3.4.26 · D5Rocket Flight Mechanics
Question bank — Terminal landing — propulsive descent, suicide burn
True or false — justify
Each claim is one line. Decide true/false, then give the mechanism — not just the label.
A rocket in a suicide burn spends most of the descent with its engine off.
True — the whole point is a single continuous burn at the end; before ignition it free-falls, so the engine is off for the bulk of the fall.
Firing earlier always gives a gentler, safer landing.
False — fire too early and makes you stop and hover high up, burning fuel every second just to fight gravity until you run dry and drop.
The braking distance depends linearly on how fast you are falling.
False — it scales as , so doubling the speed quadruples the altitude you need; speed is punished far worse than linearly.
If the engine's thrust equals the vehicle's weight exactly, a suicide burn can still barely work.
False — then so ; the rocket keeps falling at constant speed forever and can never reach zero velocity.
A suicide burn is fuel-optimal because it uses the smallest possible thrust.
False — it often uses full thrust; it is optimal because it uses thrust for the shortest possible time, minimizing the impulse wasted fighting gravity.
On the Moon the same lander at the same speed needs a higher ignition altitude than on Earth.
False — lower means a larger , so you stop in less distance; the Moon lets you ignite lower.
Treating mass as constant makes the formula predict the burn perfectly.
False — as propellant leaves, drops so rises and deceleration grows through the burn; the constant-mass answer is only a first estimate.
Once you pass the ignition altitude without firing, waiting a moment longer is fine.
False — beyond there is no altitude left to stop in, so the ground arrives before zero velocity; that "no recovery" property is exactly why it's called a suicide burn.
Spot the error
Each line contains a flawed statement. Reveal the correction.
"Use as the deceleration since the engine is what slows you down."
Gravity keeps adding downward speed during the burn, so the real braking is ; using underestimates the altitude you actually need.
"Set and solve."
The braking acceleration opposes the motion, so it enters with a minus: , giving — a positive distance.
"Fire when your altitude equals ."
That forgets gravity in the denominator; the correct trigger is , which is larger, so the naive rule fires too low and crashes.
"Since the burn is short, thrust and acceleration are both constant."
Thrust may be roughly constant, but mass falls, so rises — acceleration is not constant even if force is.
"If is negative you'll just decelerate slowly."
Negative means net force is downward; you speed up, not slow down — deceleration is impossible and you must never enter that regime.
" tells you the fall speed even while the engine is firing."
That free-fall result only holds before ignition; once the engine is on, the acceleration is (upward), not , so the relation no longer applies.
Why questions
Answer with the underlying cause, one or two sentences.
Why does the required altitude grow with the square of speed, not just proportionally?
Kinetic energy scales as and the constant braking force removes energy per unit distance, so distance must scale as to absorb it all.
Why must the engine keep fighting gravity even after it has "cancelled" the fall?
Gravity never switches off; the moment thrust stops the vehicle falls again, so the engine must provide worth of acceleration continuously just to stay stopped, plus more to decelerate.
Why is igniting too early a fuel problem rather than a timing problem?
You reach zero velocity above the ground and then hover, spending propellant every second only to counter gravity, exhausting the tank before touchdown. ::: You then fall from that height with no fuel left.
Why does a low thrust margin ( only slightly above ) force you to ignite dangerously high?
Small makes huge, so you must start braking far from the ground where errors and terrain are harder to judge.
Why is the suicide burn "all or nothing" with no margin for a late start?
The trigger altitude is exactly the minimum stopping distance, so any delay leaves less runway than the physics requires — there is no reserve to make up.
Why can you not simply hover down slowly to be safe?
Hovering means sustained for a long time, which consumes far more propellant than one brief maximal burn; a real lander lacks the fuel for a slow descent.
Why does the parent note treat as constant despite admitting it isn't?
It's an 80/20 first estimate — the constant-mass formula captures the scaling and trigger logic clearly, and the exact mass loss (Tsiolkovsky Rocket Equation) is layered on only when precision is needed.
Why does the sign of velocity (, downward) matter when writing ?
Only the magnitude enters , but the direction tells you the braking acceleration must point opposite the motion (upward), which fixes the minus sign in the energy equation.
Edge cases
Boundary and degenerate scenarios — where the naive picture breaks.
What happens at exactly (already at rest at some height)?
: you need no braking distance, so the "ignition altitude" is the ground itself — you'd just free-fall and touch at zero speed only if you're already on the ground. ::: In practice you'd hover or drop; the suicide-burn logic degenerates.
What happens when from above?
so ; the required ignition altitude diverges, meaning no finite starting height can stop you.
What if (thrust weaker than weight)?
: the vehicle accelerates downward even at full thrust, so a suicide burn is impossible — you can only slow the crash, not prevent it.
What if you begin the burn already moving upward?
The suicide-burn model assumes downward entry; if you're rising, you first coast up, gravity brings you back, and the braking analysis applies only to the subsequent fall.
What is the limiting behaviour of as thrust becomes enormous ()?
: an infinitely strong engine stops you in essentially no distance, so you could ignite right at the ground — the ideal, unreachable extreme.
On an airless world versus one with atmosphere, does the same need the same ?
Not exactly — atmospheric drag adds an extra upward force during the fall and burn, effectively increasing braking, so a real atmosphere (Terminal Velocity (atmospheric)) reduces the needed altitude below the vacuum formula.
If two landers have identical but different masses, do they need the same ignition altitude at the same ?
Yes — depends only on and , not on mass directly; mass matters only because it sets in the first place.
Connections
- Parent: Terminal landing
- Kinematics — Equations of Motion — the behind every trap here
- Thrust and Specific Impulse — why and why margin matters
- Tsiolkovsky Rocket Equation — the mass-loss correction to the constant-mass traps
- Terminal Velocity (atmospheric) — the drag edge case