Electric propulsion — thrust, power, Isp trade-off: Exercises

Before we start, one shared toolbox. Every symbol below is earned — here is the plain-word meaning of each:

Figure — Electric propulsion — thrust, power, Isp trade-off

The red curve above is the whole story: at fixed power, thrust falls as exhaust speed rises. Keep it in view as you work.

Level 1 — Recognition

Exercise 1.1

An ion engine ejects $\overset{m}{˙} = 4.0 mg/s$ of xenon at $v_{e} = 30, 000 m/s$ . What is the thrust?

Recall Solution

WHAT: We are given a mass flow and a speed, asked for a push. WHY use $F = \overset{m}{˙} v_{e}$ : thrust is momentum ejected per second, and momentum = mass × speed (this is Newton's Third Law in action). First fix the units: $4.0 mg/s = 4.0 \times 1 0^{- 6} kg/s$ (milli = one-thousandth, and a mg is a thousandth of a gram which is a thousandth of a kg → $1 0^{- 6}$ ). $F = \overset{m}{˙} v_{e} = (4.0 \times 1 0^{- 6}) (3.0 \times 1 0^{4}) = 0.12 N = 120 mN .$

Exercise 1.2

The same engine's jet power. Given $\overset{m}{˙} = 4.0 mg/s$ , $v_{e} = 30, 000 m/s$ , find $P_{j e t}$ .

Recall Solution

WHY $P_{j e t} = \frac{1}{2} \overset{m}{˙} v_{e}^{2}$ : power is kinetic energy delivered per second, and kinetic energy of mass $m$ at speed $v$ is $\frac{1}{2} m v^{2}$ . The speed is squared because energy grows with the square of speed — this is the reason fast exhaust is so power-hungry. $P_{j e t} = \frac{1}{2} (4.0 \times 1 0^{- 6}) (3.0 \times 1 0^{4})^{2} = \frac{1}{2} (4.0 \times 1 0^{- 6}) (9.0 \times 1 0^{8}) = 1800 W .$

Exercise 1.3

A thruster has $I_{s p} = 2000 s$ . Convert to exhaust speed $v_{e}$ .

Recall Solution

WHY multiply by $g_{0}$ : $I_{s p}$ is measured in seconds because it is impulse per unit weight of propellant. Weight brings in $g_{0} = 9.81 m/s^{2}$ , so undoing that division restores a real velocity. $v_{e} = g_{0} I_{s p} = 9.81 \times 2000 = 19, 620 m/s \approx 1.96 \times 1 0^{4} m/s .$

Level 2 — Application

Exercise 2.1

A Hall thruster (see Ion and Hall Thrusters) runs at $P_{in} = 4.5 kW$ with efficiency $η = 0.55$ and $I_{s p} = 1600 s$ . Find (a) $v_{e}$ , (b) thrust $F$ , (c) mass flow $\overset{m}{˙}$ .

Recall Solution

(a) $v_{e} = g_{0} I_{s p} = 9.81 \times 1600 = 1.57 \times 1 0^{4} m/s$ . (b) WHY $F = 2 η P_{in} / v_{e}$ : this is the trade-off equation — it tells you how much push you get from the wall power once you commit to a speed. With $P_{in} = 4500 W$ : $F = \frac{2 ( 0.55 ) ( 4500 )}{1.57 \times 1 0 ^{4}} = \frac{4950}{15 , 696} = 0.315 N \approx 315 mN .$ (c) $\overset{m}{˙} = F / v_{e} = 0.315/1.57 \times 1 0^{4} = 2.01 \times 1 0^{- 5} kg/s = 20.1 mg/s$ .

Exercise 2.2

How much power is wasted as heat in the Exercise 2.1 thruster, and what does that heat mean for the ship?

Recall Solution

WHAT: efficiency $η = 0.55$ means only 55% of input power reaches the beam. The rest becomes heat. $P_{j e t} = η P_{in} = 0.55 \times 4500 = 2475 W, P_{w a s t e} = P_{in} - P_{j e t} = 4500 - 2475 = 2025 W .$ That $2025 W$ must be dumped to space by radiators — see Thermal Control (radiators). WHY care: heat rejection sizes the spacecraft's radiators, a real mass/design cost.

Exercise 2.3

Two thrusters draw the same $P_{in}$ and have the same $η$ . Thruster A has $I_{s p} = 1000 s$ , thruster B has $I_{s p} = 3000 s$ . Without a calculator, what is the ratio $F_{A} / F_{B}$ ?

Recall Solution

WHY no calculator is needed: at fixed $P_{in}$ and $η$ , the trade-off equation says $F = 2 η P_{in} / v_{e} \propto 1/ v_{e} \propto 1/ I_{s p}$ . So thrust is inversely proportional to $I_{s p}$ . $\frac{F _{A}}{F _{B}} = \frac{I _{s p, B}}{I _{s p, A}} = \frac{3000}{1000} = 3.$ A pushes three times harder. This is the seesaw of the parent note in one line.

Level 3 — Analysis

Exercise 3.1

A spacecraft's solar array delivers a fixed $P_{in} = 3.0 kW$ into a thruster of efficiency $η = 0.60$ . The engineer can dial $I_{s p}$ anywhere from $1000 s$ to $5000 s$ . Plot-in-your-head: does thrust rise or fall as $I_{s p}$ increases, and by what factor does $F$ change over that whole range?

Recall Solution

Reasoning: $F = 2 η P_{in} / (g_{0} I_{s p})$ . Numerator is fixed, so $F \propto 1/ I_{s p}$ : thrust falls as $I_{s p}$ rises. At $I_{s p} = 1000 s$ : $v_{e} = 9810 m/s$ , $F = \frac{2 ( 0.60 ) ( 3000 )}{9810} = \frac{3600}{9810} = 0.367 N$ . At $I_{s p} = 5000 s$ : $v_{e} = 49, 050 m/s$ , $F = \frac{3600}{49 , 050} = 0.0734 N$ . Ratio $= 0.367/0.0734 = 5$ . Exactly the factor by which $I_{s p}$ grew ( $5000/1000$ ), confirming $F \propto 1/ I_{s p}$ . The red curve on the figure at top is this behaviour.

Exercise 3.2

For the Exercise 3.1 engine set at $I_{s p} = 2500 s$ , how much propellant does it use over a 1-year burn? (1 year $\approx 3.156 \times 1 0^{7} s$ .)

Recall Solution

Step 1 — speed: $v_{e} = 9.81 \times 2500 = 2.453 \times 1 0^{4} m/s$ . Step 2 — thrust: $F = \frac{2 ( 0.60 ) ( 3000 )}{2.453 \times 1 0 ^{4}} = \frac{3600}{24 , 525} = 0.1468 N$ . Step 3 — mass flow: $\overset{m}{˙} = F / v_{e} = 0.1468/2.453 \times 1 0^{4} = 5.99 \times 1 0^{- 6} kg/s$ . Step 4 WHY multiply by time: mass used = flow rate × burn duration. $m_{p r o p} = \overset{m}{˙} \cdot t = (5.99 \times 1 0^{- 6}) (3.156 \times 1 0^{7}) = 189 kg .$ Under 200 kg of propellant for a whole year of continuous thrust — that is why EP wins deep-space cruising (Chemical vs Electric Propulsion).

Exercise 3.3

Degenerate check: what happens to thrust as $I_{s p} \to \infty$ (imagine a perfect, infinitely economical engine) at fixed power? And as $I_{s p} \to 0$ ? Are both physical?

Recall Solution

$I_{s p} \to \infty$ : $F = 2 η P_{in} / (g_{0} I_{s p}) \to 0$ . Infinite economy gives zero push — you throw an infinitesimal mass infinitely fast, and the momentum-per-second collapses. This is the far-right tail of the red curve flattening onto the axis. $I_{s p} \to 0$ : the formula gives $F \to \infty$ , but this is unphysical: to keep $P_{in}$ fixed while $v_{e} \to 0$ you would need $\overset{m}{˙} = 2 η P_{in} / v_{e}^{2} \to \infty$ — infinite propellant flow. Real engines have a floor on $v_{e}$ set by the physics of ionisation and grid voltage. So the curve is only meaningful over a finite $I_{s p}$ band.

Level 4 — Synthesis

Exercise 4.1

A 1200 kg probe must gain $Δ v = 6.0 km/s$ . Its thruster runs at $I_{s p} = 3000 s$ , $η = 0.65$ , $P_{in} = 5.0 kW$ . Find (a) the propellant mass required, (b) the thrust, (c) roughly how long the burn lasts (treat mass as ~constant for the time estimate, using the initial mass flow).

Recall Solution

(a) WHY the rocket equation: it connects velocity gain to how much of the ship you throw away. Rearranged, $m_{f} = m_{0} e^{- Δ v / v_{e}}$ . $v_{e} = 9.81 \times 3000 = 2.943 \times 1 0^{4} m/s$ . Then $Δ v / v_{e} = 6000/29430 = 0.2039$ . $m_{f} = 1200 e^{- 0.2039} = 1200 \times 0.8156 = 978.7 kg$ . $m_{p r o p} = m_{0} - m_{f} = 1200 - 978.7 = 221.3 kg$ . (b) $F = \frac{2 ( 0.65 ) ( 5000 )}{2.943 \times 1 0 ^{4}} = \frac{6500}{29430} = 0.2208 N$ . (c) $\overset{m}{˙} = F / v_{e} = 0.2208/2.943 \times 1 0^{4} = 7.503 \times 1 0^{- 6} kg/s$ . $t \approx m_{p r o p} / \overset{m}{˙} = 221.3/7.503 \times 1 0^{- 6} = 2.95 \times 1 0^{7} s \approx 0.94 yr$ . Reading it: ~221 kg of propellant, ~0.22 N of push, and nearly a year of burning. EP trades time for propellant frugality.

Exercise 4.2

Same mission, same $Δ v = 6.0 km/s$ and $m_{0} = 1200 kg$ , but now compare a chemical stage at $I_{s p} = 450 s$ against the EP stage of 4.1 at $I_{s p} = 3000 s$ . How much propellant does each need, and what is the mass saving?

Recall Solution

Chemical: $v_{e, c} = 9.81 \times 450 = 4415 m/s$ . $Δ v / v_{e, c} = 6000/4415 = 1.359$ . $m_{f} = 1200 e^{- 1.359} = 1200 \times 0.2569 = 308.3 kg$ , so $m_{p r o p, c} = 1200 - 308.3 = 891.7 kg$ . EP (from 4.1): $m_{p r o p, e p} = 221.3 kg$ . Saving: $891.7 - 221.3 = 670.4 kg$ — the EP stage uses about one-quarter the propellant. WHY: the propellant mass sits inside an exponential of $Δ v / v_{e}$ ; multiplying $v_{e}$ by ~6.7 shrinks that exponent dramatically. This is the entire economic case for EP (Chemical vs Electric Propulsion).

Level 5 — Mastery

Exercise 5.1

Mission design. A fixed power bus gives $η P_{in} = 2.0 kW$ of beam power. The ship starts at $m_{0} = 800 kg$ and must achieve $Δ v = 8.0 km/s$ . You may pick any $I_{s p}$ . As you raise $I_{s p}$ : propellant mass falls (good) but burn time rises (bad, because thrust falls). Find the burn time as a function of $I_{s p}$ and evaluate it at $I_{s p} = 2000 s$ and $I_{s p} = 4000 s$ to show the trade quantitatively.

Recall Solution

Set up the two competing effects. Propellant: $m_{p r o p} (I_{s p}) = m_{0} (1 - e^{- Δ v / (g_{0} I_{s p})})$ . Thrust from beam power: $F = 2 (η P_{in}) / v_{e} = 2 (η P_{in}) / (g_{0} I_{s p})$ — note $P_{j e t} = η P_{in}$ is what we were given, and $F = 2 P_{j e t} / v_{e}$ . Mass flow: $\overset{m}{˙} = F / v_{e} = 2 P_{j e t} / v_{e}^{2}$ . Burn time (constant-flow estimate): $t = m_{p r o p} / \overset{m}{˙}$ .

Evaluate at $I_{s p} = 2000 s$ : $v_{e} = 9.81 \times 2000 = 1.962 \times 1 0^{4}$ . $Δ v / v_{e} = 8000/19620 = 0.4077 \Rightarrow m_{p r o p} = 800 (1 - e^{- 0.4077}) = 800 (1 - 0.6652) = 267.8 kg$ . $\overset{m}{˙} = 2 (2000) / (1.962 \times 1 0^{4})^{2} = 4000/3.849 \times 1 0^{8} = 1.039 \times 1 0^{- 5} kg/s$ . $t = 267.8/1.039 \times 1 0^{- 5} = 2.578 \times 1 0^{7} s \approx 0.82 yr$ .

Evaluate at $I_{s p} = 4000 s$ : $v_{e} = 3.924 \times 1 0^{4}$ . $Δ v / v_{e} = 8000/39240 = 0.2038 \Rightarrow m_{p r o p} = 800 (1 - e^{- 0.2038}) = 800 (1 - 0.8156) = 147.5 kg$ . $\overset{m}{˙} = 4000/ (3.924 \times 1 0^{4})^{2} = 4000/1.540 \times 1 0^{9} = 2.598 \times 1 0^{- 6} kg/s$ . $t = 147.5/2.598 \times 1 0^{- 6} = 5.677 \times 1 0^{7} s \approx 1.80 yr$ .

The trade, laid bare: doubling $I_{s p}$ (2000→4000 s) nearly halves propellant (268→147 kg, saving 120 kg) but more than doubles burn time (0.82→1.80 yr). This is the mission-design seesaw: propellant mass versus trip duration. There is no free lunch — the optimum depends on whether the mission is mass-constrained or time-constrained.

Exercise 5.2

Show algebraically why burn time grows with $I_{s p}$ in the small- $Δ v$ limit (where $Δ v ≪ v_{e}$ , so $m_{p r o p} \approx m_{0} Δ v / v_{e}$ ). Derive the scaling of $t$ with $I_{s p}$ .

Recall Solution

WHY the approximation: when $x = Δ v / v_{e}$ is small, $1 - e^{- x} \approx x$ (first term of the exponential's expansion). So $m_{p r o p} \approx m_{0} Δ v / v_{e}$ . Now assemble $t = m_{p r o p} / \overset{m}{˙}$ with $\overset{m}{˙} = 2 P_{j e t} / v_{e}^{2}$ : $t \approx \frac{m _{0} Δ v / v _{e}}{2 P _{j e t} / v _{e}^{2}} = \frac{m _{0} Δ v v _{e}}{2 P _{j e t}} .$ Since $v_{e} = g_{0} I_{s p}$ : $t \approx \frac{m _{0} Δ v g _{0}}{2 P _{j e t}} I_{s p}$ Reading it: burn time is directly proportional to $I_{s p}$ (at fixed beam power and mission). Higher economy → proportionally longer trip. This is the clean law behind the numbers of 5.1 — and the ultimate statement of the whole trade-off.

Connections

Tsiolkovsky Rocket Equation — $m_{f} = m_{0} e^{- Δ v / v_{e}}$ , used in L4–L5.
Specific Impulse — the $I_{s p}$ we dialled throughout.
Chemical vs Electric Propulsion — the 4× propellant saving of Exercise 4.2.
Ion and Hall Thrusters — the hardware in the L1–L2 problems.
Spacecraft Power Systems — sets the fixed $P_{in}$ that makes the trade-off bite.
Thermal Control (radiators) — the wasted power of Exercise 2.2.
Newton's Third Law — root of $F = \overset{m}{˙} v_{e}$ .

Recall One-line summary of the whole ladder

At fixed power: thrust $\propto 1/ I_{s p}$ , propellant $\propto$ falls with $I_{s p}$ , burn time $\propto I_{s p}$ . Pick the point on that curve your mission can afford.