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

This page is the drill floor for the parent topic. The parent built the three force families (thrust, aerodynamic axial/normal, gravity) and the equations of motion. Here we exhaust every case: every sign, every degenerate input, every limit, plus a real-world word problem and an exam twist.

Before we start, a promise: every symbol below was defined in the parent, but we re-anchor the tricky ones the moment they appear, so you never have to scroll back.

The scenario matrix

Think of this like a checklist. Each row is a way the physics can surprise you; each worked example below is tagged with the cell it covers.

#	Case class	What changes / what's tricky	Example that hits it
C1	Thrust, over-expanded ( $p_{e} < p_{a}$ )	Pressure term is a penalty (negative)	Ex 1
C2	Thrust, under-expanded at nonzero ambient ( $0 < p_{a} < p_{e}$ ) & vacuum limit	Pressure term is a bonus (positive)	Ex 1, Ex 2b
C3	Perfect expansion ( $p_{e} = p_{a}$ )	Pressure term vanishes — the degenerate "clean" case	Ex 2
C4	Aerodynamic resolve, small positive $α$	Both $A, N$ positive; lift bleeds into axis	Ex 3
C5	Aerodynamic resolve, negative $α$	Normal force flips sign — nose dips	Ex 4
C6	Zero angle of attack ( $α = 0$ )	Body axes = wind axes; $N = L$ , $A = D$	Ex 4
C6b	Large $α$ in quadrant II ( $9 0^{\circ} < α < 18 0^{\circ}$ )	$cos α$ negative — axial reverses	Ex 4b
C7	Tangential EOM, steep climb	Gravity loss dominates drag	Ex 5
C8	Tangential EOM, horizontal flight ( $γ = 0$ )	Gravity does no tangential work	Ex 5
C9	Normal EOM — gravity turn curvature	Sign of $d γ / d t$ decides pitch up/down	Ex 6
C10	Real-world word problem (uses $c$ , variable mass)	Translate messy prose into the equations	Ex 7
C11	Exam twist — solve for a hidden variable	Rearrange, don't just plug forward	Ex 8

Setup: the symbols, re-anchored

Definition Quick symbol refresher

$\overset{m}{˙}$ (read "m-dot") = mass ejected per second (kg/s). The dot means "rate of change".
$v_{e}$ = exhaust speed of gas relative to the rocket (m/s).
$c$ = effective exhaust velocity — the single number that folds the momentum thrust and the pressure term into one, so that $T = \overset{m}{˙} c$ . It equals $v_{e}$ exactly when the exhaust is perfectly matched to ambient, and is a touch larger otherwise. See Tsiolkovsky Rocket Equation.
$p_{e}$ = pressure of gas at the nozzle exit; $p_{a}$ = ambient (outside) air pressure; $A_{e}$ = area of the nozzle exit hole.
$α$ (alpha) = angle of attack — the tilt between the body's pointing direction and its actual velocity. See Angle of Attack and Stability.
$γ$ (gamma) = flight-path angle — how steeply the velocity vector points above the horizon ( $γ = 9 0^{\circ}$ straight up, $γ = 0$ level).
$L, D$ = lift and drag (wind axes); $N, A$ = normal and axial force (body axes).
$g$ = acceleration due to gravity ≈ $9.8 m/s^{2}$ near Earth's surface — the number that turns mass into weight, $W = m g$ .
$m_{0}$ = initial (launch) mass of the rocket, before any propellant is burned. As propellant leaves, the instantaneous mass is $m (t) = m_{0} - \overset{m}{˙} t$ . See Variable Mass Systems.

Ex 1 — Thrust at three pressures · covers C1, C2

Momentum thrust (the part you always earn). $\overset{m}{˙} v_{e} = 250 \times 2900 = 725000 N = 725 kN$ Why this step? This is the momentum you throw out the back every second — Newton's 3rd law pushes you forward by exactly this. It does not depend on altitude.
Pressure term at sea level. $p_{e} < p_{a}$ here (50 < 101), so the exhaust is over-expanded (squeezed by the thick air): $(p_{e} - p_{a}) A_{e} = (50000 - 101000) (0.6) = (- 51000) (0.6) = - 30600 N$ Why this step? A negative number means the outside air pushes back on the exit plane harder than the exhaust pushes out — a penalty (case C1).
Sea-level thrust. $T_{sea} = 725000 - 30600 = 694400 N = 694.4 kN$ Why this step? The full thrust law is $T = \overset{m}{˙} v_{e} + (p_{e} - p_{a}) A_{e}$ , so total thrust is literally the momentum term plus the pressure term. Here the pressure term is negative, so it subtracts from the 725 kN — the thick sea-level air is quietly stealing thrust.
Vacuum: $p_{a} = 0$ , so the term becomes a pure bonus (case C2, the limiting case): $(50000 - 0) (0.6) = 30000 N, T_{vac} = 725000 + 30000 = 755000 N = 755 kN$ Why this step? We form the vacuum thrust by taking the same momentum term (unchanged — engine flow is identical) and adding the now-positive pressure term. With no outside air pushing back, the full exit pressure $p_{e}$ acts over $A_{e}$ as a bonus, so vacuum thrust exceeds sea-level thrust.

Recall Verify

Vacuum is higher by $755 - 694.4 = 60.6$ kN, a $60600/694400 \approx 8.7%$ gain — matches the "same engine is stronger in space" rule. Units: (kg/s)(m/s) = kg·m/s² = N ✓, and (Pa)(m²) = (N/m²)(m²) = N ✓.

Ex 2 — Perfect expansion, then under-expanded at real altitude · covers C3, C2

(2a) Pressure term. $p_{e} - p_{a} = 19000 - 19000 = 0$ : $(p_{e} - p_{a}) A_{e} = 0$ Why this step? Case C3 — the degenerate "clean" point. Whatever $A_{e}$ is, multiplying by zero kills the whole pressure contribution.
(2a) Thrust = momentum thrust only. $T = \overset{m}{˙} v_{e} = 250 \times 2900 = 725000 N = 725 kN$ Why this step? Since the pressure term vanished in Step 1, the full thrust law $T = \overset{m}{˙} v_{e} + (p_{e} - p_{a}) A_{e}$ collapses to just its first piece. This is the only condition where thrust equals the pure momentum flux.
(2b) Pressure term at 25 km. Now $0 < p_{a} < p_{e}$ , so the term is a positive but not maximal bonus (case C2, non-vacuum): $(p_{e} - p_{a}) A_{e} = (19000 - 2500) (0.9) = (16500) (0.9) = 14850 N$ Why this step? Unlike Ex 1's vacuum ( $p_{a} = 0$ ), here some air still pushes back, so we subtract $p_{a}$ from $p_{e}$ before multiplying by $A_{e}$ . The bonus is real but partial — this is the everyday cruising case, between "matched" and "vacuum".
(2b) Thrust. $T = \overset{m}{˙} v_{e} + (p_{e} - p_{a}) A_{e} = 725000 + 14850 = 739850 N = 739.85 kN$ Why this step? We add the partial bonus of Step 3 to the momentum thrust, giving a value above the matched 725 kN because the ambient no longer perfectly balances the exit.

Recall Verify

(2a): $725$ kN sits between the sea-level $694.4$ kN and vacuum $755$ kN of Ex 1 — right where the pressure term crosses zero. (2b): $739.85$ kN sits above matched (725) but is a partial bonus — consistent with $0 < p_{a} < p_{e}$ . Units: (Pa)(m²) = N ✓.

Ex 3 — Resolving lift/drag into body axes, positive $α$ · covers C4

Look at the figure: the body axis (lavender) is tilted by $α$ from the velocity (slate). Drag points backward along velocity; lift is perpendicular to it. We project both onto the body axes.

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

Axial force — the component along the body: $A = D cos α + L sin α = 6 cos 1 5^{\circ} + 14 sin 1 5^{\circ}$ $= 6 (0.9659) + 14 (0.2588) = 5.795 + 3.623 = 9.42 kN$ Why this step? Because the body is tilted, part of lift ( $L sin α$ ) presses forward-along the body, adding to axial load. So $A = 9.42 > D = 6$ — the forecast trap.
Normal force — the component across the body: $N = L cos α - D sin α = 14 cos 1 5^{\circ} - 6 sin 1 5^{\circ}$ $= 14 (0.9659) - 6 (0.2588) = 13.523 - 1.553 = 11.97 kN$ Why this step? Lift mostly acts across the body (+ $N$ ), but drag being tilted steals a little from it (− $D sin α$ ).

Recall Verify

Rotations preserve length: $A^{2} + N^{2}$ must equal $L^{2} + D^{2}$ . $9.4 2^{2} + 11.9 7^{2} = 88.7 + 143.3 = 232.0 = 15.23$ kN and $6^{2} + 1 4^{2} = 36 + 196 = 232 = 15.23$ kN ✓. A pure rotation can't change magnitude — perfect check.

Ex 4 — Negative and zero angle of attack · covers C5, C6

The figure shows the body axis dropped below the velocity — a negative tilt.

Case (a), $α = - 1 5^{\circ}$ . Use $cos (- 1 5^{\circ}) = cos 1 5^{\circ} = 0.9659$ and $sin (- 1 5^{\circ}) = - 0.2588$ : $N = L cos α - D sin α = 14 (0.9659) - 6 (- 0.2588) = 13.523 + 1.553 = 15.08 kN$ $A = D cos α + L sin α = 6 (0.9659) + 14 (- 0.2588) = 5.795 - 3.623 = 2.17 kN$ Why this step? This is case C5. Cosine is even (unchanged), but sine is odd (flips sign). So the $L sin α$ term in $A$ now subtracts — axial drops sharply; and the $- D sin α$ term in $N$ now adds. Every sign must be tracked; this is exactly the "cover all quadrants" rule.
Case (b), $α = 0$ . Then $cos 0 = 1$ , $sin 0 = 0$ : $A = D (1) + L (0) = D = 6 kN, N = L (1) - D (0) = L = 14 kN$ Why this step? Case C6, the degenerate one — body axes coincide with wind axes. This is the only angle where "axial = drag" and "normal = lift". Any other $α$ breaks that identity.

Recall Verify

Magnitude check for (a): $2.1 7^{2} + 15.0 8^{2} = 4.71 + 227.4 = 232.1 = 15.23$ kN — again equals $6^{2} + 1 4^{2} = 15.23$ ✓. For (b), $N = L$ and $A = D$ trivially preserves length.

Ex 4b — Angle of attack in quadrant II ( $cos α < 0$ ) · covers C6b

At $α = 12 0^{\circ}$ we are in the second quadrant, where $cos α < 0$ but $sin α > 0$ : $cos 12 0^{\circ} = - 0.5$ , $sin 12 0^{\circ} = + 0.8660$ .

Axial force: $A = D cos α + L sin α = 6 (- 0.5) + 14 (0.8660) = - 3.0 + 12.124 = 9.12 kN$ Why this step? The drag term $D cos α$ has gone negative because $cos α < 0$ — drag now pushes the other way along the body. But the lift term $L sin α$ is still large and positive (sine is positive in quadrant II), so the net axial force stays positive. This is the case the earlier $\pm 1 5^{\circ}$ examples could not reach: a sign change in $cos α$ itself.
Normal force: $N = L cos α - D sin α = 14 (- 0.5) - 6 (0.8660) = - 7.0 - 5.196 = - 12.20 kN$ Why this step? Here both terms are negative: $cos α < 0$ flips the lift contribution, and $sin α > 0$ keeps the $- D sin α$ term negative. A negative $N$ means the normal force now presses on the opposite face of the body — exactly the loading that snaps fins on a tumbling stage.

Recall Verify

Magnitude preserved: $9.1 2^{2} + 12.2 0^{2} = 83.2 + 148.9 = 232.0 = 15.23$ kN, still equal to $6^{2} + 1 4^{2} = 15.23$ ✓. A rotation by any angle — including past $9 0^{\circ}$ — cannot change the length.

Ex 5 — Tangential EOM: steep climb vs level flight · covers C7, C8

The tangential (along-velocity) equation from the parent is $m \frac{d v}{d t} = T cos α - D - m g sin γ .$ The $sin γ$ factor is why gravity's grip depends on how steeply you climb: only the component of gravity along the flight path slows you.

Gravity's along-path pull at $γ = 7 5^{\circ}$ . $m g sin 7 5^{\circ} = 5000 (9.8) (0.9659) = 47330 N = 47.33 kN$ Why this step? Case C7 — steep. Compare to drag ( $10$ kN): gravity loss is ~5× larger. Forecast answered: gravity dominates.
Acceleration, case (a): $a = \frac{130000 - 10000 - 47330}{5000} = \frac{72670}{5000} = 14.53 m/s^{2}$ Why this step? We divide the net along-path force (thrust minus drag minus gravity's along-path pull) by the mass, which is exactly $F = m a$ solved for $a$ . The three forces are subtracted because drag and gravity both oppose the forward motion here.
Level flight, $γ = 0$ (case C8): $sin 0 = 0$ , so gravity contributes nothing to the tangential equation: $a = \frac{130000 - 10000 - 0}{5000} = \frac{120000}{5000} = 24.0 m/s^{2}$ Why this step? When level, gravity acts purely across the path — it curves the trajectory (next example) but does not slow the rocket down. This is the degenerate zero-input case for $sin γ$ .

Recall Verify

Level acceleration ( $24.0$ ) exceeds steep ( $14.53$ ) by exactly the gravity term $47330/5000 = 9.47$ m/s²: $14.53 + 9.47 = 24.0$ ✓. Units: N / kg = m/s² ✓. This "gravity loss" is why rockets do a Gravity Turn Trajectory rather than climbing vertically.

Ex 6 — Normal EOM: which way does the nose swing? · covers C9

The normal (turning) equation from the parent: $m v \frac{d γ}{d t} = T sin α + L - m g cos γ .$ The sign of the right side is the sign of $d γ / d t$ : positive → pitching up, negative → nosing over.

Thrust's turning contribution: $T sin α = 130000 sin 4^{\circ} = 130000 (0.06976) = 9069 N$ . Why this step? A tilted thrust has a small sideways component that helps steer.
Gravity's cross-path pull: $m g cos γ = 5000 (9.8) cos 8 0^{\circ} = 49000 (0.1736) = 8508 N$ . Why this step? At steep $γ = 8 0^{\circ}$ , gravity is almost along the path, so its cross component (∝ $cos γ$ ) is small.
Assemble the right side: $T sin α + L - m g cos γ = 9069 + 20000 - 8508 = 20561 N > 0$ Why this step? Positive ⇒ case C9 with $d γ / d t > 0$ : the nose pitches up. Lift and thrust-tilt beat the modest gravity cross-pull.
Turn rate: $\frac{d γ}{d t} = \frac{20561}{m v} = \frac{20561}{5000 \times 300} = \frac{20561}{1500000} = 0.01371 rad/s$ Why this step? The normal equation is $m v d γ / d t = (net cross-path force)$ , so dividing that net force by $m v$ isolates the turn rate. The factor $m v$ appears because turning bends the momentum $m v$ , not just the mass — a faster rocket at the same force curves more slowly.

Recall Verify

$0.01371$ rad/s $\times \frac{180}{π} = 0.78 5^{\circ} /$ s — a gentle upward curl, physically reasonable for a fast rocket. Units: N / (kg·m/s) = (kg·m/s²)/(kg·m/s) = 1/s = rad/s ✓.

Ex 7 — Real-world word problem · covers C10

Translate the prose: thrust $T = \overset{m}{˙} c$ (using the effective exhaust velocity $c$ from the refresher), and instantaneous mass $m (t) = m_{0} - \overset{m}{˙} t$ (from Variable Mass Systems). Vertical flight $γ = 9 0^{\circ}$ ⇒ $sin γ = 1$ , drag zero: $a (t) = \frac{T - m ( t ) g}{m ( t )} = \frac{T}{m ( t )} - g .$

Thrust (constant here): $T = \overset{m}{˙} c = 500 \times 3000 = 1, 500, 000 N = 1.5 MN$ . Why this step? Effective exhaust velocity $c$ already folds in the pressure term, so $T$ is one clean number and we never touch $p_{e}, p_{a}, A_{e}$ separately. This is exactly why $c$ was defined up front.
Liftoff ( $t = 0$ ): $m (0) = m_{0} = 60000$ kg. $a_{0} = \frac{1500000}{60000} - 9.8 = 25 - 9.8 = 15.2 m/s^{2}$ Why this step? We plug the full launch mass $m_{0}$ into $a = T / m - g$ because at $t = 0$ no propellant has yet been burned. The subtracted $g$ is gravity's full along-path pull in vertical flight ( $sin 9 0^{\circ} = 1$ ).
Burnout ( $t = 60$ s): the tank is empty exactly at the end of the 60-second burn, so $m (60) = m_{0} - \overset{m}{˙} t = 60000 - 500 (60) = 60000 - 30000 = 30000$ kg. $a_{1} = \frac{1500000}{30000} - 9.8 = 50 - 9.8 = 40.2 m/s^{2}$ Why this step? Here we insert the shrunken mass $m (60)$ into the same formula, because 60 s of burn has thrown away half the rocket. Same thrust divided by half the mass doubles the thrust-to-weight — the signature of a variable-mass system: acceleration soars near burnout.

Recall Verify

Ratio $a_{1} / a_{0} = 40.2/15.2 = 2.64$ ; the thrust term alone doubled ( $25 \to 50$ ) while $g$ stayed fixed, so the ratio must exceed 2 — consistent ✓. Units: N/kg − m/s² = m/s² ✓. See the deeper picture in the Tsiolkovsky Rocket Equation.

Ex 8 — Exam twist: solve for a hidden variable · covers C11

Start from the full thrust law $T = \overset{m}{˙} v_{e} + (p_{e} - p_{a}) A_{e}$ and isolate $v_{e}$ : $v_{e} = \frac{T - ( p _{e} - p _{a} ) A _{e}}{m ˙} .$

Pressure term (vacuum, $p_{a} = 0$ ): $(p_{e} - p_{a}) A_{e} = (60000 - 0) (0.8) = 48000 N = 48 kN$ Why this step? Case C11 — the exam wants you to subtract off the pressure bonus before dividing, otherwise you overstate $v_{e}$ . In vacuum the full exit pressure acts, so this term is a positive 48 kN slice of the measured thrust.
Solve for $v_{e}$ : $v_{e} = \frac{600000 - 48000}{200} = \frac{552000}{200} = 2760 m/s$ Why this step? We remove the 48 kN pressure contribution from the measured 600 kN to leave only the momentum thrust (552 kN), then divide by $\overset{m}{˙}$ because momentum thrust is $\overset{m}{˙} v_{e}$ . Since part of the reading came from pressure, $v_{e} = 2760 < 3000$ m/s — forecast confirmed: smaller.

Recall Verify

Plug back: $\overset{m}{˙} v_{e} + 48000 = 200 (2760) + 48000 = 552000 + 48000 = 600000$ N ✓ — recovers the measured thrust exactly. Units: N/(kg/s) = (kg·m/s²)/(kg/s) = m/s ✓.

Recall drill

Flip the sign of

α

: which force terms change?

Only the

sin α

terms — cosine is even, sine is odd.

What happens to

A

and

N

when

α

exceeds

9 0^{\circ}

cos α

goes negative, so the drag-in-

A

and lift-in-

N

contributions reverse.

γ = 0

(level flight) what is gravity's tangential effect?

Zero —

m g sin γ = 0

; gravity only curves the path.

When does the pressure thrust term vanish?

At perfect expansion,

p_{e} = p_{a}

Why does acceleration spike at burnout?

Mass

m (t) = m_{0} - \overset{m}{˙} t

shrinks while thrust stays roughly constant.

How do you get thrust from

\overset{m}{˙}

and

c

T = \overset{m}{˙} c

— effective exhaust velocity folds the pressure term in.

To find

v_{e}

from measured thrust, what must you subtract first?

The pressure term

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

, before dividing by

\overset{m}{˙}