Answer: hydrolox $\approx 2893$ kg of propellant; ion $\approx 226$ kg — the ion stage needs about 1/13th the propellant. (The catch: at ~0.15 N thrust the ion burn takes months, whereas hydrolox does it in minutes — high $I_{s p}$ trades time for propellant.)

L5 — Mastery

Recall Solution Q10

(a) $Δ v$ : $m_{0} = m_{f} + propellant = 800 + 600 = 1400$ kg. $Δ v = I_{s p} g_{0} ln \frac{m _{0}}{m _{f}} = 3000 \times 9.81 \times ln \frac{1400}{800}$ $ln (1.75) = 0.55962$ , so $Δ v = 29430 \times 0.55962 = 16, 470 m/s \approx 16.5 km/s$ . Why so large: the enormous $v_{e} = 29.4$ km/s means even a modest mass ratio of 1.75 delivers ~16 km/s — impossible for a chemical rocket at this mass ratio. (b) $\overset{m}{˙}$ : invert $F = \overset{m}{˙} I_{s p} g_{0}$ : $\overset{m}{˙} = \frac{F}{I _{s p} g _{0}} = \frac{0.15}{29430} = 5.097 \times 1 0^{- 6} kg/s \approx 5.1 mg/s .$ (c) Burn time: propellant $\div$ flow rate: $t = \frac{600}{5.097 \times 1 0 ^{- 6}} = 1.1772 \times 1 0^{8} s = \frac{1.1772 \times 1 0 ^{8}}{86400} \approx 1362 days \approx 3.7 years .$ Answer: (a) $\approx 1.65 \times 1 0^{4}$ m/s; (b) $\approx 5.1 \times 1 0^{- 6}$ kg/s; (c) $\approx 1362$ days (~3.7 yr). This is the ion-engine bargain in one problem: spectacular $Δ v$ from tiny propellant, paid for with years of patient thrust.

Recall Solution Q11

What we do: with the same $ln (m_{0} / m_{f})$ , $Δ v = I_{s p} g_{0} ln 4$ scales purely with $I_{s p}$ . Why: holding the mass ratio fixed removes the exponential; $Δ v$ then depends linearly on $I_{s p}$ , so their ratio is just $I_{s p, Y} / I_{s p, X}$ . $ln 4 = 1.38629$ . $Δ v_{X} = 260 \times 9.81 \times 1.38629 = 2550.6 \times 1.38629 = 3535.9 m/s$ . $Δ v_{Y} = 450 \times 9.81 \times 1.38629 = 4414.5 \times 1.38629 = 6120.0 m/s$ . Difference $= 6120.0 - 3535.9 = 2584.1 m/s$ . Ratio $= \frac{6120.0}{3535.9} = 1.731 = \frac{450}{260}$ . ✓ Answer: Y gains $\approx 2584$ m/s more; ratio $\approx 1.73$ (exactly $450/260$ ).

Recall Solution Q12

What we do: since $v_{e} \propto T_{c} / M$ and $I_{s p} = v_{e} / g_{0}$ , the ratio of two engines' $I_{s p}$ equals the ratio of their $T_{c} / M$ values. Why take a ratio: both $g_{0}$ and the unknown proportionality constant cancel when we divide one engine by the other, so we never need the constant's value — only the two $T_{c} / M$ numbers. $\frac{I _{s p, exp}}{450} = \frac{T _{c, exp} / M _{exp}}{T _{c, ref} / M _{ref}} = \frac{3600/22}{3400/12}$ Inner fraction: $\frac{3600/22}{3400/12} = \frac{163.636}{283.333} = 0.57754$ . Square root $= 0.75996$ . $I_{s p, exp} = 450 \times 0.75996 = 341.98 s \approx 342 s .$ Why hotter doesn't win: the experimental engine's chamber is 200 K hotter (3600 vs 3400 K), which alone would raise $v_{e}$ by only $3600/3400 = 1.029$ (~3%). But its exhaust molecules are nearly twice as heavy ( $22$ vs $12$ g/mol), which cuts $v_{e}$ by $12/22 = 0.739$ (~26%). Under a square root, the molar-mass penalty swamps the temperature gain, so $I_{s p}$ falls to ~342 s — below the lighter-exhaust hydrolox. Answer: $\approx 342$ s. Low exhaust molar mass beats high chamber temperature — the core lesson of Exhaust Velocity and Nozzle Design.