Terminal landing — propulsive descent, suicide burn: Visual walkthrough

Step 1 — Draw the situation: what is falling and where is the ground?

WHAT. A rocket (call it the lander) is dropping straight down toward flat ground. We measure height as the distance from the ground up to the lander. We call this height $h$ — a single number in metres (m). The ground is $h = 0$ .

WHY. Before any maths, we must agree on which way is positive. We choose up = positive. That one choice fixes the sign of everything that follows, so we never guess a sign later.

PICTURE. The blue arrow is the lander's velocity — how fast and in which direction it moves. Because it is falling, that arrow points down, so its value is negative. We write the falling speed (a positive number) as $v$ , so the velocity itself is $- v$ .

Figure — Terminal landing — propulsive descent, suicide burn

Step 2 — Name the two accelerations: gravity down, thrust up

WHAT. Two things change the lander's speed:

Gravity pulls it down, adding downward speed at a steady rate. That rate is $g$ (on Earth, about $9.8 m/s^{2}$ ; we often round to $10$ ).
The engine pushes it up. The push produces an upward change-of-speed rate we call $a_{T}$ ("acceleration from thrust").

WHY these symbols and not force? You could talk about force $F$ in newtons, but what actually bends the motion is acceleration — how many m/s of speed you gain each second. Since thrust gives $a_{T} = F / m$ (force divided by mass), working directly in $a_{T}$ lets us skip carrying $m$ around. One fewer symbol to track.

PICTURE. Two vertical arrows on the same lander: a red gravity arrow pointing down (length $g$ ), an orange thrust arrow pointing up (length $a_{T}$ ). Notice the orange arrow is drawn longer — that will matter in Step 4.

Step 3 — Add the arrows: the net acceleration during the burn

WHAT. While the engine is on, both arrows act at once. Up is positive, so we add the upward thrust ( $+ a_{T}$ ) and the downward gravity ( $- g$ ). The leftover is the net acceleration: $a_{n e t} = engine pushes up, + a_{T} - gravity pulls down, - g$

WHY. Gravity never turns off. The engine must fight gravity and have something left over to actually slow the fall. Only that leftover, $a_{T} - g$ , does the braking. If you forget the $- g$ you will always plan to stop too soon and crash (see the parent's mistake box).

PICTURE. The red gravity arrow is subtracted from the orange thrust arrow, leaving a short green "net" arrow pointing up. That green arrow is the only thing decelerating the fall.

Step 4 — The life-or-death condition: $a_{T}$ must beat $g$

WHAT. Look at the sign of the green arrow. Three cases:

$a_{T} > g$ → $a_{n e t} > 0$ → green arrow points up → you can slow down. ✅
$a_{T} = g$ → $a_{n e t} = 0$ → no green arrow → speed never changes; you fall at constant speed forever. ❌
$a_{T} < g$ → $a_{n e t} < 0$ → green arrow points down → the engine can't even hold gravity; you keep speeding up. ❌

WHY show all three? This is the degenerate case the reader must never hit unaware. A suicide burn is only possible when the engine out-accelerates gravity. On a heavy-gravity world or with a weak engine, there is no ignition altitude that works — you simply cannot land this way.

PICTURE. Three landers side by side, each with its thrust/gravity/net arrows. Only the first has an upward green arrow.

Step 5 — The tool: which equation of motion turns speed into distance?

WHAT. We want the distance it takes to bring the falling speed $v$ down to $0$ using the steady braking $a_{n e t}$ . We have three quantities in play — starting speed, acceleration, distance — and no time. There is exactly one equation of motion that connects those three without time: $v_{f}^{2} = v_{i}^{2} + 2 a d$

WHY this tool and not another? The other kinematic equations ( $v_{f} = v_{i} + a t$ , $d = v_{i} t + \frac{1}{2} a t^{2}$ ) all contain time $t$ . We never asked "how long?" — we asked "how far?" Choosing the time-free equation means we don't have to solve for $t$ first. Fewer steps, fewer places to slip.

PICTURE. A speed-vs-distance graph. The curve of $v^{2}$ is a straight line when plotted against distance — that straightness is why $v^{2} = 2 a d$ is the clean tool here.

Step 6 — Plug in and solve for the burn distance

WHAT. During the braking burn:

Start speed $v_{i} = v$ (the falling speed at ignition, as a positive magnitude).
End speed $v_{f} = 0$ (we want to arrive stopped).
Acceleration = the braking magnitude $a_{n e t}$ (it opposes the fall, so it reduces $v^{2}$ ).
Distance = $h_{b u r n}$ , the height we must give up while braking.

Substitute: $0 = start energy-per-mass v^{2} - taken away by braking 2 a_{n e t} h_{b u r n}$

Now rearrange — add $2 a_{n e t} h_{b u r n}$ to both sides, then divide by $2 a_{n e t}$ : $h_{b u r n} = \frac{v ^{2}}{2 a _{n e t}} = \frac{v ^{2}}{2 ( a _{T} - g )}$

WHY. This single line is the whole result. It says: to erase a fall of speed $v$ , you need a runway of height $v^{2}$ shared out over twice the braking rate. Double the speed → four times the height (because of the square). That $v^{2}$ is the reason landings are so unforgiving of extra speed.

PICTURE. The braking stretch drawn on the height axis: the lander enters at $- v$ , the green net arrow eats away the speed, and it reaches $0$ exactly at $h = 0$ . The length of that stretch is $h_{b u r n}$ .

Step 7 — The trigger: fire when altitude drops to the required burn height

WHAT. As the lander falls, its speed $v$ grows, so the required burn height $h_{b u r n} (v)$ grows too. Meanwhile its actual altitude $h$ shrinks. There is one instant where they meet: $h = h_{b u r n} (v) = \frac{v ^{2}}{2 ( a _{T} - g )} ⟹ ignite now.$

WHY. Fire before they meet ( $h > h_{b u r n}$ ) and you stop high up, then hover and waste fuel — the "too early" crash. Fire after ( $h < h_{b u r n}$ ) and you run out of runway — the "too late" crash. The crossing is the last safe instant: the "suicide" moment.

PICTURE. Two curves on one altitude-vs-speed chart: the shrinking actual altitude (blue) and the growing required burn height (orange). Their crossing point is circled red — that is the ignition instant.

Step 8 — Where does the falling speed come from? (dropping from rest)

WHAT. If the lander started at rest and fell freely a distance $H$ before igniting, gravity alone built up its speed. Using the same time-free tool with $v_{i} = 0$ , $a = g$ , $d = H$ : $v^{2} = 0 + 2 g H ⟹ v = 2 g H$

Set the fall distance to $H_{t o t} - h^{*}$ (total drop minus the ignition height $h^{*}$ above ground) and demand the fall speed equals the burn requirement: $v_{i g n} = 2 g (H_{t o t} - h^{*}), h^{*} = \frac{v _{i g n}^{2}}{2 ( a _{T} - g )}$

WHY. This closes the loop: it tells you the ignition altitude $h^{*}$ purely from the drop geometry, before you even look at a speedometer. Two equations, two unknowns ( $v_{i g n}$ , $h^{*}$ ).

PICTURE. A single vertical drop split into two coloured segments: the top gray free-fall of height $H_{t o t} - h^{*}$ (speed climbing along a curve), then the bottom green braking segment of height $h^{*}$ (speed falling back to zero).

The one-picture summary

Everything above collapses into one diagram: a full descent from rest, gray free-fall on top, the ignition crossing marked, the green braking wedge at the bottom, and the boxed formula annotated in place. If you remember only one image, remember this one.

Recall Feynman retelling — the whole walkthrough in plain words

You drop a rocket. Down is "minus," up is "plus" — we picked that once so we never argue about signs again (Step 1). Two things tug it: gravity always pulls down at rate $g$ , the engine pushes up at rate $a_{T}$ (Step 2). Add them and only the leftover, $a_{T} - g$ , actually slows you (Step 3) — and if that leftover isn't positive, forget it, you can never stop (Step 4). To find how much room you need to stop, we grab the one motion equation with no time in it, $v^{2} = 2 a d$ , because we asked "how far," not "how long" (Step 5). Feed in "start at speed $v$ , end at zero, brake at $a_{T} - g$ ," and out pops the required height $v^{2} / [2 (a_{T} - g)]$ — notice the square, so twice the speed needs four times the room (Step 6). You fire the instant your real altitude sinks down to that required height — not sooner (you'd hover and starve), not later (you'd smash) (Step 7). And if you started from a still hover way up high, gravity itself set your speed via $v = 2 g H$ , which pins down exactly the altitude to light the candle (Step 8).

Active recall

Connections

Parent topic — the full theory and worked examples
Kinematics — Equations of Motion — where $v^{2} = 2 a d$ comes from (Steps 5, 8)
Thrust and Specific Impulse — the source of $a_{T} = F / m$ (Step 2)
Tsiolkovsky Rocket Equation — corrects the constant-mass shortcut for a real burn
Powered Descent Guidance — steering this idea in real 3-D flight