A corridor a degree and a half wide

Everything that comes back from orbit arrives with the same problem: it has kinetic energy of about 32 megajoules per kilogram, which is fifteen times the energy that would melt it, and the only place to put that energy is the air. Every orbit is a conic until the atmosphere is reached, and then for four minutes it is not.

The question that decides whether the vehicle survives is not how much energy there is — that is fixed by where it came from — but over what interval it is shed. Too long, and the vehicle never slows enough to be captured. Too short, and the deceleration or the heating exceeds what the structure or its occupants can take. The interval is set by the angle at which the vehicle enters, and the band of acceptable angles is narrow.

The result that has no vehicle in it #

In 1953 H. Julian Allen and A. J. Eggers, at the NACA’s Ames laboratory, integrated a ballistic entry into an exponential atmosphere and found something that reads like an error the first time.

Assume the trajectory is a straight line at flight-path angle $\gamma$ below the horizontal, take the atmosphere as $\rho = \rho_0e^{-h/H}$, and write the deceleration as $\rho v^2/2\beta$ with $\beta = m/C_DA$ the ballistic coefficient. Then $dv/dh$ integrates in closed form,

$$v(h) = v_e\exp\left[-\frac{\rho(h)H}{2\beta\sin\gamma}\right],$$

and differentiating for the maximum gives

$$a_{\max} = \frac{v_e^2\sin\gamma}{2eH}.$$

There is no $\beta$ in it. The mass, the drag coefficient and the frontal area — everything about the vehicle — have cancelled. A feather and a cannonball entering at the same speed and angle experience the same peak deceleration; what differs is the altitude at which they experience it.

Why the peak is where it is #

The mechanism behind the cancellation is worth seeing, because it explains the shape of every curve on the previous figure.

Deceleration is proportional to $\rho v^2$. On the way down, $\rho$ is rising exponentially and $v$ is falling — slowly at first, because the air is thin, then fast. The product peaks where the fractional rate of increase of $\rho$ equals twice the fractional rate of decrease of $v$, and that condition works out to $\rho H/(\beta\sin\gamma) = 1$: the vehicle has just traversed one “scale mass” of atmosphere.

A vehicle with a large $\beta$ pushes further down before that condition is met, and meets it in denser air at higher speed; a vehicle with a small $\beta$ meets it higher up in thinner air at higher speed still. The two effects cancel exactly, and the cancellation is exact because the atmosphere is exponential and for no other reason.

Both edges of the corridor #

The steep edge is a load limit and is set by whatever is inside. Apollo’s design limit was about 12 g; an uncrewed capsule can take 50; a sample-return capsule entering at 12.8 km/s pulled 60. The Soyuz ballistic-abort mode reaches 8 to 9 g and is survivable but unpleasant.

The shallow edge is skip-out, and it exists only for entries at more than escape speed relative to the Earth.

Blunt, on purpose #

The other half of Allen and Eggers’ 1953 report is the recommendation that came out of it, and it inverted the design of every re-entry body that had been proposed until then.

The intuition of the time was that a slender, sharp-nosed shape would minimise drag and therefore minimise heating. It minimises drag and maximises heating, because low drag means the vehicle carries its speed down into dense air and sheds it there, where the heat transfer is efficient.

A blunt body does the opposite. Its high drag decelerates it at high altitude in thin air, and its detached bow shock stands well ahead of the surface, so most of the kinetic energy goes into heating the shock layer — air that then flows away — rather than the vehicle.

The heating law itself is a fit, and it omits radiative heating from the shock layer, which becomes comparable to convective heating above about 10 km/s and dominant above 12. That is exactly the regime a lunar return lives in, so the figure above is an underestimate at the very speeds the opening figure is about.

What lift does #

Every trajectory above is ballistic: no lift, no control, and a corridor of a fraction of a degree. Adding even a small lift-to-drag ratio changes the problem completely.

Apollo’s command module flew at $L/D \approx 0.3$ by offsetting its centre of mass so that it trimmed at an angle of attack, and it rolled to point the lift vector up or down. Lift up flattens the trajectory and reduces the peak load; lift down steepens it and increases the drag. The guidance system used that to fly within the corridor rather than falling through it, and the effect was to widen the acceptable entry angle from under a degree to about two and a half, and to reduce the landing dispersion from hundreds of kilometres to tens.

The Shuttle took it much further at $L/D \approx 1$, flying a controlled deceleration at nearly constant heating rate for twenty minutes and reaching a runway. Lift also converts the problem from one of aiming to one of steering, which is why an entry with lift can absorb a much larger error in the orbit determination that preceded it.

What was actually flown #

The numbers above are model numbers. It is worth setting a few measured ones beside them, because entry is one of the few parts of spaceflight where the telemetry is public and the model is testable against it.

Apollo 4, 1967. A deliberate test of the lunar-return entry at 11.14 km/s, uncrewed. Peak deceleration 7.3 g, peak heat rate about 425 W/cm², total heat load 2.3 kJ/cm². The guidance flew a double-dip profile — decelerate, pull up out of the atmosphere briefly, re-enter — which is what lift makes possible and which halves the peak load compared with a single pass.

Stardust, 2006. The fastest entry ever attempted by a returning vehicle, 12.9 km/s, uncrewed and ballistic. Peak deceleration 34 g, peak heat rate about 1,000 W/cm². The corridor was about $0.08°$ wide at half a degree of nominal, and the capsule landed within 8 km of its target.

Soyuz TMA-11, 2008. A separation failure put the vehicle into ballistic entry rather than the normal lifting one — a steeper effective angle with no lift to flatten it. Peak deceleration 8.2 g against a nominal 4, and a landing 475 km short of the target. It is a clean illustration of what the corridor costs: the vehicle was inside the survivable band, and the entire difference between a nominal entry and a frightening one was the loss of a control authority worth about two and a half degrees.

The pattern across all three is that the peak decelerations scale as the model says — with $v_e^2\sin\gamma$ — and the heat loads do not, running higher than a convective-only law gives at the highest speeds. That is the radiative term the heating figure above omits, and it is why Stardust’s shield was a different material from Apollo’s.

The four minutes with no radio #

There is a consequence of the shock layer that has nothing to do with heating and everything to do with operations, and it follows from the same physics.

Air crossing the bow shock at eleven kilometres a second arrives at several thousand kelvin, which is enough to ionise it. The vehicle is then wrapped in a plasma, and a plasma reflects radio waves below its own plasma frequency — the frequency at which the electrons can respond collectively, which depends only on the square root of the electron density. For the electron densities in an entry shock layer that frequency is of order a few gigahertz, so an S-band link at two gigahertz is simply reflected and the vehicle is unreachable.

The blackout lasts from the onset of appreciable heating to well past the peak deceleration — about four minutes for Apollo, and longer for a steeper or faster entry. It arrives at exactly the moment when the vehicle is doing the one thing that cannot be repeated, and for the whole of the Mercury, Gemini and Apollo programmes it was accepted as unavoidable.

Two things get around it, and both are worth stating because they show what the constraint actually is. Raising the frequency above the plasma frequency works, and a link in the tens of gigahertz penetrates where an S-band one does not — at the cost of atmospheric attenuation and much tighter pointing. Alternatively, the plasma is not uniform: it is densest ahead of the vehicle and thinnest in the wake behind it, so a link aimed backwards and upwards, to a relay satellite above rather than to a station on the ground, passes through the thin part. That is how the Shuttle kept communications through most of its entries, and it is why the blackout is now a solved problem for vehicles that have a relay to talk to and an unsolved one for those that do not.

The measurement the corridor is aimed with is therefore made before the entry and confirmed after it, and nothing about the trajectory can be adjusted in between. An entry is the one manoeuvre in spaceflight that is committed to entirely in advance.

What is aimed, and from where #

The corridor is stated as an angle at the entry interface, and nobody steers at the entry interface — by then the vehicle is on a ballistic arc with no time to change anything.

What is aimed is the periapsis of the incoming trajectory, days out, and the conversion is unforgiving. For a lunar return, a $0.78°$ corridor in flight-path angle corresponds to roughly 30 kilometres of periapsis altitude, and a mid-course correction made at 200,000 km has a lever arm such that a velocity error of a few centimetres per second moves the periapsis by that much.

Where the same arithmetic gets harder #

Everything above is an Earth entry, and it is worth putting the same expressions to a planet where they give a much less comfortable answer, because it explains a limitation nobody expected.

Mars has about one per cent of Earth’s surface density and a scale height half as large again. The scale height helps: peak deceleration goes as $v_e^2\sin\gamma/2eH$, so a larger $H$ is a gentler entry and a wider corridor. The density does not. The vehicle must shed its speed in air that is a hundredth as thick, which it can only do by descending further before the deceleration peaks — and on Mars there is not much further to descend.

The consequence is a hard limit that has nothing to do with heating. A blunt entry vehicle decelerating in the Martian atmosphere reaches its terminal velocity while still supersonic and still at an altitude of several kilometres, and it stays supersonic all the way down unless something else intervenes. Parachutes are the something else, and a parachute deployed above about Mach 2 is unreliable — the canopy oscillates in the wake and can fail structurally.

So a Mars entry has to arrive slow enough and high enough to deploy a parachute in a narrow band of Mach number and altitude, and the band closes as the vehicle gets heavier. A heavier vehicle at fixed size has a larger ballistic coefficient, descends further before decelerating, and reaches parachute conditions lower down — until, somewhere above a tonne or two of landed mass, there is no altitude left. Every Mars lander to date has been inside that limit, and every proposal to exceed it substantially requires either a much larger decelerator or supersonic retropropulsion, which is thrusting into a hypersonic flow and is a different problem again.

The same three parameters that make an Earth entry a targeting problem make a Mars entry a mass limit, and which of the two a planet presents depends entirely on the ratio of its atmospheric density to its gravity.

Why the corridor is narrow at all #

It is worth asking why the answer comes out at a fraction of a degree rather than at ten degrees, because nothing in the setup obviously demands it.

The narrowness comes from the exponential. Peak deceleration goes as $\sin\gamma$, which is nearly linear at small angles, so doubling the entry angle doubles the load — that alone would give a wide corridor. What tightens it is the other edge: whether the vehicle is captured depends on the total velocity lost, which is $v_e[1 - \exp(-\rho_p H/2\beta\sin\gamma)]$ at periapsis, and the argument of that exponential is inversely proportional to $\sin\gamma$. A small change in $\gamma$ is an exponentially large change in how much atmosphere is traversed.

So one edge moves linearly and the other exponentially, and the band between them is set by the ratio. Numerically: at 11 km/s, going from $5.3°$ to $6.0°$ — a thirteen per cent change in angle — takes the peak load from 4 g to 12, because the vehicle drops into air three times denser.

The corridor is narrow because an atmosphere is exponential, and the same property that makes an entry survivable at all is what makes it hard to aim. A planet with a gentler density gradient would be more forgiving in both directions and would require a longer, hotter entry.

Two consequences follow. First, corridor width scales inversely with entry speed roughly as $v_e^{-2}$, since the load limit is reached at a shallower angle the faster the arrival — which is why an interplanetary return is harder than a lunar one and a lunar one harder than an orbital one. Second, a larger scale height widens it: Mars has $H \approx 11$ km and a correspondingly more forgiving corridor, reached by a transfer whose arrival date is fixed years in advance, which is the one thing about entering Mars that is easier than entering Earth. Everything else about it is worse.

What the model leaves out #

Three things, and the first is deliberate.

The aerodynamics. Everything here treats the vehicle through one number, $\beta = m/C_DA$, and takes $C_D$ as constant. That is where hypersonic flow physics has been put — the shock structure, the chemistry of dissociating air, the boundary layer, the transition to turbulence — and it is a separate subject with its own regimes. The claim made here is only that the trajectory follows from a drag law and a scale height, and that claim is what Allen and Eggers established.

The atmosphere. A single exponential with $H = 7.2$ km is a good description over a limited altitude band and a poor one over the whole entry: the real scale height varies from about 6 km at the surface to 11 km in the thermosphere, and the density at 120 km varies by a factor of several with solar activity. The peak deceleration is insensitive to this because it occurs at 30 to 50 km where the atmosphere is well behaved; the skip-out boundary is not, because it is decided high up.

Ablation. A heat shield works by losing mass, and the mass loss changes $\beta$ during the entry — by 10 to 15 per cent for an Apollo-class shield.

Where the ladder goes next #

The rung above is aerocapture, which is the corridor problem run deliberately: enter a planet’s atmosphere on a hyperbolic approach, aim at exactly the periapsis that removes exactly the right amount of energy, and leave on a bound orbit having spent no propellant. Mars Global Surveyor and Mars Odyssey did the slow version of it — aerobraking over months, dipping repeatedly — and saved of order a kilometre per second each — a saving that, through the exponential in the rocket equation, is worth a large fraction of the launch mass. The single-pass version has never been flown, and the reason is the width of the corridor at a planet whose atmospheric density is known to a factor of two.