Equations of motion — 3DOF point mass (trajectory analysis): Question bank

Before we start, one picture we'll refer to repeatedly — the velocity frame where every force gets split into a piece along the velocity arrow and a piece across it.

Figure — Equations of motion — 3DOF point mass (trajectory analysis)

The arrow is $V$ . "Along" ( $∥$ ) changes the arrow's length (speed). "Across" ( $⊥$ ) changes its direction (turns it). Weight points straight down and splits into $m g sin γ$ along and $m g cos γ$ across. Keep this in view.

True or false — justify

Cover the answer, decide true/false, then say why.

"3DOF" means the rocket moves in three spatial dimensions

x, y, z

False — the "3" counts degrees of freedom of translation (in the planar model:

V

γ

, and position). It is about how many independent motions we track, not the number of spatial axes; here it's a 2D vertical plane.

In the tangential equation, weight's contribution is

m g sin γ

True —

γ

is measured from the horizontal, so the piece of "straight down" that lies along the velocity is

m g sin γ

; at vertical flight (

γ = 90°

) it becomes full weight, which is correct.

If thrust is exactly along the velocity and there is no lift, gravity is the only thing that can turn the trajectory.

True —

\overset{γ}{˙} = (L - m g cos γ) / (mV)

; with

L = 0

and

T

contributing nothing across

V

, only

- m g cos γ

remains to bend the path. This is precisely a Gravity Turn Trajectory.

A rocket flying perfectly vertically (

γ = 90°

) can still have its flight-path angle change on its own.

False —

\overset{γ}{˙} = - g cos 90°/ V = 0

. With zero across-component of weight, a vertical path stays vertical; it only turns if a sideways force (lift or off-axis thrust) is added.

Increasing speed

V

makes the trajectory bend faster.

False —

\overset{γ}{˙} = - g cos γ / V

has

V

in the denominator, so faster flight bends more slowly. A fast rocket resists gravity's turning; a slow one is whipped over quickly.

At the very top of a coasting arc (

γ = 0

), gravity does nothing to the speed.

True — the along-component is

m g sin 0 = 0

, so gravity neither speeds up nor slows the rocket at that instant; it acts entirely across the path (maximal turning,

\overset{γ}{˙} = - g / V

The mass

m

cancels out of all four trajectory equations.

False — it cancels from the shape equations (

\overset{γ}{˙}

\overset{x}{˙}

\dot{h}

) but not from

\dot{V} = (T - D) / m - g sin γ

, where the thrust-and-drag term depends on

m

. Heavier vehicle, smaller acceleration.

If drag

D

exceeds thrust

T

, the rocket immediately starts falling.

False —

D > T

only makes

\dot{V}

possibly negative (slowing down). "Falling" is about direction (

γ

), which is governed by the separate

\overset{γ}{˙}

equation; a slowing rocket can still be climbing.

Spot the error

Each item contains a mistaken statement or step. Name the flaw and correct it.

\dot{V} = (T - D) / m - g cos γ

, because cosine handles the along-track direction like in projectile motion."

The error is

cos γ

: weight's along-velocity piece is

m g sin γ

, not

m g cos γ

. Sanity check at

γ = 90°

— all weight should oppose the climb, and

sin 90° = 1

gives that;

cos 90° = 0

would wrongly say gravity is irrelevant when going straight up.

"Since Earth rotates, I'll add Coriolis and centrifugal terms to my flat-Earth planar 3DOF."

Those fictitious forces belong only to a rotating round-Earth model. In a flat-Earth planar analysis the frame is (approximately) inertial and non-rotating, so those terms are absent. Mixing models double-counts effects.

"Thrust always points straight up, so I'll put

+ T

in the

\dot{h}

equation."

The kinematic equations

\overset{x}{˙} = V cos γ

\dot{h} = V sin γ

contain no forces at all — they only convert speed and direction into position rates. Thrust enters via

\dot{V}

(which changes

V

), never directly into

\dot{h}

"I'll integrate

m \dot{V} = T - D - m g sin γ

treating

m

as constant since it's 'the rocket's mass'."

Propellant burns, so

\overset{m}{˙} = - T / (I_{s p} g_{0})

and

m

shrinks continuously. Holding it constant underestimates late-flight acceleration — the same thrust on a lighter vehicle gives much larger

\dot{V}

near burnout.

\overset{γ}{˙}

came out negative, so I must have a sign mistake — the rocket is climbing, angles should grow."

Negative

\overset{γ}{˙}

is correct and expected: gravity's across-component always curves the velocity arrow downward, decreasing

γ

over time. Climbing (positive

\dot{h}

) and a decreasing flight-path angle coexist happily.

"For a gravity turn I need a control system to command the pitch-over the whole way."

The whole point of a Gravity Turn Trajectory is that after a tiny initial pitch-over, gravity alone rotates

V

toward horizontal via

\overset{γ}{˙} = - g cos γ / V

— no continuous control effort, hence minimal steering losses.

"Drag is

D = ρ V^{2} S C_{D}

Missing the factor

\frac{1}{2}

D = \frac{1}{2} ρ V^{2} S C_{D}

. The half comes from the dynamic-pressure definition

q = \frac{1}{2} ρ V^{2}

; see Drag and Atmospheric Models.

Why questions

Explain the reasoning, not just the fact.

Why do we resolve forces in the velocity frame rather than the ground

x

–

h

frame?

Because a velocity vector has exactly two independent things to change — its length and its direction — and the velocity frame splits force cleanly into one piece per job (along = length, across = direction), turning one vector equation into two decoupled scalar ones. See Flight-Path Angle and Velocity Frame.

Why does

sin γ

appear in the speed equation but

cos γ

in the turning equation?

γ

is measured from horizontal, and weight points down. The angle between "down" and "backward along

V

" makes the along-piece

sin γ

; the perpendicular piece is what's left,

cos γ

. "S-along, C-across."

Why is the perpendicular acceleration written as

mV \overset{γ}{˙}

Turning the velocity vector at rate

\overset{γ}{˙}

while it has length

V

is exactly circular-motion geometry: the tip sweeps sideways at speed

V \overset{γ}{˙}

, so the across-acceleration is

V \overset{γ}{˙}

and the across-force is

mV \overset{γ}{˙}

Why can we ignore rotational (attitude) dynamics and still get a good trajectory?

Range, altitude, and burnout speed depend only on the motion of the centre of mass (translation). Rotation matters for pointing and stability — that's when you upgrade to 6DOF Rigid-Body Dynamics.

Why must these four/five ODEs be solved numerically rather than by a neat formula?

They are coupled and nonlinear —

D

depends on

V

(and on

ρ (h)

\overset{γ}{˙}

divides by

V

m

changes in time — so no closed form exists in general; we step them forward with a scheme like Numerical Integration of Trajectories (RK4).

Why does the vertical-flight case connect to the Tsiolkovsky equation?

γ = 90°

with no drag,

\dot{V} = T / m - g

, and integrating with

\overset{m}{˙} = - T / (I_{s p} g_{0})

reproduces the ideal velocity gain minus a gravity-loss term — the regime of the Tsiolkovsky Rocket Equation.

Why does the same thrust give ever-larger acceleration as the flight proceeds?

\dot{V} = (T - D) / m - \dots

, mass

m

shrinks as propellant burns. Dividing a roughly steady thrust by a smaller mass yields a growing acceleration — burnout acceleration is dramatically higher than liftoff.

Edge cases

The scenarios that break lazy formulas — state what actually happens.

V \to 0

(rocket momentarily at rest, e.g. apex of a vertical hop): what happens to

\overset{γ}{˙} = - g cos γ / V

It blows up (division by zero) — the model of "turning the velocity arrow" is undefined when there is no arrow. Physically the direction is momentarily ill-posed; near zero speed one switches to tracking velocity components directly, not

γ

γ = 0

(level flight): what do the along/across weight pieces become?

Along-piece

m g sin 0 = 0

(gravity doesn't slow you), across-piece

m g cos 0 = m g

(full weight bends the path down at maximum rate

\overset{γ}{˙} = - g / V

). Level flight is where gravity turns you hardest.

γ = - 90°

(straight-down dive): sign of

\dot{h}

and gravity's effect on speed?

\dot{h} = V sin (- 90°) = - V < 0

(descending), and

g sin (- 90°) = - g

\dot{V} = (T - D) / m + g

— gravity now adds to speed. Signs of

γ

automatically carry the direction; no special-casing needed.

T = 0

and

D = 0

(pure coast in vacuum): what shape does the trajectory take?

The equations reduce to

\dot{V} = - g sin γ

\overset{γ}{˙} = - g cos γ / V

\overset{x}{˙} = V cos γ

\dot{h} = V sin γ

— exactly a projectile parabola. The 3DOF set contains free-fall as a limiting case.

L

large enough to make

mV \overset{γ}{˙} > 0

: is that allowed?

Yes — if lift exceeds

m g cos γ

, then

\overset{γ}{˙} > 0

and the path curves upward. Nothing in the model forbids it; it's how a winged vehicle pulls up. Ballistic rockets just usually have

L \approx 0

Drag

D

V \to 0

: does drag vanish, and why does that matter at liftoff?

Yes —

D = \frac{1}{2} ρ V^{2} S C_{D} \to 0

V \to 0

, so at the instant of liftoff drag is negligible and

\dot{V} \approx T / m - g

. Drag only becomes important once speed builds; see Drag and Atmospheric Models.

High altitude where

ρ \to 0

: which term in

\dot{V}

disappears and what remains?

Air density

ρ \to 0

makes

D \to 0

, so

\dot{V} \to T / m - g sin γ

— the vacuum tangential equation. This is why upper-stage burns are so efficient: no atmospheric drag loss.

Recall One-line self-test

Cover everything and answer aloud: which equation changes speed, which changes direction, and which two change position — and which single trig function ( $sin$ or $cos$ ) of $γ$ sits in each? Answer ::: Speed: $\dot{V}$ with $g sin γ$ . Direction: $\overset{γ}{˙}$ with $g cos γ$ . Position: $\overset{x}{˙} = V cos γ$ , $\dot{h} = V sin γ$ .

Connections

Equations of motion — 3DOF point mass (trajectory analysis) (trajectory analysis) — the parent this bank drills
Flight-Path Angle and Velocity Frame — geometry behind the along/across split
Gravity Turn Trajectory — the $\overset{γ}{˙}$ edge cases in action
Tsiolkovsky Rocket Equation — the vertical/no-drag limit
Drag and Atmospheric Models — where the $ρ V^{2}$ traps live
6DOF Rigid-Body Dynamics — what the ignored rotation would add
Numerical Integration of Trajectories (RK4) — why closed forms fail