Payload fraction as function of Δv and Isp: Exercises

This page is a graded workout. Every problem builds on the parent topic, the Tsiolkovsky Rocket Equation, Specific Impulse, Structural Coefficient, and Mass Ratio. Push through the levels in order — each unlocks the next.

Definition The Symbols and Two Tools You Need on This Page

Everything below is solved with just these objects. Keep them in view.

The mission demand. $Δ v$ (read "delta-vee") is the total change in speed, in m/s, that the mission requires — how much the rocket must speed up or slow down to reach its target orbit. It is set by the destination, not by the rocket builder.

The masses. The rocket at liftoff has total mass $m_{0}$ (payload + structure + propellant). After the burn, once the fuel is gone, the leftover mass is $m_{f}$ (payload + structure only). Their ratio $m_{0} / m_{f}$ is the Mass Ratio — always $\geq 1$ because the rocket only gets lighter.

The exhaust speed. $v_{e} = I_{s p} g_{0}$ is how fast (in m/s) the engine throws its propellant out the back. It is the Specific Impulse $I_{s p}$ (in seconds) multiplied by $g_{0} = 9.81 m/s^{2}$ to convert into a real velocity. Because $v_{e} = I_{s p} g_{0}$ , the exponent below can be written either way: $\frac{Δ v}{I _{s p} g _{0}} = \frac{Δ v}{v _{e}}$ — literally "how many exhaust-speeds of velocity change we demand."

The characteristic mass ratio — the "how many times bigger is the loaded rocket than the empty one" number: $R = e^{Δ v / (I_{s p} g_{0})} = e^{Δ v / v_{e}} = \frac{m _{0}}{m _{f}}$ The payload fraction — the slice of the launch mass that is actual cargo: $λ = \frac{1 + ϵ ( 1 - R )}{R}$ where $ϵ$ (the Structural Coefficient) is the kg of tank/engine per kg of fuel.

Figure s01 (below): the solve pipeline. Three boxes across the top — Inputs ( $Δ v, I_{s p}, ϵ$ ) → Exponent ( $Δ v / (I_{s p} g_{0})$ ) → Mass ratio ( $R = e^{(\cdot)}$ ) — with an arrow curving down into a fourth box, Payload fraction $λ = (1 + ϵ (1 - R)) / R$ . It is a visual reminder that $R$ is the gateway every problem passes through before $λ$ can be computed.

Figure — Payload fraction as function of Δv and Isp

Level 1 — Recognition

Exercise 1.1 (L1)

The Tsiolkovsky equation is written $Δ v = I_{s p} g_{0} ln (m_{0} / m_{f})$ . Identify each of the four quantities in words, and state which one you would increase to build a more efficient engine (higher payload for the same mission).

Recall Solution

$Δ v$ = the total velocity change the mission demands (set by orbital mechanics, not by us).
$I_{s p}$ = specific impulse = the engine's "fuel economy", in seconds.
$g_{0} = 9.81 m/s^{2}$ = standard gravity, just a unit-conversion constant baked into $I_{s p}$ .
$m_{0} / m_{f}$ = the Mass Ratio: loaded mass $m_{0}$ over dry mass $m_{f}$ (both defined in the box above).

To improve efficiency you increase == $I_{s p}$ ==. A bigger $I_{s p}$ shrinks the exponent $Δ v / (I_{s p} g_{0})$ , which shrinks $R$ , which raises $λ$ .

Exercise 1.2 (L1)

A rocket has $R = 1$ . What does this physically say about the propellant it carries, and what is $λ$ when $ϵ = 0.10$ ?

Recall Solution

$R = e^{Δ v / (I_{s p} g_{0})} = 1$ forces the exponent to be $0$ , i.e. $Δ v = 0$ . The rocket doesn't need to change velocity, so it burns no propellant. With $R = 1$ : $λ = \frac{1 + ϵ ( 1 - 1 )}{1} = \frac{1 + 0}{1} = 1.$ Payload fraction is 100% — the entire launch mass is cargo, because there is neither fuel nor (fuel-proportional) structure. This is the degenerate limiting case.

Level 2 — Application

Exercise 2.1 (L2)

A lunar-transfer stage needs $Δ v = 3, 100 m/s$ , uses a hydrolox engine with $I_{s p} = 450 s$ , and has $ϵ = 0.08$ . Compute $R$ and $λ$ .

Recall Solution

Exponent: $\frac{3100}{450 \times 9.81} = \frac{3100}{4414.5} = 0.7022$ . $R = e^{0.7022} = 2.018.$ $λ = \frac{1 + 0.08 ( 1 - 2.018 )}{2.018} = \frac{1 - 0.0814}{2.018} = \frac{0.9186}{2.018} = 0.4552.$ Payload fraction $\approx 45.5%$ .

Exercise 2.2 (L2)

A spacecraft in LEO massing $12, 000 kg$ (total) must do a $Δ v = 1, 900 m/s$ burn with $I_{s p} = 320 s$ , $ϵ = 0.12$ . How many kilograms of actual payload does it deliver?

Recall Solution

Exponent: $\frac{1900}{320 \times 9.81} = \frac{1900}{3139.2} = 0.6053$ . $R = e^{0.6053} = 1.8318.$ $λ = \frac{1 + 0.12 ( 1 - 1.8318 )}{1.8318} = \frac{1 - 0.09982}{1.8318} = \frac{0.90018}{1.8318} = 0.4914.$ Payload mass $= λ m_{0} = 0.4914 \times 12000 = 5897 kg$ (about 5.9 tonnes).

Level 3 — Analysis

Exercise 3.1 (L3)

For a fixed engine ( $I_{s p} = 350 s$ ) and fixed $ϵ = 0.10$ , find the $Δ v$ at which the payload fraction hits exactly zero. Interpret physically.

Recall Solution

Set $λ = 0 \Rightarrow 1 + ϵ (1 - R) = 0 \Rightarrow R = 1 + \frac{1}{ϵ}$ . With $ϵ = 0.10$ : $R = 1 + 10 = 11$ . Now invert $R = e^{Δ v / (I_{s p} g_{0})}$ : $Δ v = I_{s p} g_{0} ln R = 350 \times 9.81 \times ln (11) = 3433.5 \times 2.3979 = 8234 m/s .$ Meaning: at $Δ v = 8234 m/s$ the fuel-plus-its-structure exactly consumes the whole rocket — zero cargo. Any larger $Δ v$ (like the ~9,400 m/s to LEO) gives a negative $λ$ : physically impossible in one stage. This is the mathematical origin of why chemical Staging is mandatory.

Exercise 3.2 (L3)

Using the same $I_{s p} = 350 s$ , $ϵ = 0.10$ , compute $λ$ at $Δ v = 4, 000$ and at $Δ v = 8, 000 m/s$ , and show the drop is worse than linear (compare against an explicit straight-line prediction).

Recall Solution

$I_{s p} g_{0} = 3433.5$ .

At $Δ v = 4000$ : exponent $= 1.1650$ , $R = 3.2059$ . $λ = \frac{1 + 0.10 ( 1 - 3.2059 )}{3.2059} = \frac{0.77941}{3.2059} = 0.2431 = 24.3%.$ At $Δ v = 8000$ : exponent $= 2.3300$ , $R = 10.278$ . $λ = \frac{1 + 0.10 ( 1 - 10.278 )}{10.278} = \frac{0.07220}{10.278} = 0.007025 = 0.70%.$

Building the explicit "linear guess." A linear model needs a slope. Take it from the first interval: also compute $λ$ at $Δ v = 0$ , which is $λ = 1$ (100%, from Exercise 1.2). The straight line through the two points $(0, 100%)$ and $(4000, 24.3%)$ has slope $slope = \frac{24.3% - 100%}{4000 - 0} = - 0.0189% per m/s .$ Extrapolate that same line to $Δ v = 8000$ : $λ_{linear} (8000) = 100% + (- 0.0189%/ m/s) (8000) = 100% - 151.4% = - 51.4%.$ The linear guess predicts a wildly impossible $- 51.4%$ , whereas the honest exponential curve is still barely positive at $0.70%$ . Linear thinking overshoots the fall catastrophically because the true fall is dominated by $e^{- Δ v}$ : going 4000→8000 cut $λ$ by a factor of ~35, not by a fixed amount.

Figure s02 (below): exponential tyranny. A magenta curve of $λ$ (in %) versus $Δ v$ (in km/s) for $I_{s p} = 350 s, ϵ = 0.10$ . It falls steeply from 100% at $Δ v = 0$ , passes the two violet dots marked "4 km/s → 24.3%" and "8 km/s → 0.70%", crosses the dashed zero line at the orange square marked " $λ = 0$ , 8.23 km/s", and continues negative afterwards — visibly not a straight line.

Level 4 — Synthesis

Exercise 4.1 (L4)

An engineer can spend a budget in one of two ways for a $Δ v = 5, 000 m/s$ mission starting from a baseline $I_{s p} = 340 s$ , $ϵ = 0.12$ :

Option A: improve the engine to $I_{s p} = 380 s$ (keep $ϵ = 0.12$ ).
Option B: lighten the structure to $ϵ = 0.06$ (keep $I_{s p} = 340 s$ ).

Which choice yields the larger payload fraction?

Recall Solution

Baseline ( $I_{s p} = 340, ϵ = 0.12$ ): $I_{s p} g_{0} = 3335.4$ , exponent $= 1.4991$ , $R = 4.4776$ . $λ_{0} = \frac{1 + 0.12 ( 1 - 4.4776 )}{4.4776} = \frac{0.58269}{4.4776} = 0.1301 = 13.0%.$

Option A ( $I_{s p} = 380, ϵ = 0.12$ ): $I_{s p} g_{0} = 3727.8$ , exponent $= 1.3413$ , $R = 3.8241$ . $λ_{A} = \frac{1 + 0.12 ( 1 - 3.8241 )}{3.8241} = \frac{0.66111}{3.8241} = 0.1729 = 17.3%.$

Option B ( $I_{s p} = 340, ϵ = 0.06$ ): $R = 4.4776$ (unchanged, engine same). $λ_{B} = \frac{1 + 0.06 ( 1 - 4.4776 )}{4.4776} = \frac{0.79134}{4.4776} = 0.1767 = 17.7%.$ Verdict: Option B (17.7%) narrowly beats Option A (17.3%) here. Structure lightening helps because at $R \approx 4.5$ the structural penalty $ϵ (1 - R)$ is large; halving $ϵ$ halves that penalty directly. But the margin is thin — at higher $Δ v$ the exponential dominance of $I_{s p}$ would flip the answer.

Exercise 4.2 (L4)

For the baseline of 4.1, at what $Δ v$ would the two options give equal $λ$ ? Set up the equation and solve it with a hand-runnable numerical method (bisection), showing the intermediate bracketing steps. Interpret.

Recall Solution

Let $x = Δ v$ . Define $R_{A} (x) = e^{x /3727.8}, R_{B} (x) = e^{x /3335.4},$ and the difference function $f (x) = λ_{A} \frac{1 + 0.12 ( 1 - R _{A} )}{R _{A}} - λ_{B} \frac{1 + 0.06 ( 1 - R _{B} )}{R _{B}} .$ We want the root $f (x) = 0$ . Bisection needs a bracket where $f$ changes sign. Evaluate the two payload fractions at trial points:

$x$ (m/s)	$λ_{A}$	$λ_{B}$	$f (x)$	sign
$5000$	$0.1729$	$0.1767$	$- 0.0038$	$-$
$10000$	$0.0248$	$0.0116$	$+ 0.0133$	$+$

Sign flips between 5000 and 10000, so a root sits in that interval. Bisect (test the midpoint, keep the half that still straddles zero):

step	midpoint $x$	$f (x)$	keep
1	$7500$	$+ 0.0053$	left half $[5000, 7500]$
2	$6250$	$+ 0.0011$	left half $[5000, 6250]$
3	$5625$	$- 0.0011$	right half $[5625, 6250]$
4	$5937$	$+ 0.0000$	root here

Converging to $Δ v \approx 5930 m/s$ . Interpretation: below ~5930 m/s the lighter structure (Option B) wins; above it, the better engine (Option A) wins, because the exponential exponent $\propto 1/ I_{s p}$ grows in importance as $Δ v$ climbs. This is the quantitative face of "high- $Δ v$ stages love high $I_{s p}$ ." See Optimal Staging.

Level 5 — Mastery

Exercise 5.1 (L5)

A two-stage vehicle must produce $Δ v_{total} = 9, 200 m/s$ . Each stage has $I_{s p} = 360 s$ and $ϵ = 0.09$ , and the split is equal: $Δ v_{1} = Δ v_{2} = 4, 600 m/s$ . The "payload" of stage 1 is the entire stage 2 (its own payload fraction then multiplies through). Compute the overall payload fraction $λ_{total} = λ_{1} λ_{2}$ , and compare to the single-stage result.

Recall Solution

Each stage sees $Δ v = 4600$ , $I_{s p} g_{0} = 360 \times 9.81 = 3531.6$ . Exponent $= 4600/3531.6 = 1.30253$ , so $R = e^{1.30253} = 3.6787$ . $λ_{stage} = \frac{1 + 0.09 ( 1 - 3.6787 )}{3.6787} = \frac{1 - 0.24108}{3.6787} = \frac{0.75892}{3.6787} = 0.20630.$ Because both stages are identical, $λ_{1} = λ_{2} = 0.20630$ , so $λ_{total} = 0.2063 0^{2} = 0.04256 = 4.26% .$ Compare: a single stage doing all 9,200 m/s at these numbers has exponent $9200/3531.6 = 2.6051$ , $R = 13.531$ , giving $λ_{single} = \frac{1 + 0.09 ( 1 - 13.531 )}{13.531} = \frac{- 0.12779}{13.531} = - 0.00944 < 0,$ which is impossible. Splitting the same total $Δ v$ across two stages converts an impossible mission into a viable ~4.3% payload fraction. That jump from negative to positive is the whole reason Staging exists.

Exercise 5.2 (L5)

Take the payload-fraction formula $λ (Δ v) = \frac{1 + ϵ ( 1 - R )}{R}$ with $R = e^{Δ v / (I_{s p} g_{0})}$ . Show analytically that $\frac{d λ}{d ( Δ v )} < 0$ for all $Δ v \geq 0$ (payload fraction is strictly decreasing), given $0 < ϵ < 1$ .

Recall Solution

Rewrite using $k = 1/ (I_{s p} g_{0}) > 0$ so $R = e^{k Δ v}$ and $λ = \frac{1 + ϵ}{R} - ϵ = (1 + ϵ) e^{- k Δ v} - ϵ .$ Differentiate with respect to $Δ v$ : $\frac{d λ}{d ( Δ v )} = - (1 + ϵ) k e^{- k Δ v} .$ Every factor is positive: $(1 + ϵ) > 0$ , $k > 0$ , and $e^{- k Δ v} > 0$ for all real $Δ v$ . Hence the derivative is strictly negative everywhere. $λ$ falls monotonically — never a bump, never a rise — and its shape is a pure decaying exponential shifted down by $ϵ$ . That constant $- ϵ$ floor is why $λ$ eventually goes negative: at large $Δ v$ the $e^{- k Δ v}$ term vanishes and $λ \to - ϵ$ . See Propellant Mass Fraction for the complementary view.