Terminal landing — propulsive descent, suicide burn: Worked examples

This page throws every kind of suicide-burn problem at you: normal numbers, a moving start, the "engine too weak" degenerate case, the limiting behaviour, a real-world word problem, and an exam twist that adds mass loss. Each one is fully chewed. Start from the parent note if any symbol here is unfamiliar — but I re-define everything below.

The scenario matrix

Every problem this topic can pose falls into one of these cells. The Example that covers each cell is named in the last column.

Cell	What makes it special	Sign / regime of $a_{T} - g$	Covered by
A. Plain	Given $v, a_{T}, g$ → find $h_{b u r n}$	positive, comfortable margin	Ex 1
B. Weak margin	$a_{T}$ only just above $g$	positive but small → huge height	Ex 2
C. Degenerate	$a_{T} \leq g$	zero or negative → impossible	Ex 3
D. Zero input	$v = 0$ (already stopped)	any	Ex 4
E. Free-fall start	Dropped from rest through $H$ , find ignition altitude $h^{*}$	positive	Ex 5
F. Different world	Change $g$ (Moon/Mars)	positive, large margin	Ex 6
G. Word problem	Real lander, sensor triggers	positive	Ex 7
H. Limiting behaviour	What happens as $a_{T} \to g^{+}$ or $a_{T} \to \infty$	limit	Ex 8
I. Exam twist	Mass drops during burn (Tsiolkovsky-flavoured)	positive, growing	Ex 9

The worked examples

Cell A — the plain case

Worked example Example 1 — straight substitution

A lander is falling at $v = 80 m/s$ . Engine gives $a_{T} = 25 m/s^{2}$ , and $g = 10 m/s^{2}$ . Find the ignition altitude $h_{b u r n}$ .

Forecast: The leftover push is $25 - 10 = 15$ . Guess: a few hundred metres? Write your number before reading on.

Net braking acceleration: $a_{n e t} = a_{T} - g = 25 - 10 = 15 m/s^{2}$ . Why this step? Gravity keeps pulling during the burn, so only the excess thrust actually slows you. Using $25$ would over-promise your braking.
Apply the formula: $h_{b u r n} = \frac{v ^{2}}{2 a _{n e t}} = \frac{8 0 ^{2}}{2 \cdot 15} = \frac{6400}{30}$ . Why this step? This is the kinematic relation $v_{f}^{2} = v_{i}^{2} + 2 a d$ with $v_{f} = 0$ , solved for the distance $d$ .
Divide: $\frac{6400}{30} \approx 213.3 m$ .

Verify: Units — $(m/s)^{2} / (m/s^{2}) = m$ . ✓ Good, it's a length. Sanity: falling at $80$ with a strong engine, ~ $213$ m of runway is reasonable.

Cell B — weak thrust margin

Cell C — the degenerate case (thrust too weak to brake at all)

Worked example Example 3 — when the maths refuses

Same $v = 80$ , $g = 10$ , but $a_{T} = 8 m/s^{2}$ . Find $h_{b u r n}$ .

Forecast: The engine pushes up at $8$ , gravity pulls down at $10$ . Which wins? What does that do to the formula?

Net acceleration: $a_{n e t} = 8 - 10 = - 2 m/s^{2}$ . Why this step? Negative means the net acceleration still points down — even at full throttle you keep speeding up toward the ground.
Try the formula: $h_{b u r n} = \frac{6400}{2 \cdot ( - 2 )} = - 1600 m$ . Why this step? A negative height is physically meaningless — it's the maths screaming "there is no altitude that saves you."

Verify: No suicide burn exists when $a_{T} \leq g$ . The vehicle cannot decelerate; it will crash regardless of ignition timing. The requirement is a hard inequality: $a_{T} > g$ . See figure below for the sign map.

Look at the figure: only the region right of the vertical line ( $a_{T} > g$ ) gives a real, positive braking distance. On the line ( $a_{T} = g$ ) the height blows up to infinity; left of it, no solution.

Cell D — zero input

Cell E — free-fall start, find ignition altitude

Worked example Example 5 — dropped from rest through a known height

A vehicle is released from rest and free-falls. Total drop available to the ground is $H_{t o t} = 2000 m$ . Engine: $a_{T} = 30$ , $g = 10$ . At what altitude $h^{*}$ above the ground must it ignite?

Forecast: As it falls, speed grows and the needed burn height grows too. They must meet somewhere. Higher or lower than the halfway mark, $1000$ m?

Speed at ignition, having fallen $(H_{t o t} - h^{*})$ : free-fall from rest gives $v_{i g n} = 2 g (H_{t o t} - h^{*})$ , so $v_{i g n}^{2} = 2 g (H_{t o t} - h^{*})$ . Why this step? Energy / kinematics: falling from rest through distance $d$ builds speed $v = 2 g d$ . Here $d = H_{t o t} - h^{*}$ (the part already fallen).
Set ignition altitude = required burn height: $h^{*} = \frac{v _{i g n}^{2}}{2 ( a _{T} - g )}$ . Why this step? We ignite exactly when current altitude $h^{*}$ equals the distance we'll need to stop.
Substitute step 1 into step 2: $h^{*} = \frac{2 g ( H _{t o t} - h ^{*} )}{2 ( a _{T} - g )} = \frac{g ( H _{t o t} - h ^{*} )}{a _{T} - g}$ . Why this step? Eliminates $v$ ; now one unknown $h^{*}$ .
Solve algebraically: multiply out: $h^{*} (a_{T} - g) = g H_{t o t} - g h^{*}$ , so $h^{*} (a_{T} - g + g) = g H_{t o t}$ , giving $h^{*} = \frac{g H _{t o t}}{a _{T}}$ . Why this step? The $g$ 's recombine neatly — the ignition altitude is just $g / a_{T}$ of the total drop.
Plug numbers: $h^{*} = \frac{10 \cdot 2000}{30} = \frac{20000}{30} \approx 666.7 m$ .

Verify: Below the halfway mark $1000$ m, as forecast — you fall most of the way, then brake hard low down. Cross-check speed: fell $2000 - 666.7 = 1333.3$ m, so $v_{i g n} = 2 \cdot 10 \cdot 1333.3 = 26666.7 \approx 163.3 m/s$ ; then $h_{b u r n} = 163. 3^{2} / (2 \cdot 20) = 26666.7/40 = 666.7$ m. ✓ Same number both ways.

Cell F — a different world

Cell G — real-world word problem

Worked example Example 7 — the radar-triggered landing

A Mars lander (treat $g = 3.6 m/s^{2}$ ) descends at a steady $v = 60 m/s$ under a parachute it just cut away. Its engine can deliver $a_{T} = 15 m/s^{2}$ . Its landing radar reads altitude directly. (a) At what altitude should the flight computer light the engine? (b) If radar has $\pm 5 m$ error, is that dangerous here?

Forecast: Mars gravity is between Moon and Earth. Guess the ignition altitude to the nearest 50 m.

Net braking: $a_{n e t} = 15 - 3.6 = 11.4 m/s^{2}$ . Why this step? Subtract Mars gravity — it keeps pulling during the burn.
Ignition altitude: $h_{b u r n} = \frac{6 0 ^{2}}{2 \cdot 11.4} = \frac{3600}{22.8} \approx 157.9 m$ . Why this step? Standard formula; this is the trigger the computer waits for.
(b) Radar error impact: $5 m$ out of $\approx 158 m$ is about $3.2%$ . A late trigger by $5$ m eats into stopping distance; whether that's fatal depends on thrust reserve — here margin is comfortable, so a small percentage error is survivable but should be padded. Why this step? Always translate an absolute error into a fraction of the quantity that matters.

Verify: $5/157.9 = 0.0317 = 3.17%$ . ✓ Units on $h_{b u r n}$ : $(m/s)^{2} / (m/s^{2}) = m$ . ✓ Answer sits between Moon-like and Earth-like values, as expected for Mars.

Cell H — limiting behaviour

Worked example Example 8 — pushing the knobs to extremes

Keep $v = 100$ , $g = 10$ fixed. Describe $h_{b u r n}$ as (a) $a_{T} \to g^{+} = 1 0^{+}$ (engine barely beats gravity) and (b) $a_{T} \to \infty$ (infinitely strong engine). Give the numeric value at $a_{T} = 10.1$ for (a).

Forecast: One limit blows up, one shrinks to nothing. Which is which?

Limit (a), $a_{T} \to g^{+}$ : $a_{n e t} = a_{T} - g \to 0^{+}$ , so $h_{b u r n} = \frac{v ^{2}}{2 a _{n e t}} \to + \infty$ . Why this step? Dividing a fixed number by something shrinking to zero → unbounded. You'd need infinite altitude — matches Cell C's boundary.
Numeric taste at $a_{T} = 10.1$ : $a_{n e t} = 0.1$ , $h_{b u r n} = \frac{10000}{0.2} = 50000 m = 50 km$ . Why this step? Shows how brutally the height explodes near the boundary.
Limit (b), $a_{T} \to \infty$ : $a_{n e t} \to \infty$ , so $h_{b u r n} = \frac{v ^{2}}{2 a _{n e t}} \to 0$ . Why this step? An arbitrarily strong engine stops you in essentially no distance — ignite at the last metre.

Verify: At $a_{T} = 10.1$ : $10000/ (2 \cdot 0.1) = 10000/0.2 = 50000$ . ✓ Confirms the blow-up near $g$ ; and $h_{b u r n} \to 0$ as thrust grows, matching physical intuition (huge deceleration ⇒ tiny stopping distance).

Cell I — exam twist: mass drops during the burn

Worked example Example 9 — constant-mass estimate vs. mass loss

A lander enters the burn at $v = 200 m/s$ , mass $m_{0} = 1000 kg$ , constant thrust $F = 30000 N$ , $g = 10$ . (a) Give the constant-mass burn-height estimate using the initial mass. (b) Explain qualitatively why the true height is less, invoking Tsiolkovsky Rocket Equation.

Forecast: As fuel leaves, mass falls, so $a_{T} = F / m$ rises. Does that make the real stopping distance longer or shorter than the constant-mass guess?

Initial thrust acceleration: $a_{T} = F / m_{0} = 30000/1000 = 30 m/s^{2}$ . Why this step? $a_{T} = F / m$ from Thrust and Specific Impulse — the acceleration thrust alone provides at that instant.
Constant-mass net braking: $a_{n e t} = 30 - 10 = 20 m/s^{2}$ . Why this step? Subtract gravity from the initial thrust acceleration — only the leftover push brakes you, exactly as in every earlier example.
Constant-mass height: $h_{b u r n} = \frac{20 0 ^{2}}{2 \cdot 20} = \frac{40000}{40} = 1000 m$ . Why this step? This uses the weakest $a_{T}$ (biggest mass), so it's an over-estimate of the required height.
(b) Why true height is less: as propellant burns, $m$ shrinks, so $a_{T} = F / m$ grows through the burn; deceleration gets stronger the longer you brake. Stronger braking than the constant- $m$ model assumed ⇒ you stop in less distance. The exact answer comes from integrating $a_{T} (t) = F / (m_{0} - \overset{m}{˙} t)$ , the same mass-varying idea behind the Tsiolkovsky Rocket Equation. Why this step? The exam wants you to know the constant-mass formula is a safe upper bound, and to name the correct tool for exactness.

Verify: $20 0^{2} / (2 \cdot 20) = 40000/40 = 1000$ . ✓ The reasoning " $m ↓\Rightarrow a_{T} ↑\Rightarrow h_{b u r n} ↓$ " is monotonic and consistent with the parent note's "mass constant is the 80/20 estimate."

Recall Which cell was hardest for you?

Cover the table and re-derive Ex 5's clean result $h^{*} = g H_{t o t} / a_{T}$ from scratch. Derivation of $h^{*}$ for free-fall start ::: Set $h^{*} = 2 g (H_{t o t} - h^{*}) / [2 (a_{T} - g)]$ , cross-multiply, collect $h^{*}$ terms → $h^{*} = g H_{t o t} / a_{T}$ . Sign condition for any real solution ::: $a_{T} > g$ (strictly), else $h_{b u r n} \leq 0$ or infinite — no burn saves you. Behaviour of $h_{b u r n}$ as $a_{T} \to g^{+}$ ::: It diverges to $+ \infty$ — you'd need unlimited altitude.

Connections

Terminal landing — propulsive descent, suicide burn (index 3.4.26) — parent note & core derivation
Kinematics — Equations of Motion — the $v^{2} = 2 a d$ used in every cell
Thrust and Specific Impulse — where $a_{T} = F / m$ (Ex 9) comes from
Tsiolkovsky Rocket Equation — exact mass-loss treatment (Cell I)
Powered Descent Guidance — how a real computer schedules the trigger (Ex 7)
Gravity Turn Ascent — the ascent counterpart of this vertical model