Exercises — Specific impulse Isp = v_e - g₀ — definition, physical meaning, units

WHAT: apply $I_{sp}=v_e/g_0$ directly. WHY: the definition converts an exhaust speed into the size-independent efficiency number. $I_{sp}=\frac{4413}{9.80665}\approx 450\text{s}$

(b) specific impulse. Thrust is in newtons (N), mass-flow in kg/s. $I_{sp}=\frac{\text{N⋅s}}{\text{N}}=\text{s}$: the newtons cancel, seconds survive.

Still $300\text{s}$. The formula uses the constant $g_0=9.80665$, never local $g$. $I_{sp}$ is a property of the engine, not the planet.

WHAT: invert the definition, $v_e=I_{sp}{g}_0$. WHY: we want a speed back out of the efficiency number. $v_e=320\times 9.80665\approx 3138\text{m/s}$

WHAT: use $F=\dot{m}{g}_0{I}_{sp}$ (thrust needs both efficiency and flow). WHY: thrust is momentum ejected per second; $g_0{I}_{sp}=v_e$ is the ejection speed. $F=300\times 9.80665\times 250\approx 7.36\times 10^5\text{N}\approx 0.74\text{MN}$

WHAT: $J=I_{sp}{g}_0{m}_p$. WHY: total impulse is the whole "push-package," equal to $I_{sp}$ times the propellant weight $W_p=g_0{m}_p$. $J=340\times 9.80665\times 15000\approx 5.00\times 10^7\text{N⋅s}$

WHAT: first get $v_e=F/\dot{m}$, then $I_{sp}=v_e/g_0$. WHY: thrust and flow give us the ejection speed via $F=\dot{m}{v}_e$; then convert to seconds. $v_e=\frac{1.8\times 10^6}{420}\approx 4285.7\text{m/s},I_{sp}=\frac{4285.7}{9.80665}\approx 437\text{s}$

Compute $F=\dot{m}{g}_0{I}_{sp}$ for each. $F_A=500\times 9.80665\times 450\approx 2.21\times 10^6\text{N}$ $F_B=0.006\times 9.80665\times 3000\approx 176.5\text{N}$ A wins by a factor $\frac{F_A}{F_B}\approx \frac{2.21\times 10^6}{176.5}\approx 1.25\times 10^4$ — about 12,500× more thrust, even though B is 6.7× more efficient. See the figure below: efficiency (height) and thrust (a different combination) are not the same axis.

$J=I_{sp}{g}_0{m}_p$, so with $m_p$ and $g_0$ fixed, $\Delta J=(I_{spY}-I_{spX})g_0{m}_p$. $\Delta J=(350-280)\times 9.80665\times 10000=70\times 9.80665\times 10000\approx 6.86\times 10^6\text{N⋅s}$

WHAT: get $v_e=I_{sp}{g}_0$, then feed it into Tsiolkovsky $\Delta v=v_e\ln (m_0/m_f)$. WHY: $I_{sp}$ sets the exhaust speed that the rocket equation multiplies by the log of the mass ratio. $v_e=340\times 9.80665\approx 3334.3\text{m/s}$ $\Delta v=3334.3\times \ln \frac{50000}{20000}=3334.3\times \ln (2.5)\approx 3334.3\times 0.91629\approx 3055\text{m/s}$

Propellant burned $m_p=m_0-m_f=30000\text{kg}$. Burn time: $t_b=m_p/\dot{m}=30000/250=120\text{s}$. Thrust: $F=\dot{m}{g}_0{I}_{sp}=250\times 9.80665\times 340\approx 8.34\times 10^5\text{N}\approx 0.83\text{MN}$.

Way 1 (force × time): $J=F{t}_b=8.3357\times 10^5\times 120\approx 1.0003\times 10^8\text{N⋅s}$. Way 2 (definition): $J=I_{sp}{g}_0{m}_p=340\times 9.80665\times 30000\approx 1.0003\times 10^8\text{N⋅s}$. They match (any tiny gap is rounding). WHY they must: $F=\dot{m}{g}_0{I}_{sp}$ and $m_p=\dot{m}{t}_b$, so $F{t}_b=\dot{m}{g}_0{I}_{sp}\cdot (m_p/\dot{m})=g_0{I}_{sp}{m}_p$. The $\dot{m}$ cancels — the same cancellation that makes $I_{sp}$ size-independent.

WHAT: invert Tsiolkovsky for $v_e$, then $I_{sp}=v_e/g_0$. WHY: the mission fixes $\Delta v$ and the tankage fixes the mass ratio, leaving $I_{sp}$ as the unknown the engine team must hit. $v_e=\frac{\Delta v}{\ln (m_0/m_f)}=\frac{4000}{\ln 3.2}=\frac{4000}{1.16315}\approx 3438.8\text{m/s}$ $I_{sp}=\frac{3438.8}{9.80665}\approx 350.7\text{s}$ So the engine needs $I_{sp}\gtrsim 351\text{s}$ — a good cryogenic engine.

Both have the same mass ratio $m_0/m_f=(m_f+m_p)/m_f=12000/4000=3$. $\Delta v_{\text{chem}}=330\times 9.80665\times \ln 3\approx 3236.2\times 1.09861\approx 3555\text{m/s}$ $\Delta v_{\text{ion}}=3200\times 9.80665\times \ln 3\approx 31381.3\times 1.09861\approx 3.448\times 10^4\text{m/s}$ The ion engine gives ~9.7× more $\Delta v$ from the same propellant — that is exactly what high $I_{sp}$ buys you. (The catch, from L3.2: it delivers this over months of tiny thrust, not seconds.)

$I_{sp}=v_e/g_0$ is unchanged — it depends only on the speed $v_e$, not on how much gas flows. A perfectly efficient engine can have a well-defined $I_{sp}$ even as flow vanishes. But thrust $\to 0$: $F=\dot{m}{v}_e\to 0\cdot v_e=0$. This is the mathematical heart of the L3 lesson: efficiency ($I_{sp}$) and force ($F$) are decoupled. In the degenerate zero-flow limit you keep all the efficiency and lose all the thrust.

Given $v_e$, want $I_{sp}$? ::: $I_{sp}=v_e/g_0$. Given $I_{sp}$, want $v_e$? ::: $v_e=I_{sp}{g}_0$. Given $I_{sp}$ and $\dot{m}$, want thrust? ::: $F=\dot{m}{g}_0{I}_{sp}$. Given $I_{sp}$ and $m_p$, want total impulse? ::: $J=I_{sp}{g}_0{m}_p$. Given $I_{sp}$ and mass ratio, want $\Delta v$? ::: $\Delta v=I_{sp}{g}_0\ln (m_0/m_f)$. Does $I_{sp}$ change with location or with $\dot{m}$? ::: No — it depends only on $v_e$ (and the constant $g_0$).