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

Throughout, we reuse these tools from the parent note (nothing new is assumed):

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

Here $\overset{m}{˙}$ = 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

The figure above is our compass: the red arrow is the velocity $v$ ; the black arrow along the nose is the body axis. The gap between them is $α$ . Keep this picture in mind — every problem below lives on it.

Level 1 — Recognition

Recall Solution

WHAT: We are just labelling the two additive pieces.

$\overset{m}{˙} v_{e}$ = momentum thrust — the push you earn by throwing mass out the back.
$(p_{e} - p_{a}) A_{e}$ = pressure thrust — the bonus or penalty from the exhaust plane not being at ambient pressure. Matched nozzle means $p_{e} = p_{a}$ , so $(p_{e} - p_{a}) A_{e} = 0$ and the pressure thrust vanishes. Momentum thrust never vanishes while $\overset{m}{˙} > 0$ .

Recall Solution

WHAT: Substitute $α = 0$ , so $cos 0 = 1$ and $sin 0 = 0$ . $A = D (1) + L (0) = D, N = L (1) - D (0) = L .$ WHY it matters: At $α = 0$ body axes and wind axes coincide — axial force equals drag and normal force equals lift. This is the ONLY angle where the two names mean the same thing.

Level 2 — Application

Recall Solution

WHAT & WHY: Plug straight into $T = \overset{m}{˙} v_{e} + (p_{e} - p_{a}) A_{e}$ ; the pressure term will be negative because the exhaust is below ambient (over-expanded). $T = 250 (3000) + (50000 - 101000) (0.6)$ $= 750000 + (- 51000) (0.6) = 750000 - 30600 = 719.4 kN .$ Sanity: The pressure penalty costs $30.6$ kN, about $4%$ of the momentum thrust — a modest but real sea-level loss.

Recall Solution

WHAT: Only $p_{a}$ changes, from $101$ kPa to $0$ . $T_{vac} = 750000 + (50000 - 0) (0.6) = 750000 + 30000 = 780 kN .$ Percentage gain relative to $719.4$ kN: $\frac{780 - 719.4}{719.4} \times 100 = 8.42%.$ WHY: As $p_{a} \to 0$ the pressure term flips from a $- 30.6$ kN penalty to a $+ 30$ kN bonus — the same engine is stronger in space.

Recall Solution

WHAT: Rotate from wind axes to body axes by $α = 8^{\circ}$ ( $cos 8^{\circ} = 0.9903$ , $sin 8^{\circ} = 0.1392$ ). $A = 6 (0.9903) + 15 (0.1392) = 5.942 + 2.088 = 8.03 kN .$ $N = 15 (0.9903) - 6 (0.1392) = 14.854 - 0.835 = 14.02 kN .$ WHY: Even a small $8^{\circ}$ tilt already sends $2.1$ kN of lift into the axial direction — a load the airframe must carry that a naive " $A = D$ " estimate would miss.

Level 3 — Analysis

Recall Solution

WHAT: Use the tangential equation $a = \frac{T cos α - D - m g sin γ}{m}$ with $cos α \approx 1$ . Gravity's along-path drag: $m g sin γ = 3500 (9.8) (sin 5 5^{\circ}) = 3500 (9.8) (0.8192) = 28098 N \approx 28.1 kN$ . $a = \frac{120000 - 9000 - 28098}{3500} = \frac{82902}{3500} = 23.69 m/s^{2} .$ WHICH resists more: Gravity loss $28.1$ kN $>$ drag $9$ kN. On a steep climb gravity is the dominant enemy — this is why rockets pitch over toward the horizontal early (gravity turn).

Recall Solution

WHAT: Use $m v \frac{d γ}{d t} = T sin α + L - m g cos γ$ , then solve for $\frac{d γ}{d t}$ .

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

Net across-path force $= 12543 + 10000 - 19674 = 2869 N$ . $\frac{d γ}{d t} = \frac{2869}{m v} = \frac{2869}{3500 ( 400 )} = \frac{2869}{1.4 \times 1 0 ^{6}} = 2.049 \times 1 0^{- 3} rad/s .$ WHY the sign matters: The result is positive, so $γ$ is increasing — the rocket is still pitching up. Gravity's across-path pull ( $19.7$ kN) isn't yet enough to bend the path down. Once thrust and lift can't beat $m g cos γ$ , $\frac{d γ}{d t}$ goes negative and the gravity turn begins.

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. $a = \frac{T cos α - D - m g sin γ}{m}$ .

$T cos α = 90000 cos 5^{\circ} = 90000 (0.99619) = 89657 N$ .
$m g sin γ = 2800 (9.6) (sin 4 0^{\circ}) = 2800 (9.6) (0.64279) = 17278 N$ . $a = \frac{89657 - 17010 - 17278}{2800} = \frac{55369}{2800} = 19.77 m/s^{2} .$ WHY chain it this way: You cannot skip $q$ — drag isn't a given constant, it emerges from air density, speed, area and shape. Only after $D$ is built does the equation of motion make sense.

Recall Solution

WHAT: Total thrust equals momentum thrust when the pressure term is zero: $(p_{e} - p_{a}) A_{e} = 0 \Rightarrow p_{a} = p_{e} = 45 kPa .$ INTERPRET: At $p_{a} = 45$ kPa the nozzle is perfectly expanded (matched). Below that altitude ( $p_{a} > 45$ kPa) the engine is over-expanded → pressure penalty. Above it ( $p_{a} < 45$ kPa) it is under-expanded → pressure bonus. The break-even is exactly the design-match altitude, independent of $\overset{m}{˙} v_{e}$ .

Level 5 — Mastery

Recall Solution

WHAT: Write $T (p_{a}) = 750000 + (50000 - p_{a}) (0.6)$ . This is linear in $p_{a}$ with a negative slope $- 0.6$ , so $T$ is largest when $p_{a}$ is smallest — i.e. $p_{a} = 0$ (vacuum). $T_{m a x} = T (0) = 750000 + (50000) (0.6) = 780 kN .$ Largest bonus: the pressure term at $p_{a} = 0$ is $(50000) (0.6) = 30000 N = 30$ kN. No altitude can beat this because $p_{a}$ cannot drop below $0$ . Physically the exhaust simply pushes on the exit plane against literally nothing behind it.

Recall Solution

WHAT: At $α = 9 0^{\circ}$ , $cos 9 0^{\circ} = 0$ and $sin 9 0^{\circ} = 1$ . $A = D (0) + L (1) = L, N = L (0) - D (1) = - D .$ So the axial (along-body) load becomes the lift, and the normal load becomes $- D$ — the drag now presses fully across the body. Thrust: $T cos α = T cos 9 0^{\circ} = 0$ . A sideways rocket gets zero forward drive from its engine — all the thrust is now perpendicular ( $T sin 9 0^{\circ} = T$ ), trying to shove it broadside. This is exactly why flying at large $α$ is catastrophic: thrust stops accelerating you along the path and the airframe eats the full aerodynamic load across its weakest axis.

Recall Solution

WHAT: As $γ$ falls from $9 0^{\circ}$ to $0^{\circ}$ : $sin γ$ drops $1 \to 0$ , so along-path gravity loss shrinks; $cos γ$ rises $0 \to 1$ , so across-path gravity grows.

$γ$	$m g sin γ$ (along, slows you)	$m g cos γ$ (across, turns you)
$9 0^{\circ}$	$20 \times 1 = 20$ 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: Straight up, gravity is a pure brake (all $20$ kN opposes motion, none turns you). Horizontal, gravity does no braking but pulls fully sideways — it is entirely a turning force. The gravity turn exploits exactly this: by pitching over, the rocket converts gravity from a wasteful brake into a free steering force.

Recall Quick self-check (cloze)

Momentum thrust is == $\overset{m}{˙} v_{e}$ and pressure thrust is $(p_{e} - p_{a}) A_{e}$ ==. At $α = 0$ , axial force equals ==drag $D$ and normal force equals lift $L$ ==. Thrust is maximum when $p_{a} =$ 0 (vacuum). The pressure term is a penalty when == $p_{e} < p_{a}$ (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.