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

This page is the practice ground for the parent topic. Before we compute anything, we lay out every kind of question the specific-impulse idea can throw at you. Then we solve one example per box, so no scenario ever surprises you in an exam.

Everything we use was built in the parent note. As a one-line reminder of the three tools we lean on:

The scenario matrix

Every question about $I_{s p}$ falls into one of these case classes. The last column names the example that covers it.

#	Case class	What's special about it	Covered by
A	Forward: $I_{s p} \to v_{e}$	Straight multiply	Ex 1
B	Backward: $v_{e} \to I_{s p}$	Divide (inverse direction)	Ex 2
C	Thrust from flow rate	Combine two relations	Ex 3
D	Same- $Δ v$ comparison across engines	Ratio / exponential, two engines	Ex 4
E	Degenerate input: $\overset{m}{˙} \to 0$ (ion limit)	Tiny thrust despite huge $I_{s p}$	Ex 5
F	Degenerate input: no fuel burned ( $m_{0} = m_{f}$ )	$Δ v \to 0$ , the "zero" case	Ex 6
G	Limiting/ceiling case — chemical cap near 450 s	Why $M$ (molar mass) sets the wall	Ex 7
H	Real-world word problem — staging choice	Pick the engine for the mission	Ex 8
I	Exam twist — the $g_{0}$ trap (Moon)	Constant vs local gravity	Ex 9

How to read the figure below. It is the map we keep pointing back to. The horizontal axis is specific impulse $I_{s p}$ (the efficiency knob) and the vertical axis is a representative thrust $F$ (the force knob, on a log scale so the huge range fits). Each of the four coloured dots is one engine type from the parent table. The dashed arrow traces the fundamental trade-off: as you move right (more efficient, higher $I_{s p}$ ) you also move down (less thrust). Examples 3, 5 and 8 are all really about which axis the question cares about — so glance at this map before each of them.

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

(Alt text: a log–log scatter of four rocket engines. Magenta "Solid ~260 s" sits top-left (high thrust, low efficiency); orange "LOX/RP-1 ~311 s" just below and right; violet "LOX/LH2 ~450 s" lower and further right; navy "Ion ~3000 s" at far right and bottom (huge efficiency, near-zero thrust). A dashed violet arrow runs from top-left down to bottom-right, labelled with the trade-off.)

Case A — Forward direction ( $I_{s p} \to v_{e}$ )

Worked example Example 1 — Exhaust speed of a hydrolox engine

An LOX/LH2 engine is rated at $I_{s p} = 450$ s. Find its effective exhaust velocity $v_{e}$ . (On the map: the violet dot at the top-left of the chemical cluster.)

Forecast: the table says hydrolox exhaust leaves at roughly $4, 400$ m/s. Guess before computing — will multiplying $450$ by about $10$ land you near $4, 500$ ? (Yes — this is a good habit: $g_{0} \approx 10$ .)

Write the master relation: $v_{e} = I_{s p} g_{0}$ . Why this step? $I_{s p}$ was defined as $v_{e} / g_{0}$ , so multiplying by $g_{0}$ simply undoes the definition and hands back the physical speed.
Substitute numbers: $v_{e} = 450 \times 9.81$ . Why this step? We plug the given $I_{s p}$ and the fixed constant $g_{0}$ so only a number remains to compute.
Compute: $v_{e} = 4414.5 m/s \approx 4.41 km/s$ . Why this step? Multiplying out turns the symbolic relation into the physical answer we were asked for.

Verify: Units check — $[s] \times [m/s^{2}] = [m/s]$ , a velocity. ✓ And $4414.5$ matches the parent table's $\sim 4410$ m/s. ✓

Case B — Backward direction ( $v_{e} \to I_{s p}$ )

Worked example Example 2 — A new engine measured on the test stand

Engineers measure an effective exhaust velocity of $v_{e} = 3050$ m/s. What $I_{s p}$ is this, and which propellant does it match? (On the map: this lands on the orange dot.)

Forecast: $3050$ m/s is between kerolox ( $\sim 3050$ ) and solid ( $\sim 2550$ ). Predict the answer is around $311$ s.

Rearrange the master relation to isolate $I_{s p}$ : $I_{s p} = \frac{v _{e}}{g _{0}}$ . Why this step? Now we know $v_{e}$ and want $I_{s p}$ — the inverse of Example 1. Dividing undoes the multiplying.
Substitute: $I_{s p} = \frac{3050}{9.81}$ . Why this step? We drop in the measured $v_{e}$ and the fixed $g_{0}$ so the division is purely numeric.
Compute: $I_{s p} = 310.9 s \approx 311 s$ . Why this step? Carrying out the division gives the seconds figure we can match to the benchmark table.

Verify: Multiply back — $311 \times 9.81 = 3050.9$ m/s, recovering our input. ✓ $311$ s is exactly the LOX/RP-1 (kerolox) benchmark. ✓

Case C — Thrust from flow rate

Worked example Example 3 — Force from a big kerolox engine

An LOX/RP-1 engine burns propellant at $\overset{m}{˙} = 500$ kg/s with $I_{s p} = 311$ s. What thrust does it produce? (On the map: the orange dot, but now we read the vertical (thrust) axis, not the horizontal.)

Forecast: A first-stage engine. Guess: is this closer to $1$ N, $1, 000$ N, or $1, 000, 000$ N? (A launch engine must lift many tonnes — so millions of newtons.)

Chain the two relations: $F = \overset{m}{˙} v_{e} = \overset{m}{˙} (I_{s p} g_{0})$ . Why this step? Thrust is momentum thrown per second, $\overset{m}{˙} v_{e}$ — exactly the relation we built in the intuition callout above. We don't have $v_{e}$ directly, but $v_{e} = I_{s p} g_{0}$ supplies it.
Substitute: $F = 500 \times 311 \times 9.81$ . Why this step? All three quantities are known, so the product is a single number.
Compute: $F = 1, 525, 455 N \approx 1.53 \times 1 0^{6} N = 1.53$ MN. Why this step? Multiplying out converts the formula into the newton figure asked for.

Verify: Units — $[kg/s] \times [s] \times [m/s^{2}] = [kg \cdot m/s^{2}] = [N]$ . ✓ $1.5$ MN is a realistic large first-stage thrust. ✓

Case D — Same- $Δ v$ comparison across engines

Worked example Example 4 — Solid vs ion for the same maneuver

A maneuver requires $Δ v = 3000$ m/s. Find the mass ratio $m_{0} / m_{f}$ for a solid ( $I_{s p} = 260$ s) and for an ion engine ( $I_{s p} = 3000$ s). Which needs less propellant? (On the map: the leftmost magenta dot versus the far-right navy dot.)

Forecast: The ion engine is $\sim 11 \times$ more efficient per kg. Predict its mass ratio sits just above $1$ (barely any fuel), while the solid needs a fat ratio above $3$ .

Rearrange Tsiolkovsky for the ratio: $\frac{m _{0}}{m _{f}} = e^{Δ v / (I_{s p} g_{0})}$ . Why this step? We want mass ratio given $Δ v$ , so we invert $Δ v = I_{s p} g_{0} ln (m_{0} / m_{f})$ . Undoing $ln$ requires the exponential $e^{x}$ — that is why $e$ appears here. See Tsiolkovsky Rocket Equation and Staging and Mass Ratio.
Solid: exponent $= \frac{3000}{260 \times 9.81} = 1.176$ , so $\frac{m _{0}}{m _{f}} = e^{1.176} = 3.243$ . Propellant fraction $ζ = 1 - 1/3.243 = 0.692 \Rightarrow 69%$ . Why this step? We plug the solid's low $I_{s p}$ into the exponent, then raise $e$ to it; $ζ$ (just defined above) turns the mass ratio into the fuel share.
Ion: exponent $= \frac{3000}{3000 \times 9.81} = 0.1019$ , so $\frac{m _{0}}{m _{f}} = e^{0.1019} = 1.107$ . Propellant fraction $ζ = 1 - 1/1.107 = 0.097 \Rightarrow 9.7%$ . Why this step? Same formula, but the ion's huge $I_{s p}$ makes the exponent tiny, so the ratio barely exceeds 1 and $ζ$ is small.

Verify: Take $ln$ back: $260 \times 9.81 \times ln (3.243) = 3000$ m/s ✓ and $3000 \times 9.81 \times ln (1.107) = 3003$ m/s ✓ (rounding of the ratio to 3 d.p. leaves a $\sim 3$ m/s residue) — both recover the required $Δ v$ . The ion engine uses $\sim 7 \times$ less propellant. ✓

Case E — Degenerate input: $\overset{m}{˙} \to 0$ (the ion trap)

Worked example Example 5 — Huge

$I_{s p}$ , vanishing thrust An ion engine has $I_{s p} = 3000$ s but flows only $\overset{m}{˙} = 5$ mg/s $= 5 \times 1 0^{- 6}$ kg/s. Find its thrust. Could it lift its own $1$ kg mass off the ground on Earth? (On the map: the navy dot — far right for $I_{s p}$ , yet way down at the bottom for thrust.)

Forecast: $I_{s p}$ is enormous — but $\overset{m}{˙}$ is nearly zero. Predict the thrust is a tiny fraction of a newton, far below the $\sim 9.81$ N needed to lift $1$ kg.

Use $F = \overset{m}{˙} I_{s p} g_{0}$ again. Why this step? Same thrust formula as Ex 3 — but now $\overset{m}{˙}$ is the degenerate (near-zero) input, showing thrust follows $\overset{m}{˙}$ , not $I_{s p}$ . Note the $g_{0}$ here is the unit-conversion constant inside $I_{s p}$ , not a gravity field.
Substitute: $F = 5 \times 1 0^{- 6} \times 3000 \times 9.81$ . Why this step? We drop in the tiny flow rate to see how it dominates the product.
Compute: $F = 0.1472 N$ . Why this step? Multiplying out reveals the actual force — a fraction of a newton.
Weight to lift $1$ kg on Earth: $W = m g_{local} = 1 \times 9.81 = 9.81$ N, using Earth's local gravity $g_{local} = 9.81 m/s^{2}$ . Why this step? Here we use the genuine local gravitational field $g_{local}$ , because lifting off is a real weight comparison — this is a different role from the constant $g_{0}$ inside $I_{s p}$ , even though both happen to equal $9.81$ at Earth's surface (see the "Two different g" definition above and Example 9). Since $0.147 ≪ 9.81$ , it cannot lift off.

Verify: $F / W = 0.147/9.81 = 0.015$ — thrust is $1.5%$ of the weight. As $\overset{m}{˙} \to 0$ , $F \to 0$ no matter how big $I_{s p}$ is. ✓ This is the quantitative reason ion engines are cruise-only (Ion and Electric Propulsion).

Case F — Degenerate input: no propellant burned

Worked example Example 6 — What if

$m_{0} = m_{f}$ ? A stage carries full tanks but the engine never fires, so its start and end mass are equal: $m_{0} = m_{f} = 20, 000$ kg. What $Δ v$ does it gain? (This is the boundary "zero" case of Tsiolkovsky.)

Forecast: No fuel is expelled → no momentum change → predict $Δ v = 0$ , for any $I_{s p}$ .

Apply $Δ v = I_{s p} g_{0} ln \frac{m _{0}}{m _{f}}$ . Why this step? We test the formula at its degenerate edge to confirm it behaves sensibly.
Substitute the ratio: $\frac{m _{0}}{m _{f}} = \frac{20000}{20000} = 1$ , so $Δ v = I_{s p} g_{0} ln (1)$ . Why this step? Equal masses collapse the ratio to exactly 1, which is the key that unlocks the whole result.
Since $ln (1) = 0$ , we get $Δ v = 0$ regardless of $I_{s p}$ . Why this step? $ln (1) = 0$ because $e^{0} = 1$ — "no growth needed to stay at 1." No mass change means no velocity change. Propellant fraction $ζ = 1 - 1/1 = 0$ confirms nothing was burned.

Verify: Physically, if nothing is thrown backward, conservation of momentum forbids any speed gain. ✓ The math and the physics agree at the boundary. ✓ (Sanity check: even an ion engine at $I_{s p} = 3000$ gives $3000 \times 9.81 \times 0 = 0$ .)

Case G — Limiting case: the chemical ceiling

Intuition Where does

$v_{e} \propto T_{c} / M$ come from? This is not magic — it is energy bookkeeping for a hot gas rushing through a nozzle (Exhaust Velocity and Nozzle Design). Each mole of exhaust leaves the hot chamber with thermal energy roughly proportional to $T_{c}$ (temperature is stored molecular motion). The nozzle converts that thermal energy into directed kinetic energy $\frac{1}{2} M v_{e}^{2}$ (here $M$ is the mass of one mole of exhaust). Setting "thermal energy in" $\propto T_{c}$ equal to "kinetic energy out" $= \frac{1}{2} M v_{e}^{2}$ gives $v_{e}^{2} \propto T_{c} / M$ , i.e. $v_{e} \propto T_{c} / M$ . The hidden proportionality constant bundles the gas constant, the specific-heat ratio, and the nozzle expansion — a fixed number for a given gas, which is why we can cancel it by taking a ratio below. The full constant is derived in Exhaust Velocity and Nozzle Design; for comparing two exhausts at equal $T_{c}$ we don't need its value.

Worked example Example 7 — Why chemistry caps out near 450 s

Using $v_{e} \propto T_{c} / M$ (with $T_{c}$ = chamber temperature in kelvin, K, and $M$ = mean molar mass of the exhaust in kg/mol), suppose an engine runs at a fixed $T_{c} = 3600$ K. Compare the exhaust speed with heavy exhaust ( $M = 0.030$ kg/mol, CO $_{2}$ -rich) versus light exhaust ( $M = 0.010$ kg/mol, hydrogen-rich). Which way must you push $M$ to reach $\sim 450$ s?

Forecast: Since $M$ is under the square root in the denominator, lighter exhaust (smaller $M$ ) means faster $v_{e}$ . Predict the light case is faster by a factor of $3$ .

Form the ratio to kill the hidden proportionality constant: $\frac{v _{e, light}}{v _{e, heavy}} = \frac{M _{heavy}}{M _{light}}$ . Why this step? Same $T_{c}$ and same gas-type constant, so both cancel; only the molar-mass ratio survives. Taking a ratio lets us compare without ever needing the constant's value — that's why we use a ratio here rather than absolute numbers.
Substitute: $0.030/0.010 = 3 = 1.732$ . Why this step? Plugging the two molar masses turns the symbolic ratio into a plain number we can interpret.
Interpret: the light-exhaust engine gets $73%$ more $v_{e}$ from the same temperature. Hydrogen-rich exhaust (lowest possible $M$ ) is what pushes chemical $I_{s p}$ up to the $\sim 450$ s wall. Why this step? You cannot lower $M$ below hydrogen's, and you cannot raise $T_{c}$ without melting the chamber (Combustion Chamber Temperature) — both limits together cap chemical $I_{s p}$ .

Verify: $3 = 1.732$ , and $1.73 2^{2} = 3$ recovers the molar-mass ratio. ✓ Direction is correct: smaller $M$ → larger $v_{e}$ , matching why LOX/LH2 beats every other chemical propellant. ✓

Case H — Real-world word problem: choosing the engine

Worked example Example 8 — Mission designer's dilemma

A probe must first launch off Earth (needs $\sim 9$ MN thrust for a few minutes) and later perform a slow $Δ v = 6000$ m/s deep-space cruise where fuel mass is scarce. You have solids ( $260$ s, huge thrust) and ion engines ( $3000$ s, tiny thrust). Which engine for which phase, and what mass ratio does the ion cruise need? (On the map: launch pulls you to the high-thrust magenta dot, cruise pulls you to the high- $I_{s p}$ navy dot — you literally travel across the chart.)

Forecast: Launch needs brute force → solids. Cruise needs efficiency, has time → ion. Predict the ion mass ratio is small (near $1.2$ ).

Launch phase: thrust dominates. $F = \overset{m}{˙} I_{s p} g_{0}$ — to get MN-scale $F$ you need huge $\overset{m}{˙}$ , which solids provide. Choose solid boosters. Why this step? From Ex 5 we saw thrust tracks $\overset{m}{˙}$ , not $I_{s p}$ ; liftoff is a thrust problem.
Cruise phase: propellant is precious, so choose the high- $I_{s p}$ ion engine. Mass ratio: $\frac{m _{0}}{m _{f}} = e^{6000/ (3000 \times 9.81)} = e^{0.2039}$ . Why this step? Cruise is a $Δ v$ /efficiency problem — exactly where $I_{s p}$ pays off; we reuse the inverted Tsiolkovsky from Ex 4.
Compute: $e^{0.2039} = 1.226$ . Propellant fraction $ζ = 1 - 1/1.226 = 0.184 \Rightarrow 18.4%$ . Why this step? Raising $e$ to the exponent gives the mass ratio, and $ζ$ gives how much of the probe is fuel.
Compare: a solid doing the same cruise would need $e^{6000/ (260 \times 9.81)} = e^{2.352} = 10.5$ — over $90%$ propellant. Impossible for a small probe. Why this step? Running the same numbers with the solid's low $I_{s p}$ shows why the mission forces the ion choice for cruise.

Verify: Ion back-check: $3000 \times 9.81 \times ln (1.226) = 6001$ m/s ✓. Solid back-check: $260 \times 9.81 \times ln (10.5) = 6001$ m/s ✓. The staged choice (solid boost + ion cruise) is the standard real design pattern. ✓

Case I — Exam twist: the $g_{0}$ Moon trap

Worked example Example 9 — Does

$I_{s p}$ change on the Moon? The same hydrolox engine from Example 1 ( $I_{s p} = 450$ s on Earth) is flown near the Moon, where local gravity is $g_{local} = 1.62 m/s^{2}$ . A student "corrects" $I_{s p}$ by rescaling with the Moon's gravity: $450 \times (9.81/1.62) = 2725$ s. Is the student right? What is the true $I_{s p}$ and $v_{e}$ near the Moon? (On the map: the student is trying to slide the violet dot rightward toward the ion region — but the dot cannot move, because the engine hardware is unchanged.)

Forecast: $I_{s p}$ is an engine/propellant property. Predict it stays $450$ s and $v_{e}$ stays $4414.5$ m/s — the student is wrong.

Recall the two-g distinction: the $g_{0}$ inside $I_{s p} = v_{e} / g_{0}$ is the fixed unit-conversion constant ( $9.81$ m/s²). It is not $g_{local}$ , the real field where you fly. Why this step? The entire "trap" is swapping the constant $g_{0}$ for the local field $g_{local}$ ; naming that swap is what dissolves the confusion (see the "Two different g" definition above).
True exhaust velocity near the Moon: $v_{e} = I_{s p} g_{0} = 450 \times 9.81 = 4414.5$ m/s — unchanged. Why this step? The engine physics (chamber temperature, molar mass, nozzle) is identical wherever you fly, so nothing in $v_{e} = I_{s p} g_{0}$ changes with location. $g_{0}$ stays $9.81$ by definition.
The student's $2725$ s is meaningless: substituting $g_{local}$ for the constant $g_{0}$ breaks the definition of $I_{s p}$ . Why this step? If $I_{s p}$ really changed with location, one fixed engine would have infinitely many "efficiencies" — impossible for a single piece of hardware.

Verify: $450 \times (9.81/1.62) = 2725$ s is what the student computed, but it is not a valid $I_{s p}$ ; the correct value stays $450$ s and $v_{e}$ stays $4414.5$ m/s. ✓ Conclusion: $I_{s p}$ depends on the engine, not on the sky above it — so the exam answer is "no, it does not change on the Moon." ✓

Recall checkpoint

Recall Which case does each cue belong to?

"Given $v_{e}$ , find $I_{s p}$ " — which direction and operation? ::: Backward (Case B): divide by $g_{0}$ . " $m_{0} = m_{f}$ , find $Δ v$ " — what's the answer and why? ::: $Δ v = 0$ because $ln (1) = 0$ ; no fuel expelled (Case F). "Ion engine can't lift off" — which quantity is degenerate? ::: $\overset{m}{˙} \to 0$ , so thrust $F = \overset{m}{˙} I_{s p} g_{0}$ is tiny (Case E). "Same $Δ v$ , two engines, less fuel?" — which tool? ::: Mass ratio $m_{0} / m_{f} = e^{Δ v / (I_{s p} g_{0})}$ ; higher $I_{s p}$ → smaller ratio (Case D). "Does $I_{s p}$ change on the Moon?" — the trap and the fix? ::: No; $g_{0}$ is a fixed constant (not $g_{local}$ ), $I_{s p}$ is an engine property (Case I). "Why can't chemical $I_{s p}$ exceed ~450 s?" ::: $v_{e} \propto T_{c} / M$ ; $T_{c}$ is limited by melting and $M$ can't go below hydrogen exhaust (Case G). What does propellant fraction $ζ$ mean? ::: The share of starting mass that was fuel, $ζ = 1 - m_{f} / m_{0} = 1 - 1/ (m_{0} / m_{f})$ .