Terminal landing — propulsive descent, suicide burn

WHAT is a suicide burn?

WHY the name? If you mis-time ignition even slightly late, there is no recovery — the ground arrives before you can stop. It's "all or nothing," hence suicide.

WHY do it at all (the 80/20 reason)? Rockets have low mass fraction budget. Every second of thrust burns propellant $\propto$ thrust. Any hovering or early braking wastes fuel fighting gravity longer than necessary. The single-continuous-burn-at-the-end minimizes total impulse against gravity → maximum payload / minimum landing fuel.

HOW to derive the ignition altitude (from first principles)

Set up a 1-D vertical model. Let up be positive. Vehicle mass $m$ (treat as ~constant over the short burn), gravitational acceleration $g$, engine gives thrust producing upward acceleration $a_T = F/m$.

Phase 1 — free fall from rest (or given speed) through altitude, gaining speed. Phase 2 — braking burn: net upward acceleration $a_{net} = a_T - g$ (must be positive to decelerate!).

Step 1: Net deceleration during burn

Net force while thrusting: $F - mg$ upward. $a_{net} = \frac{F}{m} - g = a_T - g$ Why this step? Gravity never switches off — the engine must overcome both the downward momentum and keep fighting gravity, so the useful braking is only the excess thrust above weight.

Step 2: Distance needed to kill velocity $v$

Enter the burn moving downward at speed $v$ (so velocity $=-v$). We want it to reach $0$. Using $v_f^2 = v_i^2 + 2a\,d$ with the braking acceleration magnitude $a_{net}$ acting over braking distance $h_{burn}$: $0 = v^2 - 2\,a_{net}\,h_{burn}$ $\boxed{h_{burn} = \frac{v^2}{2\,a_{net}} = \frac{v^2}{2(a_T - g)}}$ Why this step? This is pure kinematics (energy form of the equations of motion). It says: the faster you're falling, the much longer runway you need — distance grows as $v^2$.

Step 3: The trigger condition

If the vehicle is falling with speed $v$ at altitude $h$, you must ignite the moment $h \le h_{burn}(v) = \frac{v^2}{2(a_T-g)}$ Since both $h$ and $v$ change as you fall, you continuously compute $h_{burn}(v)$ and fire when current altitude drops to it.

Step 4: How the falling speed relates to altitude (if starting from rest)

If you fell freely from rest through height $H$ before the burn, energy gives $v = \sqrt{2gH}$. Substitute to find at what altitude the trigger fires. Equating fall speed to burn requirement at ignition altitude $h^*$ measured from ground, with total drop height $H_{tot}$: $v_{ign} = \sqrt{2g(H_{tot}-h^*)}, \qquad h^* = \frac{v_{ign}^2}{2(a_T-g)}$

Steel-manned mistakes

Worked examples

Active recall

Connections

• Tsiolkovsky Rocket Equation — exact mass loss during the burn
• Kinematics — Equations of Motion — source of $v^2=2ad$
• Thrust and Specific Impulse — where $a_T=F/m$ comes from
• Gravity Turn Ascent — the launch analogue of trajectory optimization
• Powered Descent Guidance — real-world closed-loop version (G-FOLD)
• Terminal Velocity (atmospheric) — when drag, not thrust, limits descent

Concept Map

Hinglish (regional understanding)