Specific impulse Isp = v_e - g₀ — definition, physical meaning, units: Worked examples

3.3.4 · D3 · Physics › Rocket Propulsion › Specific impulse Isp = v_e - g₀ — definition, physical meani

Numbers touch karne se pehle, chaliye poora toolbox ek jagah pin karte hain, kyunki neeche diye har example in paanch relations ki sirf ek re-arrangement hai.

Har symbol, saral shabdon mein:

$F$ = thrust, aage ki taraf ka push, newtons (N) mein measure hota hai. Ek newton woh force hai jo 1 kg ko har second $1 m/s$ se accelerate kare.
$\overset{m}{˙}$ = mass-flow rate, propellant ke kilograms jo har second nozzle se bahar jaate hain, $kg/s$ mein. Chota dot matlab "per second."
$v_{e}$ = effective exhaust velocity, gas kitni tez rocket ke relative se bahar jaati hai, $m/s$ mein.
$m_{p}$ = poori firing mein jalaya gaya kul propellant mass, kg mein.
$J$ = total impulse, poore burn ka pura "push-package", $N\cdots$ mein. Yeh thrust ko time ke saath add karke milta hai.

Scenario matrix

Neeche case-classes ka poora grid hai jo yeh topic produce kar sakta hai. Har example jो follow hota hai, us cell ke saath tagged hai jisme woh fit hota hai, aur sab milke har cell fill karte hain.

Cell	Case class	Kya diya hai → kya poochha hai	Example
A	Forward	$v_{e} \to I_{s p}$	Ex 1
B	Backward	$I_{s p} \to v_{e} \to F$	Ex 2
C	Total impulse	$m_{p}, I_{s p} \to J$	Ex 3
D	Zero / degenerate	$\overset{m}{˙} = 0$ or $m_{p} = 0$	Ex 4
E	Limiting behaviour	$I_{s p} \to \infty$ (photon-like)	Ex 5
F	Unit-swap trap	$I_{s p}$ in m/s vs seconds	Ex 6
G	Real-world word problem	burn time from $m_{p}, \overset{m}{˙}$ ; check $J$	Ex 7
H	Exam twist — wrong $g$	koi local Mars $g$ use kare	Ex 8
I	Two engines compared	high- $I_{s p}$ vs high-thrust	Ex 9

Hum poore time $g_{0} = 9.80665 m/s^{2}$ rakhenge. Local gravity kabhi mat substitute karo — yeh sabse zyada punish hone wali galti hai, jo explicitly Ex 8 mein handle ki gayi hai.

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

Upar ka figure hamaara master map hai: jo tumhare paas given hai (top row) woh pick karo, arrow follow karo, aur jaha jaana hai wahan land karo. Har example is map se ek path hai.

Worked examples

Cell A — seedha forward calculation

Worked example Example 1 — exhaust velocity se

$I_{s p}$ Ek kerosene/oxygen engine gas ko $v_{e} = 3100 m/s$ par eject karti hai. Iska specific impulse nikalo.

Forecast: Chemical engines $250$ – $450 s$ ke aas-paas hote hain. Compute karne se pehle guess karo — yeh low end ke paas hoga ya high end ke?

Relation likho. $I_{s p} = \frac{v _{e}}{g _{0}}$ . Yeh step kyun? Yeh constant exhaust speed ke liye defining shortcut hai; yeh ek speed ko size-independent "seconds" figure of merit mein convert karta hai.
Substitute karo. $I_{s p} = \frac{3100}{9.80665}$ . Yeh step kyun? $g_{0}$ fixed reference constant hai, isliye hum exactly $9.80665$ se divide karte hain, kisi bhi local $g$ se nahi.
Evaluate karo. $I_{s p} \approx 316.1 s$ .

Verify: Units: $\frac{m/s}{m/s ^{2}} = s$ ✓. Number ( $\approx 316 s$ ) expected chemical band mein hai — ek sensible kerolox value.

Cell B — backward jao, phir thrust tak pahuncho

Worked example Example 2 —

$I_{s p}$ se wapas thrust tak Ek engine $I_{s p} = 380 s$ rated hai aur $\overset{m}{˙} = 250 kg/s$ par propellant consume karta hai. Iska thrust nikalo.

Forecast: Bada mass-flow, decent $I_{s p}$ . Meganewtons ya kilonewtons? Order of magnitude guess karo.

Exhaust speed recover karo. $v_{e} = I_{s p} g_{0} = 380 \times 9.80665 \approx 3726.5 m/s$ . Yeh step kyun? Thrust ko ek speed chahiye, "seconds" rating nahi. $g_{0}$ se multiply karke pehle ki division undo hoti hai aur humein $v_{e}$ milti hai.
Momentum-thrust relation apply karo. $F = \overset{m}{˙} v_{e} = 250 \times 3726.5$ . Yeh step kyun? Newton ke 3rd law se, thrust us rate ke barabar hai jis rate par momentum peeche pheka jata hai — mass-per-second times iska speed.
Evaluate karo. $F \approx 9.316 \times 1 0^{5} N \approx 0.93 MN$ .

Verify: Units: $kg/s \times m/s = kg\cdotm/s^{2} = N$ ✓. Mid-size engine ke liye sub-meganewton — reasonable hai.

Cell C — propellant load se total impulse

Cell D — degenerate / zero input

Worked example Example 4 — kya hota hai jab kuch bhi flow nahi karta?

Wohi $I_{s p} = 340 s$ engine band kar di gayi hai, isliye $\overset{m}{˙} = 0$ . Ab thrust aur total impulse kya hain, aur kya $I_{s p}$ abhi bhi defined hai?

Forecast: Kya ek well-defined $I_{s p}$ koi bhi push guarantee karta hai? Padhne se pehle guess karo.

Thrust. $F = \overset{m}{˙} v_{e} = 0 \times v_{e} = 0 N$ . Yeh step kyun? Thrust = momentum thrown per second; kuch mat phenko, kuch mat push karo.
Total impulse with $m_{p} = 0$ . $J = I_{s p} g_{0} m_{p} = I_{s p} g_{0} \times 0 = 0 N\cdots$ . Yeh step kyun? Koi propellant consume nahi → koi impulse nahi, chahe engine rating kitni bhi "efficient" ho.
Kya $I_{s p}$ abhi bhi defined hai? Haan — $I_{s p} = v_{e} / g_{0}$ sirf design exhaust speed par depend karta hai, valve open hai ya nahi iss par nahi. Yeh $340 s$ rehta hai. Yeh step kyun? Yeh crucial split hai: $I_{s p}$ ek engine ki property hai; $F$ aur $J$ firing ki properties hain. Ek chup engine ka $I_{s p}$ abhi bhi hota hai.

Verify: $F = 0$ aur $J = 0$ consistent hain (kisi bhi time mein zero thrust → zero impulse) ✓. $I_{s p}$ non-zero phir bhi $F = 0$ — yahi reason hai "high $I_{s p}$ ≠ high thrust."

Cell E — limiting case

Worked example Example 5 —

$I_{s p}$ ko ceiling ki taraf push karna Ek idealized photon rocket "gas" ko almost speed of light par exhaust karta hai, $v_{e} \to c = 3.00 \times 1 0^{8} m/s$ . Limiting $I_{s p}$ estimate karo.

Forecast: Ion drives $\sim 3000 s$ tak pahunchte hain. Guess karo photon limit kitne orders of magnitude badi hai.

Extreme par forward relation apply karo. $I_{s p} = \frac{v _{e}}{g _{0}} = \frac{3.00 \times 1 0 ^{8}}{9.80665}$ . Yeh step kyun? Same formula — limit sirf ek bahut bada $v_{e}$ hai. Ek formula ko uske extreme par dekhna reveal karta hai ki woh sensible rehta hai ya nahi.
Evaluate karo. $I_{s p} \approx 3.06 \times 1 0^{7} s$ .
Trend interpret karo. Jaise $v_{e} \to \infty$ , $I_{s p} \to \infty$ : unbounded efficiency. Lekin ek photon stream ke saath $F = \overset{m}{˙} v_{e}$ mein microscopic $\overset{m}{˙}$ hoti hai, isliye thrust tiny rehti hai. Yeh step kyun? Limiting case poore page ke theme ko confirm karta hai — efficiency knob aur thrust knob independently ghoomte hain.

Verify: Ion drive se ratio: $\frac{3.06 \times 1 0 ^{7}}{3000} \approx 1 0^{4}$ — lagbhag chaar orders of magnitude zyada efficient ✓. Units abhi bhi seconds ✓.

$I_{s p} = 4500$ " seconds hai ya metres-per-second? Do datasheets ek hi engine describe karte hain. Sheet A kehta hai $I_{s p} = 458.9 s$ . Sheet B kehta hai " $I_{s p} = 4500 m/s$ ." Kya yeh consistent hain?

Forecast: Same engine, do numbers — contradiction ya convention? Guess karo.

Do conventions recognize karo. Sheet A $I_{s p}$ seconds mein deta hai ( $g_{0}$ se divide karke). Sheet B directly $v_{e}$ quote kar raha hai, jise specific impulse by mass kehte hain, $m/s$ mein. Yeh step kyun? Clash ek unit convention hai, physics ka disagreement nahi — ise spot karna ek fake contradiction ko rokta hai.
Sheet A ko speed mein convert karo. $v_{e} = I_{s p} g_{0} = 458.9 \times 9.80665 \approx 4500 m/s$ . Yeh step kyun? "Seconds" value ko $g_{0}$ se multiply karna use Sheet B ki language mein turn karta hai.
Compare karo. $4500 m/s$ (A se) $= 4500 m/s$ (Sheet B). Identical engine.

Verify: Round-trip check: $\frac{4500}{9.80665} \approx 458.9 s$ Sheet A recover karta hai ✓. Dono same $v_e$ describe karte hain.

Cell G — real-world word problem

Worked example Example 7 — yeh kitni der burn karega, aur kitna total push milega?

Ek booster mein $m_{p} = 8000 kg$ propellant hai, use $\overset{m}{˙} = 400 kg/s$ par burn karta hai, aur $I_{s p} = 300 s$ hai. (a) burn time, (b) thrust, (c) total impulse — do tareekon se — nikalo, aur check karo ki dono agree karte hain.

Forecast: Burn seconds mein chalega, ya minutes mein? Pehle (a) guess karo.

Burn time. $t_{b} = \frac{m _{p}}{m ˙} = \frac{8000}{400} = 20 s$ . Yeh step kyun? Burn time = fuel mass ÷ kitni tez kharch karo. Note karo $t_{b} = 20 s \neq = I_{s p} = 300 s$ — $I_{s p}$ ke "seconds" burn duration nahi hain (ek classic trap).
Exhaust speed phir thrust. $v_{e} = 300 \times 9.80665 \approx 2942.0 m/s$ ; $F = \overset{m}{˙} v_{e} = 400 \times 2942.0 \approx 1.177 \times 1 0^{6} N$ . Yeh step kyun? Rating ko speed mein convert karo, phir thrust relation use karo.
Total impulse — tarika 1 (force × time, constant thrust). $J = F t_{b} = 1.177 \times 1 0^{6} \times 20 \approx 2.353 \times 1 0^{7} N\cdots$ .
Total impulse — tarika 2 (propellant weight se). $J = I_{s p} g_{0} m_{p} = 300 \times 9.80665 \times 8000 \approx 2.353 \times 1 0^{7} N\cdots$ . Yeh step kyun? Do independent routes agree karne chahiye — ek built-in consistency check.

Verify: Dono routes $\approx 2.35 \times 1 0^{7} N\cdots$ dete hain ✓. Burn time $20 s \neq = 300 s$ confirm karta hai ki $I_{s p}$ -in-seconds ek figure of merit hai, clock nahi ✓.

Cell H — exam twist (galat gravity)

Worked example Example 8 — ek dost Mars gravity use karta hai

Ek dost $v_{e} = 2942 m/s$ engine ke liye $I_{s p}$ compute karta hai "Mars par operate ho raha hai" aur $g_{Mars} = 3.71 m/s^{2}$ se divide karta hai, answer aata hai $792.9 s$ . Kya yeh sahi hai? Correct value batao.

Forecast: Kya same engine sirf Mars par fly karne se $2.6 \times$ zyada efficient ho jaata hai? Guess karo.

Error spot karo. $I_{s p}$ defined constant $g_{0} = 9.80665 m/s^{2}$ use karta hai, kabhi local gravity nahi. $g_{0}$ sirf ek mass↔weight conversion factor hai, planet ka pull nahi. Yeh step kyun? Local $g$ use karne se ek engine ki rating location ke saath change ho jaati — ek fixed hardware ke liye yeh nonsense hai.
Correct computation. $I_{s p} = \frac{2942}{9.80665} \approx 300.0 s$ .
Compare karo. Dost ka $792.9 s$ factor $\frac{9.80665}{3.71} \approx 2.64$ se galat hai; sahi value har jagah $\approx 300 s$ hai. Yeh step kyun? Error size quantify karna dikhata hai ki galat- $g$ slip answer ko kitna distort karta hai.

Verify: Sanity: wohi engine Earth, Mars, ya deep space mein $I_{s p} \approx 300 s$ rakhta hai ✓ kyunki $g_{0}$ constant hai. Dost ka error ratio $\approx 2.64$ ✓.

Cell I — do engines compare kiye

Worked example Example 9 — efficiency versus brute force

Engine X (chemical): $I_{s p} = 350 s$ , $\overset{m}{˙} = 300 kg/s$ . Engine Y (ion): $I_{s p} = 3000 s$ , $\overset{m}{˙} = 5 \times 1 0^{- 6} kg/s$ . Kaun zyada efficient hai? Kaun zyada thrust deta hai?

Forecast: Ion drive ka $I_{s p}$ $\sim 9 \times$ zyada hai. Kya woh thrust mein bhi jeetta hai? Compute karne se pehle guess karo.

Efficiency = seedha $I_{s p}$ . Y jeetta hai: $3000 s ≫ 350 s$ . Yeh step kyun? $I_{s p}$ hi efficiency figure hai — efficiency rank karne ke liye koi computation nahi chahiye.
X ka thrust. $v_{e, X} = 350 \times 9.80665 \approx 3432.3 m/s$ ; $F_{X} = 300 \times 3432.3 \approx 1.030 \times 1 0^{6} N$ .
Y ka thrust. $v_{e, Y} = 3000 \times 9.80665 \approx 29419.95 m/s$ ; $F_{Y} = 5 \times 1 0^{- 6} \times 29419.95 \approx 0.147 N$ . Yeh step kyun? Thrust ke liye $\overset{m}{˙}$ bhi chahiye. Y ki speed huge hai lekin mass-flow minuscule hai, isliye force collapse ho jaati hai.
Compare karo. $F_{X} \approx 1.0 MN$ vs $F_{Y} \approx 0.15 N$ — $\sim 7 \times 1 0^{6}$ ka factor.

Verify: Efficiency winner (Y) aur thrust winner (X) alag-alag engines hain ✓ — page ka central lesson, quantified. Dono $F$ ki units: N ✓.

Recall

Recall Kya ek well-defined

$I_{s p}$ koi bhi thrust guarantee karta hai? Nahi — Ex 4 dikhata hai $\overset{m}{˙} = 0$ se $F = 0$ milta hai jabki $I_{s p}$ $340 s$ rehta hai. $I_{s p}$ engine ki property hai; thrust firing ki property hai. ::: Does a well-defined Isp guarantee thrust? ::: No — with zero mass-flow $F = 0$ though $I_{s p}$ is unchanged.

Recall Burn time

$I_{s p}$ ke seconds ke barabar kyun nahi hoti? Burn time $m_{p} / \overset{m}{˙}$ hai (Ex 7: $20 s$ ); $I_{s p}$ (wahan $300 s$ ) ek normalized figure of merit hai, clock nahi. ::: Is burn time equal to Isp in seconds? ::: No — burn time is propellant mass over mass-flow, a different quantity.

Connections

Rocket Propulsion
Parent: Specific impulse
Thrust and Mass Flow Rate — $F = \overset{m}{˙} v_{e}$ Ex 2, 7, 9 mein kaam aata hai
Effective Exhaust Velocity — woh $v_{e}$ jo Ex 2, 6 mein recover hoti hai
Standard Gravity g0 — woh constant jo Ex 8 mein misuse hoti hai
Impulse-Momentum Theorem — total impulse $J$ Ex 3, 7 mein
Tsiolkovsky Rocket Equation — jahan yeh $v_{e}$ values $Δ v$ mein feed hoti hain

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