Terminal landing — propulsive descent, suicide burn: Exercises

Definition Sign convention used on this whole page

We work in 1-D vertical motion and choose up = positive, down = negative (the same convention as the parent note).

Altitude $h \geq 0$ : height above the ground.
Velocity while falling is negative (pointing down); its magnitude — the plain "speed" — we call $v > 0$ . So a lander "falling at $v = 200 m/s$ " has velocity $- 200 m/s$ .
$g > 0$ is the magnitude of gravity; gravity's acceleration is $- g$ (points down).
$a_{T} = F / m > 0$ is the upward acceleration the engine alone would produce. Here $F$ is the engine thrust force (newtons) and $m$ the vehicle mass (kg); dividing force by mass gives an acceleration, from Newton's second law $F = ma$ (see Thrust and Specific Impulse for where $F$ comes from).
During the burn the net acceleration is $a_{n e t} = a_{T} - g$ (up positive minus the downward pull), which must be positive to actually slow a fall.

Formula Where

$v^{2} = 2 a d$ comes from (full steps, so this page stands alone) We want a relation between speed and distance that never mentions time — because in the burn problems we care about "how much altitude", not "how many seconds". Start from the two constant-acceleration facts (derived in Kinematics — Equations of Motion): $v_{f} = v_{i} + a t (1)$ $d = v_{i} t + \frac{1}{2} a t^{2} (2)$ WHY eliminate $t$ ? Because $t$ is a quantity we do not know and do not care about; we only have speeds and a distance. So we remove it. Step A — solve (1) for time: $t = \frac{v _{f} - v _{i}}{a}$ . Step B — substitute this $t$ into (2): $d = v_{i} (\frac{v _{f} - v _{i}}{a}) + \frac{1}{2} a (\frac{v _{f} - v _{i}}{a})^{2} .$ Step C — expand each piece over a common denominator $a$ : $d = \frac{v _{i} ( v _{f} - v _{i} )}{a} + \frac{( v _{f} - v _{i} ) ^{2}}{2 a} = \frac{2 v _{i} v _{f} - 2 v _{i}^{2} + v _{f}^{2} - 2 v _{i} v _{f} + v _{i}^{2}}{2 a} .$ Step D — the $2 v_{i} v_{f}$ terms cancel and $- 2 v_{i}^{2} + v_{i}^{2} = - v_{i}^{2}$ , leaving $d = \frac{v _{f}^{2} - v _{i}^{2}}{2 a} ⟹ v_{f}^{2} = v_{i}^{2} + 2 a d .$ WHAT it means: distance grows with the square of speed — this is the seed of every " $v^{2}$ " you will see below.

Level 1 — Recognition

Problem 1.1

A lander is falling. Its engine, on its own, would push it upward at $a_{T} = 25 m/s^{2}$ . Local gravity is $g = 10 m/s^{2}$ . What is the net braking acceleration the vehicle actually feels once the engine is lit?

Recall Solution

Gravity does not switch off when the engine fires — it keeps pulling down the whole time. So the useful deceleration is only the leftover push above what gravity steals: $a_{n e t} = a_{T} - g = 25 - 10 = 15 m/s^{2} .$ Answer: $15 m/s^{2}$ .

Problem 1.2

Two landers, A and B, both approaching Mars ( $g = 3.7 m/s^{2}$ ). A has $a_{T} = 5 m/s^{2}$ , B has $a_{T} = 3 m/s^{2}$ . Which one is physically able to perform a suicide burn, and which one cannot stop at all?

Recall Solution

A suicide burn only works if the net acceleration points upward (positive), i.e. if $a_{T} > g$ .

Lander A: $a_{T} - g = 5 - 3.7 = + 1.3 m/s^{2} > 0$ → can brake.
Lander B: $a_{T} - g = 3 - 3.7 = - 0.7 m/s^{2} < 0$ → net acceleration is still downward; the engine cannot even hold it up, let alone slow it. Cannot stop. Answer: A can, B cannot.

Level 2 — Application

Problem 2.1

A booster is falling straight down at $v = 200 m/s$ . Engine gives $a_{T} = 30 m/s^{2}$ ; $g = 10 m/s^{2}$ . At what altitude must it ignite so velocity and altitude hit zero together?

Recall Solution

Net braking: $a_{n e t} = 30 - 10 = 20 m/s^{2}$ . Burn altitude: $h_{b u r n} = \frac{v ^{2}}{2 a _{n e t}} = \frac{20 0 ^{2}}{2 \cdot 20} = \frac{40000}{40} = 1000 m .$ Answer: ignite at $1000 m$ .

Problem 2.2

Same booster, same engine, but now it is falling twice as fast, $v = 400 m/s$ . Predict first whether the required altitude doubles, then compute.

Recall Solution

Forecast: height grows as $v^{2}$ , so doubling $v$ should quadruple the height → expect $4000 m$ , not $2000$ . Verify: $h_{b u r n} = \frac{40 0 ^{2}}{2 \cdot 20} = \frac{160000}{40} = 4000 m . ✓$ Answer: $4000 m$ — four times, not two.

Problem 2.3

A lander must ignite no higher than $600 m$ (any higher and it clips a ridge). Its engine gives $a_{n e t} = 12 m/s^{2}$ . What is the maximum falling speed it may have at ignition?

Recall Solution

Read the problem carefully: the ridge sets the largest allowed burn altitude, so the ceiling $600 m$ is the burn altitude at the fastest permitted speed. We therefore set $h_{b u r n} = 600 m$ and solve for $v$ . Rearrange the burn formula for $v$ : $h_{b u r n} = \frac{v ^{2}}{2 a _{n e t}} ⟹ v = 2 a_{n e t} h_{b u r n} .$ Substitute $h_{b u r n} = 600 m$ and $a_{n e t} = 12 m/s^{2}$ : $v = 2 \cdot 12 \cdot 600 = 14400 = 120 m/s .$ Answer: $120 m/s$ — faster than this and 600 m of runway is not enough to stop in time.

Level 3 — Analysis

Problem 3.1

A rocket falls from rest through a total height $H_{t o t} = 2000 m$ . Engine gives $a_{T} = 30 m/s^{2}$ , $g = 10$ . Find the altitude $h^{*}$ above the ground at which the suicide burn ignites.

The figure below is your map for this problem — refer to it as we go. The vertical axis (white arrow, labelled "altitude h") is height above ground, up positive. The amber ▼ at the top marks the start point ( $h = 2000 m$ , at rest). The long cyan arrow is Phase 1, the free fall where speed grows. The amber dashed line is the ignition altitude $h^{*}$ we are solving for, marked with an amber ●. Below it the amber arrow is Phase 2, the braking burn down to the cyan ground line at $h = 0$ , $v = 0$ .

Figure — Terminal landing — propulsive descent, suicide burn

Recall Solution

Look at the figure: the fall splits into the two arrows, and each gives one equation.

Fall speed as a function of drop (the cyan Phase-1 arrow). Falling freely from rest through a drop of $(H_{t o t} - h^{*})$ — the length of the cyan arrow — the speed reached at the amber ignition line is (from $v^{2} = 2 a d$ with $a = g$ , $v_{i} = 0$ ): $v_{i g n} = 2 g (H_{t o t} - h^{*}) .$
Burn requirement (the amber Phase-2 arrow). To stop from $v_{i g n}$ over the remaining height $h^{*}$ — the length of the amber arrow down to the ground line: $h^{*} = \frac{v _{i g n}^{2}}{2 ( a _{T} - g )} .$

Substitute (1) into (2). The $v_{i g n}^{2} = 2 g (H_{t o t} - h^{*})$ term goes on top: $h^{*} = \frac{2 g ( H _{t o t} - h ^{*} )}{2 ( a _{T} - g )} = \frac{g ( H _{t o t} - h ^{*} )}{a _{T} - g} .$ Put numbers: $\frac{g}{a _{T} - g} = \frac{10}{20} = 0.5$ . $h^{*} = 0.5 (2000 - h^{*}) = 1000 - 0.5 h^{*} ⟹ 1.5 h^{*} = 1000 ⟹ h^{*} = 666.7 m .$ That is exactly where the amber dashed line sits in the figure. Check the ignition speed (the value beside the amber ● dot): $v_{i g n} = 2 \cdot 10 \cdot (2000 - 666.7) = 26667 \approx 163.3 m/s$ . Answer: ignite at $h^{*} \approx 667 m$ , falling at $\approx 163 m/s$ .

Problem 3.2

For a fall from rest, show that the ignition altitude is a fixed fraction of the total drop height, and find that fraction in terms of $a_{T}$ and $g$ .

Recall Solution

From Problem 3.1's substituted equation: $h^{*} = \frac{g}{a _{T} - g} (H_{t o t} - h^{*}) .$ Let $r = \frac{g}{a _{T} - g}$ . Then $h^{*} = r H_{t o t} - r h^{*} ⟹ h^{*} (1 + r) = r H_{t o t}$ , so $\frac{h ^{*}}{H _{t o t}} = \frac{r}{1 + r} = \frac{g}{a _{T}} .$ The ignition fraction is simply $g / a_{T}$ . Higher thrust → smaller fraction → you wait longer (fire lower). Check 3.1: $g / a_{T} = 10/30 = 1/3$ , and $2000/3 = 666.7 m$ ✓. Answer: $h^{*} / H_{t o t} = g / a_{T}$ .

Level 4 — Synthesis

Problem 4.1

A lander has dry-plus-payload mass that gives $a_{T} = 24 m/s^{2}$ at ignition ( $g = 10$ ). It falls at $v = 180 m/s$ . The engine's exhaust velocity gives it a specific-impulse-based mass-flow such that it burns $8%$ of its mass per second of full thrust (from Thrust and Specific Impulse). (a) Find the burn altitude with the constant-mass formula. (b) Estimate the burn duration using the constant- $a_{n e t}$ approximation. (c) Estimate the propellant fraction consumed.

Recall Solution

(a) $a_{n e t} = 24 - 10 = 14 m/s^{2}$ . $h_{b u r n} = \frac{18 0 ^{2}}{2 \cdot 14} = \frac{32400}{28} \approx 1157.1 m .$ (b) With constant deceleration, time to kill speed $v$ is $t = v / a_{n e t}$ : $t = \frac{180}{14} \approx 12.86 s .$ (c) At $8% = 0.08$ per second for $\approx 12.86 s$ (linear first estimate): $Δ m / m \approx 0.08 \times 12.86 \approx 1.03.$ This exceeds $100%$ — a red flag. It means the constant-mass model breaks down: you cannot burn that long at this flow without running dry. In reality mass drops, $a_{T} = F / m$ rises, so $a_{n e t}$ grows, the burn is shorter than 12.86 s, and less propellant is used. See Problem 5.1 for the honest exact treatment. Answers: (a) $\approx 1157 m$ , (b) $\approx 12.9 s$ (over-estimate), (c) linear estimate $\approx 103%$ — signals the constant-mass model is invalid here.

Problem 4.2

Two candidate engines for the same $v = 150 m/s$ , $g = 10$ descent: Engine X gives $a_{T} = 20 m/s^{2}$ , Engine Y gives $a_{T} = 40 m/s^{2}$ . Which lets you ignite lower, and by what factor is Y's burn altitude smaller than X's?

Recall Solution

$h_{b u r n} \propto 1/ (a_{T} - g)$ .

X: $a_{n e t} = 10$ , $h_{X} = \frac{15 0 ^{2}}{2 \cdot 10} = \frac{22500}{20} = 1125 m$ .
Y: $a_{n e t} = 30$ , $h_{Y} = \frac{22500}{2 \cdot 30} = \frac{22500}{60} = 375 m$ . Ratio $h_{Y} / h_{X} = 375/1125 = 1/3$ . Engine Y ignites lower (safer margin) and needs only one-third the altitude. Note doubling $a_{T}$ from 20→40 did not halve the height — it tripled the net braking (10→30), so the height fell by a factor of 3. Answer: Y ignites lower; $h_{Y} = h_{X} /3$ .

Level 5 — Mastery

Problem 5.1

Exact-ish check on the mass-varying burn. A lander enters its burn at speed $v_{0} = 100 m/s$ with thrust producing an initial $a_{T} = 20 m/s^{2}$ ; $g = 10$ . Because propellant leaves, the true deceleration grows through the burn. Compare the constant-mass burn altitude against a two-segment estimate that uses $a_{T} = 20$ for the first half of the speed loss and $a_{T} = 25$ for the second half (mimicking mass dropping so thrust-accel rises). Which needs less altitude, and does that match the physics?

Recall Solution

Constant-mass baseline: $a_{n e t} = 20 - 10 = 10$ , $h_{co n s t} = \frac{10 0 ^{2}}{2 \cdot 10} = \frac{10000}{20} = 500 m .$ Two-segment estimate. Split the speed drop $100 \to 0$ into two halves at $v = 50$ . For each segment use $v_{f}^{2} = v_{i}^{2} + 2 a d$ rearranged as $d = (v_{i}^{2} - v_{f}^{2}) / (2 a_{n e t})$ (here $v_{i}$ = faster speed entering the segment, $v_{f}$ = slower speed leaving it).

Segment 1: slow from $100$ to $50 m/s$ , mass still high so $a_{T} = 20$ , $a_{n e t} = 20 - 10 = 10$ : $h_{1} = \frac{10 0 ^{2} - 5 0 ^{2}}{2 \cdot 10} = \frac{10000 - 2500}{20} = \frac{7500}{20} = 375 m .$
Segment 2: slow from $50$ to $0$ , mass has dropped so $a_{T} = 25$ , $a_{n e t} = 25 - 10 = 15$ : $h_{2} = \frac{5 0 ^{2} - 0 ^{2}}{2 \cdot 15} = \frac{2500}{30} \approx 83.3 m .$
Total burn altitude: add the two segment heights: $h_{2 se g} = h_{1} + h_{2} = 375 + 83.3 \approx 458.3 m .$

Conclusion: the mass-varying estimate needs less altitude ( $\approx 458 m$ vs $500 m$ constant-mass). This matches the physics: as fuel burns, $a_{T} = F / m$ rises, net braking strengthens, and the vehicle stops in less distance. So the constant-mass formula is a conservative over-estimate of the runway you need — which is good, because over-estimating your braking room errs on the safe side. Answer: mass-varying $\approx 458 m < 500 m$ constant-mass; the growing thrust-acceleration shortens the runway.

Problem 5.2

A booster falls from rest on Earth ( $g = 9.81$ ) and must reach zero speed at the ground. Its engine gives $a_{T} = 29.43 m/s^{2}$ (exactly $3 g$ ). Total drop $H_{t o t} = 3000 m$ . (a) Find the ignition altitude fraction and altitude. (b) Find the ignition speed. (c) Compare the ignition speed to the ground-impact speed the vehicle would have had with no burn, and comment.

Recall Solution

(a) From Problem 3.2, ignition fraction $= g / a_{T} = 1/3$ . So $h^{*} = \frac{1}{3} \times 3000 = 1000 m .$ (b) Ignition speed after falling through $H_{t o t} - h^{*} = 2000 m$ : $v_{i g n} = 2 g (2000) = 2 \cdot 9.81 \cdot 2000 = 39240 \approx 198.09 m/s .$ Cross-check with burn formula: $h^{*} = \frac{v _{i g n}^{2}}{2 ( a _{T} - g )} = \frac{39240}{2 \cdot 19.62} = \frac{39240}{39.24} = 1000 m$ ✓. (c) With no burn the impact speed after the full $3000 m$ would be $v_{g r o u n d} = 2 \cdot 9.81 \cdot 3000 = 58860 \approx 242.6 m/s .$ So the vehicle ignites at $\approx 198 m/s$ (already very fast) and, in the last kilometre, bleeds off all of that plus the extra it would have gained — the burn works hardest right at the end. That "let it fall almost the whole way, then slam the brakes once" is exactly the fuel-optimal logic of the suicide burn. Answers: (a) fraction $1/3$ , $h^{*} = 1000 m$ ; (b) $v_{i g n} \approx 198.1 m/s$ ; (c) no-burn impact $\approx 242.6 m/s$ .

Recall One-line summary of the whole page

$h_{b u r n} = v^{2} / [2 (a_{T} - g)]$ (valid only for $a_{T} > g$ ); from rest, ignite at fraction $g / a_{T}$ of the drop; height scales as $v^{2}$ ; forgetting $g$ is dangerous, ignoring mass loss is conservative.

Connections

3.4.26 Terminal landing — propulsive descent, suicide burn (Hinglish) — the parent note these exercises drill
Kinematics — Equations of Motion — the $v^{2} = 2 a d$ engine behind every problem
Tsiolkovsky Rocket Equation — the exact mass-loss treatment invoked in L4/L5
Thrust and Specific Impulse — source of $a_{T} = F / m$ (with $F$ = thrust force) and the mass-flow in Problem 4.1
Powered Descent Guidance — real-time recomputation of $h_{b u r n} (v)$ in flight