Forces on a rocket in flight — thrust, aerodynamic (normal, axial), gravity: Exercises

3.4.3 · D4 · Physics › Rocket Flight Mechanics › Forces on a rocket in flight — thrust, aerodynamic (normal,

Poore page mein, hum parent note ke yeh tools reuse karte hain (koi nayi cheez assume nahi ki gayi hai):

Thrust $T = \overset{m}{˙} v_{e} + (p_{e} - p_{a}) A_{e}$ — momentum push plus pressure push.
Aerodynamic split $A = D cos α + L sin α$ (body ke along), $N = L cos α - D sin α$ (body ke across).
Dynamic pressure $q = \frac{1}{2} ρ v^{2}$ , isliye $D = q S C_{D}$ , $L = q S C_{L}$ .
Equations of motion $m \frac{d v}{d t} = T cos α - D - m g sin γ$ aur $m v \frac{d γ}{d t} = T sin α + L - m g cos γ$ .

Yahan $\overset{m}{˙}$ = mass ejected per second, $v_{e}$ = exhaust speed relative to rocket, $p_{e}$ = exhaust pressure, $p_{a}$ = outside air pressure, $A_{e}$ = nozzle exit area, $α$ = angle of attack (body-axis vs. velocity), $γ$ = flight-path angle (velocity above horizontal), $ρ$ = air density, $S$ = reference area, $g$ = gravity.

Figure — Forces on a rocket in flight — thrust, aerodynamic (normal, axial), gravity

Upar ki figure hamaara compass hai: red arrow velocity $v$ hai; naak ke saath wala black arrow body axis hai. Dono ke beech ka gap $α$ hai. Yeh picture dimag mein rakho — neeche ke har problem isi par based hai.

Level 1 — Recognition

Recall Solution

KYA HAI: Hum sirf do additive pieces ko label kar rahe hain.

$\overset{m}{˙} v_{e}$ = momentum thrust — yeh push jo tumhe mass ko peeche phenk ke milta hai.
$(p_{e} - p_{a}) A_{e}$ = pressure thrust — yeh bonus ya penalty jo exhaust plane ke ambient pressure par na hone se aati hai. Matched nozzle ka matlab hai $p_{e} = p_{a}$ , isliye $(p_{e} - p_{a}) A_{e} = 0$ aur pressure thrust vanish ho jaata hai. Momentum thrust kabhi vanish nahi hota jab tak $\overset{m}{˙} > 0$ hai.

Recall Solution

KYA HAI: $α = 0$ substitute karo, isliye $cos 0 = 1$ aur $sin 0 = 0$ . $A = D (1) + L (0) = D, N = L (1) - D (0) = L .$ YEH KYU MAAYENE RAKHTA HAI: $α = 0$ par body axes aur wind axes coincide karte hain — axial force equals drag aur normal force equals lift. Yeh SIRF wahi angle hai jahan dono names ka matlab ek hi hota hai.

Level 2 — Application

Recall Solution

KYA AUR KYU: Seedha $T = \overset{m}{˙} v_{e} + (p_{e} - p_{a}) A_{e}$ mein daalo; pressure term negative hoga kyunki exhaust ambient se neeche hai (over-expanded). $T = 250 (3000) + (50000 - 101000) (0.6)$ $= 750000 + (- 51000) (0.6) = 750000 - 30600 = 719.4 kN .$ Sanity check: Pressure penalty $30.6$ kN kaati hai, jo momentum thrust ka lagbhag $4%$ hai — yeh ek modest lekin real sea-level loss hai.

Recall Solution

KYA HAI: Sirf $p_{a}$ badla hai, $101$ kPa se $0$ par. $T_{vac} = 750000 + (50000 - 0) (0.6) = 750000 + 30000 = 780 kN .$ Percentage gain $719.4$ kN ke relative mein: $\frac{780 - 719.4}{719.4} \times 100 = 8.42%.$ KYU: Jaise $p_{a} \to 0$ , pressure term $- 30.6$ kN penalty se $+ 30$ kN bonus mein flip ho jaata hai — wahi engine space mein zyada powerful hota hai.

Recall Solution

KYA HAI: Wind axes se body axes par $α = 8^{\circ}$ se rotate karo ( $cos 8^{\circ} = 0.9903$ , $sin 8^{\circ} = 0.1392$ ). $A = 6 (0.9903) + 15 (0.1392) = 5.942 + 2.088 = 8.03 kN .$ $N = 15 (0.9903) - 6 (0.1392) = 14.854 - 0.835 = 14.02 kN .$ KYU: Sirf $8^{\circ}$ ke chhote tilt se bhi $2.1$ kN lift axial direction mein aa jaata hai — yeh ek aisa load hai jo airframe ko carry karna padta hai, lekin naive " $A = D$ " estimate ise miss kar deta.

Level 3 — Analysis

Recall Solution

KYA HAI: Tangential equation $a = \frac{T cos α - D - m g sin γ}{m}$ use karo, $cos α \approx 1$ ke saath. Along-path gravity drag: $m g sin γ = 3500 (9.8) (sin 5 5^{\circ}) = 3500 (9.8) (0.8192) = 28098 N \approx 28.1 kN$ . $a = \frac{120000 - 9000 - 28098}{3500} = \frac{82902}{3500} = 23.69 m/s^{2} .$ KAUN ZYADA RESIST KARTA HAI: Gravity loss $28.1$ kN $>$ drag $9$ kN. Steep climb mein gravity sabse bada dushman hai — isi liye rockets early pitch over karte hain horizontal ki taraf (gravity turn).

Recall Solution

KYA HAI: $m v \frac{d γ}{d t} = T sin α + L - m g cos γ$ use karo, phir $\frac{d γ}{d t}$ ke liye solve karo.

$T sin α = 120000 sin 6^{\circ} = 120000 (0.10453) = 12543 N$ .
$L = 10000 N$ .
$m g cos γ = 3500 (9.8) (cos 5 5^{\circ}) = 3500 (9.8) (0.57358) = 19674 N$ .

Net across-path force $= 12543 + 10000 - 19674 = 2869 N$ . $\frac{d γ}{d t} = \frac{2869}{m v} = \frac{2869}{3500 ( 400 )} = \frac{2869}{1.4 \times 1 0 ^{6}} = 2.049 \times 1 0^{- 3} rad/s .$ SIGN KYU MAAYENE RAKHTA HAI: Result positive hai, isliye $γ$ badh raha hai — rocket abhi bhi upar pitch kar raha hai. Gravity ka across-path pull ( $19.7$ kN) abhi tak path ko neeche bend karne ke liye kaafi nahi hai. Jab thrust aur lift $m g cos γ$ ko beat nahi kar paate, tab $\frac{d γ}{d t}$ negative ho jaata hai aur gravity turn shuru hota hai.

Level 4 — Synthesis

Recall Solution

Step 1 — dynamic pressure. $q = \frac{1}{2} ρ v^{2} = \frac{1}{2} (0.9) (30 0^{2}) = 0.45 (90000) = 40500 Pa$ . Step 2 — drag. $D = q S C_{D} = 40500 (1.2) (0.35) = 17010 N \approx 17.01 kN$ . Step 3 — tangential EoM. $a = \frac{T cos α - D - m g sin γ}{m}$ .

$T cos α = 90000 cos 5^{\circ} = 90000 (0.99619) = 89657 N$ .
$m g sin γ = 2800 (9.6) (sin 4 0^{\circ}) = 2800 (9.6) (0.64279) = 17278 N$ . $a = \frac{89657 - 17010 - 17278}{2800} = \frac{55369}{2800} = 19.77 m/s^{2} .$ IS TARAH CHAIN KARNE KI KYU: Tum $q$ skip nahi kar sakte — drag koi diya hua constant nahi hai, yeh air density, speed, area aur shape se emerge hota hai. Sirf $D$ build hone ke baad hi equation of motion kaam karta hai.

Recall Solution

KYA HAI: Total thrust momentum thrust ke barabar hoti hai jab pressure term zero hoti hai: $(p_{e} - p_{a}) A_{e} = 0 \Rightarrow p_{a} = p_{e} = 45 kPa .$ INTERPRETATION: $p_{a} = 45$ kPa par nozzle perfectly expanded (matched) hai. Us altitude se neeche ( $p_{a} > 45$ kPa) engine over-expanded hai → pressure penalty. Usse upar ( $p_{a} < 45$ kPa) yeh under-expanded hai → pressure bonus. Break-even exactly design-match altitude hai, $\overset{m}{˙} v_{e}$ se independent.

Level 5 — Mastery

Recall Solution

KYA HAI: $T (p_{a}) = 750000 + (50000 - p_{a}) (0.6)$ likho. Yeh $p_{a}$ mein linear hai negative slope $- 0.6$ ke saath, isliye $T$ tab sabse bada hota hai jab $p_{a}$ sabse chhota ho — yani $p_{a} = 0$ (vacuum). $T_{m a x} = T (0) = 750000 + (50000) (0.6) = 780 kN .$ Sabse bada bonus: $p_{a} = 0$ par pressure term hai $(50000) (0.6) = 30000 N = 30$ kN. Koi altitude isse beat nahi kar sakta kyunki $p_{a}$ $0$ se neeche nahi ja sakta. Physically, exhaust exit plane par push karta hai literally kuch bhi nahi ke against.

Recall Solution

KYA HAI: $α = 9 0^{\circ}$ par, $cos 9 0^{\circ} = 0$ aur $sin 9 0^{\circ} = 1$ . $A = D (0) + L (1) = L, N = L (0) - D (1) = - D .$ Isliye axial (along-body) load lift ban jaata hai, aur normal load $- D$ ban jaata hai — drag ab body ke across fully press karta hai. Thrust: $T cos α = T cos 9 0^{\circ} = 0$ . Ek sideways rocket ko apne engine se zero forward drive milti hai — saari thrust ab perpendicular hai ( $T sin 9 0^{\circ} = T$ ), use broadside dhakka dene ki koshish kar rahi hai. Isi liye bade $α$ par fly karna catastrophic hai: thrust path ke along accelerate karna band kar deti hai aur airframe poora aerodynamic load apne sabse kamzor axis ke across khata hai.

Recall Solution

KYA HAI: Jaise $γ$ $9 0^{\circ}$ se $0^{\circ}$ par jaata hai: $sin γ$ $1 \to 0$ girta hai, isliye along-path gravity loss shrink hoti hai; $cos γ$ $0 \to 1$ badhta hai, isliye across-path gravity grow hoti hai.

$γ$	$m g sin γ$ (along, tumhe slow karta hai)	$m g cos γ$ (across, tumhe turn karta hai)
$9 0^{\circ}$	$20 \times 1 = 20$ kN	$20 \times 0 = 0$ kN
$4 5^{\circ}$	$20 \times 0.7071 = 14.14$ kN	$20 \times 0.7071 = 14.14$ kN
$0^{\circ}$	$20 \times 0 = 0$ kN	$20 \times 1 = 20$ kN
PHYSICS: Seedhe upar, gravity ek pure brake hai (saare $20$ kN motion ko oppose karte hain, koi nahi turn karta). Horizontal par, gravity koi braking nahi karta lekin poori taraf sideways kheenchta hai — yeh entirely ek turning force hai. Gravity turn exactly isi ko exploit karta hai: pitch over karke, rocket gravity ko ek wasteful brake se ek free steering force mein convert karta hai.

Recall Quick self-check (cloze)

Momentum thrust hai == $\overset{m}{˙} v_{e}$ aur pressure thrust hai $(p_{e} - p_{a}) A_{e}$ ==. $α = 0$ par, axial force equals ==drag $D$ aur normal force equals lift $L$ ==. Thrust maximum hoti hai jab $p_{a} =$ 0 (vacuum). Pressure term ek penalty hoti hai jab == $p_{e} < p_{a}$ (over-expanded)==.

Related deeper reads: Tsiolkovsky Rocket Equation, Angle of Attack and Stability, Gravity Turn Trajectory, Drag Coefficient and Mach Number, Variable Mass Systems, Nozzle Expansion and Pressure Matching.