Coming back through the atmosphere

Every flight of Aster brings two vehicles back through the air. The first stage never reaches orbit: it separates at 70 km, climbs to 145 km, and falls back into the atmosphere at about 2 km/s. The upper stage goes to orbit and comes home from it, meeting the air at 7.4 km/s. Both have to lose almost all of that speed before a landing burn a kilometre or so above the ground, and both give nearly all of it to the air, for nothing.

The air does not take it gently. What it takes, it takes as a force on the structure and as heat, and both depend on the same few things: how fast the vehicle arrives, how steeply, and how much mass it carries behind each square metre of drag. This lesson is about those three numbers — the theory that ties them together, where it holds and where it does not, and why the booster, which could leave the whole job to the air, fires three engines for twenty seconds on the way down instead.

Two ways back into the air

The energy to be removed is the kinetic energy relative to the air, $\frac{1}{2}{v}^2$ per kilogram. On the default mission the booster passes 100 km on the way down at 2.00 km/s: 2.0 MJ in every kilogram. The upper stage passes 100 km at 7.40 km/s relative to the air, which turns with the Earth: 27.4 MJ per kilogram, fourteen times as much. Its inertial speed is 7.82 km/s, but the air near Starbase is already moving east at 0.42 km/s, and it is the speed through the air that the air has to take away.

The two arrive differently, too. The booster is on a short, high arc and comes down steeply, 26° below the horizon at 100 km and 35° by 55 km. The upper stage comes out of an orbit and meets the air almost level, at 1.5°. Those two differences, fourteen times the energy at a seventeenth of the angle, make them two different problems.

One number for the vehicle

The air decelerates a vehicle at

where $D$ is the drag, $\rho$ the air density, $v$ the speed, $q=\frac{1}{2}\rho v^2$ the dynamic pressure, $C_D$ the drag coefficient, $A$ the frontal area and $m$ the mass. The ballistic coefficient $\beta$ is mass per unit of effective drag area, and it is everything about the vehicle that matters to an entry without lift. A low $\beta$ decelerates high up, where the air is thin; a high $\beta$ falls deep before the air takes hold.

The simulator's stages are dense. Aster's booster is 3.66 m across — 10.52 m² of frontal area — and, flying engines-first, has an axial force coefficient near 1.3 at hypersonic speed, plus the drag of four grid fins. Dividing the dynamic pressure by the drag deceleration in the reference flight gives its ballistic coefficient directly: about 3,000 kg/m² at Mach 7, with 50 t aboard. After its entry burn has taken 18 t out of it, about 1,800 kg/m². The upper stage, 15.4 t behind the same diameter, is about 950 kg/m². The Mars landers of course 16 were designed the other way, as light as possible behind as wide a shield as would fit: Viking entered at 64 kg/m², Curiosity at 146.

A ballistic entry

Treat the Earth as a non-rotating sphere of radius $R$, the vehicle as a point of constant $\beta$ with no lift, and measure the flight-path angle $\gamma$ below the local horizontal. At altitude $h$, radius $r=R+h$ and local gravity $g=\mu /r^2$,

The first equation is drag against the pull of gravity along the path. The second is the path bending: down under gravity, up under its own curvature. For the booster, far below orbital speed, gravity wins and the path steepens all the way down. For the upper stage, at nearly orbital speed, the two almost cancel at first, and the path steepens only as the air takes the speed away.

The straight-line solution

H. Julian Allen and Alfred Eggers, working on ballistic missile warheads at NACA's Ames laboratory in the 1950s, found the solution that still organises the subject. Suppose drag is so much larger than gravity that $g$ can be dropped, so the path is a straight line at constant $\gamma$, and suppose the density falls exponentially, $\rho ={\rho }_0{e}^{-h/H}$, with ${\rho }_0$ the density at the ground and $H$ the scale height. Dividing the first equation by the third turns time into altitude, and with $dh=-Hd\rho /\rho$ it integrates at once:

with $v_E$ the entry speed, reached where the density is effectively zero. For Earth take ${\rho }_0=1.225$ kg/m³ and $H=7.2$ km: between 15 and 75 km, where entries happen, that exponential stays within about 20 % of the standard atmosphere.

Peak deceleration. The deceleration $\rho v^2/2\beta$ is, along this solution, $(\rho v_E^2/2\beta )\exp (-\rho H/\beta \sin \gamma )$. As the vehicle descends, the growing density raises it and the falling speed lowers it; it peaks where $\rho =\beta \sin \gamma /H$, with the speed down to $v_E{e}^{-1/2}=0.61v_E$, at

— independent of $\beta$. A dense vehicle does not feel a harder peak than a light one. It feels the same peak lower down, at

Peak dynamic pressure. At that point $q=\frac{1}{2}\rho v^2=\beta a_{\text{max}}$. The deceleration does not depend on $\beta$, so the dynamic pressure that produces it grows in proportion: a vehicle three times as dense meets the same peak deceleration at three times the dynamic pressure. That is the booster's problem, and it is worth putting numbers on it.

Where it holds, and where it does not

Take the booster as it passes 55 km with nothing done: 2.20 km/s at 34.8° below the horizon, $\beta =3,000$ kg/m². The straight-line solution predicts a peak of 7.2 g at 11.8 km and a peak dynamic pressure of 212 kPa. The simulator, flying the same booster on the same mission with its entry burn switched off, gives 12.1 g at 13.0 km and 310 kPa. The theory has the right altitude and the wrong size, and the reasons are the things it dropped. The booster is not a warhead: at 2 km/s gravity keeps adding speed and steepening the path, from 35° at 55 km to 42° at the peak. The peak falls near the tropopause, where the real atmosphere is not the exponential and its scale height is nearer 6.4 km than 7.2. And the booster's drag coefficient grows as it slows through Mach 5 towards Mach 2, and its legs come out just before the peak.

The upper stage breaks the theory in the other direction. Entering at 7.40 km/s and 1.5°, the formula gives 3.8 g at 42 km. The stage feels 8.5 g at 32 km. An entry that starts level is not a straight line: as the air takes the speed away, the curvature of the path stops holding it up, gravity bends it down, and by the time of the peak the stage is descending at 5.6°. An orbital entry steepens itself, and that makes its peak remarkably insensitive to the angle it starts at. Integrating the full equations above for the upper stage's $\beta$ from 7.8 km/s, an entry at 0.5° peaks at 8.9 g and one at 2° at 9.8 g; only steeper than that does the peak climb quickly — 11.6 g at 3°, 16.7 g at 5°. Soyuz capsules that have fallen back to a ballistic entry after a guidance fault have pulled 8 g or so on the way down; the normal lifting entry, which holds the capsule up in thinner air for longer, keeps it to about 4.

The theory is still the right way to think. It says which way everything moves: the steepness and the speed set the load, the ballistic coefficient sets the altitude and the dynamic pressure. For the numbers, integrate.

Terminal speed

Once the pulse has passed, the vehicle slows until drag balances its weight. That is its terminal speed,

and it depends only on the ballistic coefficient and the local density — not on how fast or how steeply the vehicle arrived. Every entry, whatever its start, ends up near the same curve. It falls with height, because the air thickens: a vehicle near its terminal speed is still slowing as it descends. The hop demo shows this cleanly. The Hopper coasts over the top at 19.7 km, falls tail-first, and is at its fastest, 265 m/s, at 14 km; by the time it lights its landing burn, less than 400 m above the pad, it has slowed to 151 m/s without any help from its engine.

The terminal speed is also what the landing burn has to stop. Aster's booster, with its legs out and 32 t aboard, has a subsonic ballistic coefficient of about 2,050 kg/m², a terminal speed of 200 m/s at 2 km, and is doing 233 m/s there, still decelerating at 1.3 g. The upper stage, lighter behind the same diameter, is at 147 m/s by 1 km. Of the 7.4 km/s it arrived with, the air has taken away 98 %. Powered descent guidance is about the last 2 %.

Heating

The air in front of a hypersonic vehicle is stopped by a shock wave and heated, at the stagnation point, to thousands of kelvin. Most of that heat stays in the air and is carried away round the body. A thin boundary layer lets some of it through to the wall. For the stagnation point, Sutton and Graves fitted the boundary-layer solutions into a simple form,

with $r_n$ the nose radius and $k=1.7415\times 10^{-4}$ in SI units for Earth air. The cube of the speed is the reason entry is hard: halve the speed and the heat flux falls eightfold. The nose radius under the square root is the other half of Allen and Eggers' work. A blunt body stands its shock off further, leaves more of the heat in the air, and receives less of it: the heating falls as the body gets blunter, the opposite of what a streamlined intuition suggests. That is why every entry vehicle is blunt, and why a rocket stage comes home engines-first. The simulator gives a stage flying base-first an effective nose radius of 1.2 times its diameter, 4.39 m for Aster, and a pointed nose a fifth of its diameter, which takes 2.4 times the heat flux at the same speed.

Along the straight-line solution, $\sqrt{\rho }{v}^3$ peaks where the density is a third of the deceleration peak's, $\rho =\beta \sin \gamma /3H$ — about 7.9 km higher up, $H\ln 3$ — with the speed still at $v_E{e}^{-1/6}=0.85v_E$:

The vehicle is hottest before it is most heavily loaded. Unlike the peak deceleration, the peak heating does depend on $\beta$: a denser vehicle goes deeper, into thicker air, while still fast. Integrating the flux over the whole entry gives the heat load per square metre,

which runs the opposite way in the angle: a steep entry is short and sharp, a shallow one gentle and long, and the long one takes in more heat altogether. The upper stage shows the order. Its heating peaks at 80 W/cm² at 44.9 km, while it is still doing 5.92 km/s, 80 % of its entry speed; its deceleration peaks 12.8 km lower, at 32.1 km. The theory, for the record, puts the heating peak at 70 W/cm² and 50 km, and the whole entry's heat load at 130 MJ/m².

That load is far more than the heat shield could soak up and survive. The simulator models it as a single mass of skin, 45,000 J/m²K, which warms as heat comes in and cools by radiating. At 1,800 K a surface with an emissivity of 0.85 radiates 51 W/cm², most of what arrives at the peak. On the default mission the shield reaches 1,805 K at 36.9 km, below the flux peak — a heat sink lags its source — and 595 K short of its 2,400 K limit.

The figure starts on the booster as it would fall with nothing done, from 100 km. The integration peaks at 11.2 g and 329 kPa at 11.6 km; the straight-line theory, dashed, peaks near the same height at only 4.6 g, because it has no gravity to steepen the fall. Now choose the upper stage from orbit, and switch the right panel to heating: the peak, 92 W/cm² at 44.5 km, comes well above the deceleration's, 9.2 g at 31.9 km. (The figure's Earth does not turn, so the stage meets its air at the full 7.8 km/s, and its heating comes out about 15 % above the simulator's.) Move the angle between 0.5° and 2° and the peak deceleration hardly moves; move it to 6° and it passes the simulator's 15 g limit. Drag $\beta$ down to 300 and the whole pulse lifts by 7 km, much gentler in heating and almost as hard in deceleration. Then choose the booster after its entry burn, starting at 35 km and 0.95 km/s: 3.5 g, 62 kPa and 1.9 W/cm².

Why a booster lights its engines before the thick air

A booster coming back is not in the upper stage's position. It is slower, and it still has propellant. At 55 km on the way down, where the air is starting to matter — 1.5 kPa of dynamic pressure, a tenth of a g of deceleration — the reference program lights three of its nine engines, points them into the oncoming flow, and burns for twenty seconds. The entry burn takes the booster from 2.20 km/s to 0.96 km/s by 35 km, and uses 18.1 t of propellant: two thirds of everything it kept for the return.

Here is what it buys, on the default mission, against the same booster flown with the burn switched off:

The dynamic pressure is the heart of it. The ascent went through max-Q at 31.5 kPa; without the burn the booster comes down through ten times that. The structural limit the simulator enforces is on the bending load $q\alpha$, 250 kPa·°, so at 310 kPa the stage can fly at most 0.8° off the relative wind before it breaks up. The grid fins steer the booster by tilting it off the wind; at that dynamic pressure the reference program allows them half a degree. The booster falls through the densest part of its descent all but unable to steer, and it comes down 28 m from the centre of the deck. What it cannot correct there it has to correct in its landing burn, which starts six kilometres up at Mach 2 and lasts 37 s.

The heating is the other half. On a real booster the engines, their plumbing and the wiring round them are what the heat reaches first, and the entry burn is usually described as protecting them. The simulator treats a bare engine bay as one lumped skin that it destroys above 900 K. At Aster's 2.2 km/s the bay survives without the burn, at 499 K, so in the simulator the load is what binds. Arrive faster and heating catches up: the figure below finds the 15 g limit reached, with no burn, at 2.7 km/s through 55 km, and the 900 K limit at 3.7 km/s.

The burn works in two ways. It removes speed before the air can, which is where the propellant goes. And while it burns, the exhaust plume pushes the shock ahead of the stage and fills the region behind it, and the air's drag and heating almost vanish. The simulator models both: the axial drag coefficient is divided by $1+\frac{1}{2}{C}_T$, with $C_T$ the thrust over $q$ times the frontal area, and the heat flux is cut to 0.3 of its value. At 55 km, where $C_T$ is about 180, the air's push on the stage all but disappears; the booster is braking on its engines alone, and the hardest push of the whole descent is their own — 8.9 g as the burn ends, when three engines are pushing a stage that is 18 t lighter. Entry, descent and landing on Mars makes the same plume the centre of a whole landing strategy.

It is also a trade of propellant for structural margin, because the air would have done the work for nothing. Without the burn, the landing burn took 10.1 t and the booster touched down with 11.3 t left; with it, entry and landing together took 21.6 t and it touched down with 0.4 t. The air would have done 11.5 t's worth of braking for free. The booster pays for its margins instead.

Start at the reference program's cut, 950 m/s: 18.2 t, a peak of 58 kPa at 13 km and an engine bay at 320 K — the simulator's flight gave 18.1 t, 61 kPa and 321 K. Slide the cut up to 1,400 m/s and nearly a third of the propellant comes back — the burn takes 12.7 t — for twice the dynamic pressure, 114 kPa. Slide it to "no burn" and the path follows the dashed line through the 300 kPa contour. Then leave the cut at 950 m/s and raise the arrival speed. At 2.7 km/s a booster with no burn would pass 15 g; at 3.7 km/s its bay would pass 900 K. With the burn it is fine, until the propellant runs out: Aster's booster has 22 t left at 55 km, enough for the reference cut from up to about 2.5 km/s; above that the tanks run dry first.

The grid fins

Four grid fins sit at the top of the booster — the back, when it falls engines-first — like the feathers on an arrow. They keep it pointed into the flow, add drag, and steer. Tilt the stage off the relative wind and its body and fins push it sideways, towards the side its nose leans to; the program uses that to walk its landing point onto the ship. Lattice fins have a known weakness near Mach 1: the cells choke, and in the simulator their normal-force slope drops from 2.6 per radian to 1.4, while their drag rises to 1.6 times the subsonic value between Mach 1 and 1.2. How the program steers with them is in Coming back down.

Suborbital and orbital

Put the two entries of the default mission side by side:

The booster could never afford to leave its entry to a heat shield: it has none, and at 2 km/s it does not need one if it spends propellant instead. The upper stage could never afford the propellant: slowing from 7.4 km/s on its engines would take about ten times its own mass in propellant. It carries a shield instead — 699 kg of its 8.45 t — and enters in a way the shield can take. A booster staged much faster, on a heavier mission or a bigger rocket, is pushed towards the upper stage's side of the table: a longer entry burn, or a shield, or both.

In Vivapse

The air is src/sim/atmosphere.ts: the 1976 US Standard Atmosphere, modified by the day's weather, co-rotating with the Earth. The aerodynamic tables and the Sutton–Graves constant, K_SG = 1.7415e-4, are in src/sim/aero.ts, with the limits the simulator enforces: LIMITS = { qAlpha: 250000, g: 15 }. The heating itself is in src/sim/vehicle.ts. It takes the flux at the effective nose radius, reduces it to the share a hot wall receives, $(h_0-c_p{T}_w)/h_0$ with $h_0$ the total enthalpy of the flow, cuts it to 0.3 while engines fire into the flow, and warms two lumped skins: the base and the forward end. A base behind a heat shield holds 45,000 J/m²K and is destroyed above 2,400 K; a bare engine bay holds 15,000 and fails above 900 K. Both radiate at an emissivity of 0.85. The physics audit checked the constant, the radii, the hot-wall term and the plume factor (finding T-1). What is not modelled is ablation: the shield is a heat sink that radiates, not a surface that chars and recedes (T-2). The structural limits are more generous than a real stage's, which would be built for 5–8 g, and are kept that way on purpose (fidelity and its limits).

A program sees all of it. fc.dynamicPressure, fc.gForce and fc.qAlpha are the loads; fc.heatFlux is the stagnation-point flux in W/m²; fc.baseTemperature and fc.noseTemperature are the two skins, and fc.maxBaseTemperature the limit that applies — 900 K or 2,400 K. For planning, fc.predict({ entryBurn, landingBurn }) flies a whole return ahead: an air-retrograde entry burn lit descending through entryBurn.altitude and cut at untilSpeed, then the latest landing burn that still stops at the surface.

The reference program, src/programs/full-mission.js, lights its entry burn at ENTRY_ALT = 55e3 on three engines and plans it around EB_PLAN = { altitude: ENTRY_ALT, engines: 3, throttle: 1, untilSpeed: 1150 }. On real sites it then budgets it: every re-plan of the return nudges the planned cut speed until the planned landing burn would leave 0.075 % of the stage's load — 300 kg on Aster — and the burn is cut on target, never above ENTRY_CUT_SPEED = 1400 m/s and always by ENTRY_MIN_SPEED = 600. The upper stage has no entry burn. It enters tail-first on fc.airRetrograde, behind its shield, and leaves the rest to the air.

Try it

Choose the Aster preset and the Full mission — orbit & return from anywhere example, and keep the default mission: Starbase, random weather, seed 7. Lock the seed in the Mission panel, or every launch rolls a new day. Add a log to the booster's program, as the first lines inside function booster(fc) {:

if (fc.verticalSpeed < 0 && fc.altitude < 60e3 && fc.t >= (fc.mem.logAt || 0)) {
  fc.mem.logAt = fc.t + 2;
  fc.log(fc.mem.phase, (fc.altitude / 1000).toFixed(1), 'km', fc.surfaceSpeed.toFixed(0), 'm/s',
    'q', (fc.dynamicPressure / 1000).toFixed(0), 'kPa', fc.gForce.toFixed(1), 'g',
    (fc.heatFlux / 1e4).toFixed(1), 'W/cm²', fc.baseTemperature.toFixed(0), 'K');
}

Every two seconds on the way down the console reports the booster's phase and its loads. At 55 km it reads coast 55.0 km 2199 m/s q 1 kPa 0.1 g 2.2 W/cm² 300 K and the entry burn starts. Near its end, entryBurn 36.5 km 1110 m/s q 5 kPa 8.5 g 0.3 W/cm² 306 K: the load is the engines', and the heating is a seventh of what it was. The air's own peak comes at descent 15.0 km 734 m/s q 61 kPa 3.6 g 1.6 W/cm² 320 K. The booster lands on the drone ship, bullseye.

Now find the line

const ENTRY_MIN_SPEED = 600;        // m/s — and always stop it by this speed

and change 600 to 3000. The booster is slower than that at 55 km, so the program skips the entry burn and goes straight to its descent. Fly again. The heating climbs through descent 20.1 km 1983 m/s q 188 kPa 6.4 g 20.0 W/cm² 429 K, and the dynamic pressure peaks at descent 13.2 km 1471 m/s q 309 kPa 10.8 g 14.1 W/cm² 483 K. The booster survives, lights its landing burn at 6.8 km and 703 m/s, and comes down on the deck 28 m from its centre, graded hard, with 11 t of propellant it did not need. Put the line back before flying anything else.

What carries forward

The entry burn is one of three burns a returning booster makes, and the propellant for all three has to be held back from the climb, where it costs payload. Bringing the booster home follows the whole return, and what each way of doing it costs. The dynamic pressure of this lesson is the same quantity as the ascent's in Dynamic pressure and max-Q, met from the other side; and at Mars, where the air is too thin to do the whole job, entry, descent and landing is the same theory with the opposite problem.