Electric propulsion — thrust, power, Isp trade-off: Question bank

3.3.40 · D5 · Physics › Rocket Propulsion › Electric propulsion — thrust, power, Isp trade-off

Shuru karne se pehle, har symbol ko clearly define kar lete hain taaki neeche koi bhi line bina samjhe use na ho:

Definition Symbol glossary (har symbol ko earn karo)

$F$ = thrust, forward push in newtons (N). Ye har second backward throw ki hui momentum hai.
$\overset{m}{˙}$ = mass flow rate, propellant ka kilograms per second mein nikalna (kg/s). Dot ka matlab hai "per second".
$v_{e}$ = exhaust speed, propellant kitni tezi se (m/s) ship ke relative bahar nikalta hai.
$P_{in}$ = input power jo solar array ya reactor se li jaati hai (watts, W).
$P_{j e t}$ = jet/beam power, kinetic energy per second jo actually exhaust mein carry hoti hai.
$η$ (eta) = efficiency $= P_{j e t} / P_{in}$ , hamesha 0 aur 1 ke beech: wall power ka wo fraction jo beam mein pahunchta hai.
$I_{s p}$ = specific impulse seconds mein, ek fuel-economy score $= v_{e} / g_{0}$ .
$g_{0} = 9.81 m/s^{2}$ , standard-gravity constant jo sirf $I_{s p}$ (seconds) ko real speed mein convert karne ke liye use hota hai.
$Δ v$ = mission mein achieve hone wala total velocity change, Tsiolkovsky se: $Δ v = v_{e} ln (m_{0} / m_{f})$ .

True or false — justify karo

True or false: Fixed input power par, exhaust speed

v_{e}

double karne se thrust bhi double ho jaata hai.

False.

F = 2 η P_{in} / v_{e}

se, fixed power par thrust

v_{e}

ke inversely proportional hai —

v_{e}

double karne se thrust half ho jaata hai. Faster exhaust matlab har kilogram zyaada energy leta hai, isliye tum unhe kam throw kar sakte ho.

True or false: Zyaada

I_{s p}

wala thruster hamesha target pe faster pahunchta hai.

False. Zyaada

I_{s p}

matlab zyaada

v_{e}

, isliye fixed power par kam thrust, isliye chhoti acceleration aur zyaada lamba trip time. Ye propellant bachata hai, time nahi.

True or false: Efficiency

η

bahut acche thruster ke liye 1 ke barabar ya zyaada ho sakti hai.

False.

η = P_{j e t} / P_{in}

input power ka wo fraction hai jo beam tak pahunchta hai; baaki heat ban jaata hai, isliye

η < 1

hamesha. Energy conservation ye forbid karta hai ki input se zyaada beam power nikale.

True or false:

I_{s p}

ki units speed hoti hain.

False.

I_{s p}

seconds mein measure hota hai — ye impulse per unit weight of propellant hai,

F / (\overset{m}{˙} g_{0})

. Isse speed

v_{e}

mein convert karne ke liye

g_{0} = 9.81 m/s^{2}

se multiply karna padta hai.

True or false: Thrust equation

F = \overset{m}{˙} v_{e}

Newton's third law ka consequence hai.

True. Har second momentum

v_{e} d m

backward throw karne ka matlab hai ship ko equal-and-opposite forward push milta hai, isliye

F = \overset{m}{˙} v_{e}

. Dekho Newton's Third Law.

True or false: Fixed thrust

F

ke liye,

v_{e}

badhane se required jet power kam ho jaati hai.

False.

F

fixed hone par,

P_{j e t} = \frac{1}{2} F v_{e}

v_{e}

ke saath linearly badhti hai. Thrust hold karte hue exhaust speed badhane par zyaada power chahiye, kam nahi.

True or false: Electric propulsion ek spacecraft ko Earth ki surface se lift kar sakti hai.

False. EP thrust millinewtons mein hota hai jabki launch ke liye vehicle ka weight beat karne ke liye meganewtons chahiye; thrust-to-weight ratio 1 se kaafi neeche hai. EP sirf wahan kaam karta hai jahan

Δ v

accumulate karne ka time ho, yaani space mein. Dekho Chemical vs Electric Propulsion.

True or false: Agar tum

\overset{m}{˙}

double karo aur

v_{e}

fixed rakho, to thrust double hoga aur required power bhi double hogi.

True.

F = \overset{m}{˙} v_{e}

double hota hai, aur

P_{j e t} = \frac{1}{2} \overset{m}{˙} v_{e}^{2}

bhi double hoti hai (

\overset{m}{˙}

mein linear hai). Tum basically same engine ko "twice as hard" chala rahe ho.

Spot the error

Spot the error: "Kyunki

P_{j e t} = \frac{1}{2} \overset{m}{˙} v_{e}^{2}

, solar array ko exactly

\frac{1}{2} \overset{m}{˙} v_{e}^{2}

watts supply karne chahiye."

Array ko

P_{in} = P_{j e t} / η

supply karna padta hai, jo

P_{j e t}

se zyaada hai. Jet power sirf wo fraction hai jo beam mein pahunchti hai; hamesha pehle poochho "input ya jet power?"

Spot the error: "

v_{e} = I_{s p}

, isliye 3000 s ka engine

v_{e} = 3000

m/s rakhta hai."

Tumne

g_{0}

drop kar diya. Sahi hai

v_{e} = g_{0} I_{s p} = 9.81 \times 3000 \approx 2.9 \times 1 0^{4}

m/s — lagbhag das guna bada.

g_{0}

bhoolna classic

I_{s p}

blunder hai.

Spot the error: "Wasted power

P_{in} - P_{j e t}

bas gayab ho jaati hai; 1 se kam efficiency kuch cost nahi karti."

Ye gayab nahi hoti — energy conserved hoti hai, isliye

(1 - η) P_{in}

heat ban jaati hai jo spacecraft ko radiate karni padti hai. Ye radiator sizing drive karta hai; dekho Thermal Control (radiators).

Spot the error: "Kyunki zyaada

v_{e}

se har kg mein zyaada

Δ v

milta hai, hamesha maximum possible

v_{e}

choose karna chahiye."

Rocket equation high

v_{e}

ko reward karta hai, lekin fixed power par yahi choice thrust slash kar deti hai aur mission lamba kar deti hai (extra gravity/spiral losses ke saath). Optimal

I_{s p}

propellant mass aur trip time ko balance karta hai, maximize nahi karta.

Spot the error: "

F = 2 η P_{in} / v_{e}

dikhata hai thrust efficiency ke saath badhti hai, isliye 100% efficient engine infinite thrust rakhta hai."

Thrust

η

ke proportional hai, aur

η

1 par top out karta hai — ye 100% exceed nahi kar sakta.

η = 1

par tumhe

F = 2 P_{in} / v_{e}

milta hai, jo ek finite value hai. Efficiency help karta hai, lekin bounded hai.

Spot the error: "Fixed power par

I_{s p}

half karne se thrust bhi half ho jaati hai."

Bilkul ulta:

F \propto 1/ I_{s p}

, isliye

I_{s p}

half karne se thrust double ho jaati hai. Kam economy → har kilogram energy mein sasta → tum unhe zyaada per second throw karte ho → zyaada push. Dekho Specific Impulse.

Spot the error: "Vacuum mein thrust ko pressure term

(p_{e} - p_{a}) A_{e}

chahiye, isliye

F = \overset{m}{˙} v_{e}

incomplete hai."

Space ke vacuum mein ambient pressure

p_{a} \approx 0

hai aur electric thrusters ke liye exhaust pressure term negligible hai, isliye momentum term

F = \overset{m}{˙} v_{e}

poori kahani hai. Pressure correction atmospheric chemical nozzles ke liye matter karta hai, yahan nahi.

Why questions

Kyun fixed power par zyaada exhaust speed hamesha kam thrust deti hai?

Kyunki power split hoti hai

P_{j e t} = \frac{1}{2} F v_{e}

ke roop mein: fixed energy budget ke saath, har kilogram mein zyaada energy spend karne (zyaada

v_{e}

) par tum kam kilograms per second eject kar sakte ho, isliye momentum-per-second (thrust)

1/ v_{e}

ki tarah girta hai.

Electric propulsion ko "energy aur reaction mass ko decouple karna" kyun kaha jaata hai?

Chemical rocket mein propellant hi energy store hota hai,

v_{e} \sim 4.5

km/s cap karta hai. EP alag electrical source se energy leta hai, isliye propellant sirf mass supply karta hai, jisse

v_{e}

20–50 km/s tak chadh sakta hai. Dekho Chemical vs Electric Propulsion.

EP fuel saalon tak kyun chalta hai jabki chemical fuel minutes mein?

Kyunki

\overset{m}{˙} = F / v_{e}

bahut chhota (milligrams per second) hota hai jab

v_{e}

bahut bada ho — engine propellant ko sip karta hai. Ye ek strong brief push ko ek feeble lekin almost endless push ke saath trade karta hai.

I_{s p}

v_{e}

mein convert karne ke liye local gravity

g

ki jagah

g_{0}

se kyun multiply karte hain?

g_{0} = 9.81

ek fixed standard constant hai jo definition

I_{s p} = F / (\overset{m}{˙} g_{0})

mein baked in hai; ye ek bookkeeping conversion hai, na ki actual gravity jahan ship fly kar rahi hai. Ship ka real

Δ v

kabhi is par depend nahi karta ki wo kahan fly karti hai.

1 se kam efficiency thermal-design problem kyun ban jaati hai?

Input power ka

(1 - η)

fraction jo kabhi beam tak nahi pahunchta wo thruster mein heat ke roop mein deposit hota hai. Space mein tum heat sirf radiate karke shed kar sakte ho, isliye ye waste heat radiator area set karta hai — ek genuine power aur thermal driver.

Same input power wale do thrusters wildly different thrusts kyun deliver kar sakte hain?

Kyunki thrust chosen

v_{e}

(hence

I_{s p}

) aur

η

par bhi depend karta hai:

F = 2 η P_{in} / v_{e}

. Same

P_{in}

slow-heavy exhaust mein split karo to bada thrust; fast-light exhaust mein to chhota thrust.

Edge cases

Edge case: Fixed input power par

v_{e} \to \infty

hone par thrust ka kya hota hai?

F = 2 η P_{in} / v_{e} \to 0

. Infinitely fast exhaust ki limit mein engine vanishingly little mass throw karta hai, isliye thrust zero par mar jaata hai — tumhe spectacular economy aur koi push nahi milegi.

Edge case: Fixed thrust par

v_{e} \to 0

hone par required power ka kya hota hai?

P_{j e t} = \frac{1}{2} F v_{e} \to 0

: bahut slow exhaust se push karne mein almost koi power nahi lagti per newton. Lekin phir thrust maintain karne ke liye enormous

\overset{m}{˙} = F / v_{e} \to \infty

chahiye, isliye tum impossibly fast propellant burn karte — trade ka doosra side.

Edge case:

η = 0

physically kya mean karta hai?

Koi bhi input power beam tak nahi pahunchti — har watt heat ban jaata hai,

P_{j e t} = 0

, isliye

F = 0

. Completely inefficient thruster sirf ek heater hai jo koi push nahi karta.

Edge case: Agar

\overset{m}{˙} = 0

(valve band) lekin power on rahe, to

F

aur

P_{j e t}

kya hain?

Dono zero hain:

F = \overset{m}{˙} v_{e} = 0

aur

P_{j e t} = \frac{1}{2} \overset{m}{˙} v_{e}^{2} = 0

. Kuch bhi eject nahi ho raha to koi reaction force nahi — electrical power bas heat mein dump ho jaati hai.

Edge case: Deep space mein negligible external forces ke saath, kya tiny thrust kabhi "effect se bahar" run ho sakta hai?

Nahi — yahan tak ki ek millinewton, mahino tak continuously apply hone par, ek bada

Δ v

integrate karta hai kyunki

Δ v = v_{e} ln (m_{0} / m_{f})

total propellant used par depend karta hai, na ki tum kitni tezi se use karte ho. Patience weak thrust ko badi velocity change mein convert karta hai.

Edge case: Ek mission ke liye "optimal"

I_{s p}

par, kya thrust maximize hota hai, minimize hota hai, ya kuchh aur?

Kuchh aur. Optimum total mission cost (propellant mass plus trip time se penalties) minimize karta hai; ye ek moderate

I_{s p}

par land karta hai jahan thrust deliberately max nahi hai aur economy deliberately max nahi hai.

Recall Quick self-test

Ek governing idea ::: Fixed power par, thrust aur exhaust speed inversely trade off karte hain, $F = 2 η P_{in} / v_{e}$ ; tum dono ko maximize nahi kar sakte. Ek unit trap ::: $I_{s p}$ seconds mein hota hai; $v_{e} = g_{0} I_{s p}$ . Ek energy trap ::: $P_{in} = P_{j e t} / η$ , aur $(1 - η) P_{in}$ heat hai jo radiate karni padti hai.

6. Connections

3.3.40 Electric propulsion — thrust, power, Isp trade-off (Hinglish) — parent topic.
Tsiolkovsky Rocket Equation — kyun continuous weak thrust bhi huge $Δ v$ build karta hai.
Specific Impulse — in traps ke peeche ka fuel-economy score.
Chemical vs Electric Propulsion — launch-vs-cruise edge cases.
Ion and Hall Thrusters — wo hardware jin numbers ko describe karte hain.
Spacecraft Power Systems — $P_{in}$ kahan se aati hai.
Thermal Control (radiators) — wasted power kahan jaati hai.
Newton's Third Law — $F = \overset{m}{˙} v_{e}$ ki jad.