The speed that is left over

Escape is a threshold, and a threshold is a strange thing to aim at — the more so since the threshold itself is not the clean ratio it looks once the mass being escaped from is spread out. A vehicle that leaves at exactly escape speed is on a parabola, the one conic with no scale in it: it arrives at infinity with precisely zero speed, having taken infinitely long to get there. Nothing has ever been launched onto one and nothing ever will be.

What a mission is actually specified by is the speed the vehicle still has when the planet’s pull has run out — the hyperbolic excess, $v_\infty$ — and the relation between that and what the launch has to supply is the whole of this essay.

Why the two speeds do not add #

The temptation is to suppose that arriving at infinity with three kilometres a second requires three kilometres a second more than escaping. It requires four hundred metres a second, and the reason is that energies add and speeds do not.

The specific energy of an orbit is $v^2/2 - GM/r$, and escape is the case where it is zero. A vehicle that leaves with a residual speed $v_\infty$ has specific energy $v_\infty^2/2$, so at the departure radius

$$\frac{v_{\rm dep}^2}{2} - \frac{GM}{r} = \frac{v_\infty^2}{2} \quad\Longrightarrow\quad v_{\rm dep} = \sqrt{\frac{2GM}{r} + v_\infty^2} = \sqrt{v_{\rm esc}^2 + v_\infty^2}.$$

The two speeds combine as the legs of a right triangle. Adding a small square to a large one barely moves the root: from a low Earth orbit the escape speed is 11.0 kilometres a second, and adding three gives $\sqrt{121 + 9} = 11.4$.

That is why the quantity on a launch contract is not a speed but its square. $C_3 = v_\infty^2$ — the characteristic energy, twice the specific orbital energy, in kilometres squared per second squared — and it is the natural currency because it is what adds.

The exchange rate, and where it collapses #

Differentiating the departure relation gives the price of one more kilometre a second at infinity:

$$\frac{\mathrm{d}v_{\rm dep}}{\mathrm{d}v_\infty} = \frac{v_\infty}{v_{\rm dep}},$$

which is small when the excess is small and approaches one when it is large.

The steepness at the left is the reason the outer solar system is reached the way it is. A gravity assist that reduces the required departure excess by one kilometre a second is worth several hundred metres a second of propellant — far more than the same kilometre would be worth if the mission were already at fourteen. Stealing speed from a planet is valuable in inverse proportion to how much has already been spent.

It is also the reason a mission’s difficulty is so badly conveyed by its destination. Mars asks for $C_3$ of about 9 and Jupiter for about 80, which sounds like nine times as hard; the departure costs are 3.63 and 6.40 kilometres a second, which is a factor of 1.8. And a launch vehicle’s performance is quoted as a mass against $C_3$ rather than against a speed, because it is the energy that the upper stage’s own propellant load converts into linearly.

Four destinations, priced #

The abstraction is easier to hold with the numbers in it, and the numbers are short.

Leaving a two-hundred-kilometre circular orbit takes 3.22 kilometres a second to reach escape with nothing left over. Venus, on a minimum-energy transfer, asks for a characteristic energy of about 6.8 — an excess of 2.6 kilometres a second — and costs 3.52. Mars asks for about 8.9, an excess of 3.0, and costs 3.63. Jupiter direct asks for 80, an excess of 8.9, and costs 6.40. The highest-energy launch ever flown asked for 158, an excess of 12.6, and cost about 8.6.

So the four destinations span a factor of twenty-three in the energy and a factor of 2.4 in the velocity change. The first two, which are the most-flown missions in the catalogue, differ from one another by a hundred metres a second — about one per cent of a launch vehicle’s total capability — which is why the choice between them is made on almost anything other than energy.

The comparison also prices the alternative to paying. A mission to Jupiter that accepts a flyby of Venus and two of Earth can depart at an excess of about 4 rather than 8.9, which drops the cost from 6.40 to about 3.87: two and a half kilometres a second saved, for six years of flight time. That trade is the entire reason the outer planets have been visited by vehicles small enough to have been launched at all, and its size is read straight off the flatness of the cost curve.

A negative characteristic energy #

That last point is worth a paragraph of its own, because the quantity is usually introduced as though it only existed above escape.

$C_3$ is twice the specific orbital energy, and specific orbital energy is defined for every orbit. A bound orbit has $C_3 = -GM/a$, which is negative; a parabola has zero; a hyperbola has $v_\infty^2$. So the same number describes a lunar transfer at about $-1.9$, a parking orbit at $-60$, a geostationary orbit at $-9.4$ and a departure to Jupiter at $+80$, and the whole of orbital mechanics near a planet is one axis with a zero on it.

The reason this matters practically is that a launch vehicle’s performance chart runs continuously across that zero. A stage that can put six tonnes on a lunar trajectory and four on a Mars one is not doing two different things; it is delivering the same quantity of energy per kilogram to two points on one line, and the curve through them is smooth. A threshold that looks like a qualitative change in the physics is a single value of a continuous parameter, and the only thing that changes at it is which conic section the trajectory is called.

A parking orbit is a choice, and it is not free #

Everything above is priced from a circular orbit two hundred kilometres up, and the number depends on that choice more than it looks.

The trade is real and it goes both ways. A high parking orbit makes the departure burn small, which is useful when the departure stage is small; a low one exploits the well, which is useful when the departure stage is large. What settles it is usually neither — it is that a vehicle cannot loiter in a low orbit for long, so the parking orbit’s altitude is chosen by how long the mission needs to wait for its departure geometry.

The direction, and the part that is not a number #

The excess is not only a speed. It is a vector: the departure asymptote has a direction, and the mission needs that direction to point where the planet will be.

Two consequences follow, and both are what turn a scalar into a mission design.

The departure hyperbola’s asymptote must lie in the plane of the target’s orbit, which constrains where round the parking orbit the burn happens and therefore when it can happen. That is the reason a departure window is a few tens of minutes long rather than a day, even when the interplanetary window is a month.

And the excess vector adds to the planet’s own motion. The Earth carries 29.8 kilometres a second round the Sun, so a departure with $v_\infty = 3$ aimed forwards gives a heliocentric speed of 32.8 and aimed backwards 26.8 — which is the difference between going out and going in, from the same launch energy. One trajectory stitched from three two-body problems is how that bookkeeping is done, and the excess is precisely the quantity handed from one conic to the next.

What was actually measured #

Nothing on these curves is an observation; they are two lines of algebra with a gravitational parameter in them. What is measured is $GM$ for the Earth, which is known to nine significant figures from decades of laser ranging to satellites, and the radius of the parking orbit, which the vehicle’s own navigation supplies.

What is specified rather than measured is the required $C_3$, and it comes from solving the interplanetary problem backwards: given a departure date and an arrival date, the trajectory between two positions is determined, and the excess at each end falls out of it. So a mission’s $C_3$ is not a property of the destination but of a pair of dates, and the plots that show it against those dates are the ones a launch is actually scheduled from.

The number that gets checked against reality is the one at the other end. A departure’s excess is verified by tracking: after the burn, the spacecraft’s Doppler shift and range are measured over hours to days, an orbit is fitted, and the energy is compared with the intended one. Errors of a few metres a second in a departure burn are normal and are corrected within days, when correcting them is cheap — the same arithmetic as everything above, run in reverse, since a small error in $v_\infty$ made early costs little to remove and a great deal to leave.

The record, and why it still needed a flyby #

The highest characteristic energy ever launched was about 158 kilometres squared per second squared, achieved by a three-stage vehicle with a solid upper stage, delivering a 478-kilogram spacecraft onto a Jupiter-crossing trajectory in 2006.

Three things about that launch illustrate the whole argument.

The mass was tiny and the stage was enormous. A launch vehicle’s capability falls steeply with characteristic energy, because the quantity that has to be added scales with the energy while the propellant needed to add it scales exponentially through the rocket equation. Pushing from a Mars-class $C_3$ of 9 to one of 158 costs about five kilometres a second, which on a hydrogen upper stage is a factor of four in mass fraction.

It escaped the solar system at launch and could not reach its target. At 16.3 kilometres a second relative to Earth it was already unbound from the Sun, and it still needed a flyby of Jupiter to arrive at Pluto within a working lifetime — the flyby adding about four kilometres a second heliocentric and cutting five years off a journey of nine.

And the record has stood. No launch since has asked for more, not because vehicles have not improved but because nothing has needed to: the exchange-rate curve says that buying speed this way is the most expensive option available, and every mission since has bought it from a planet instead.

The one exception is the mission going the other way. Reaching the Sun’s immediate neighbourhood requires removing the Earth’s 29.8 kilometres a second of orbital motion, which is far more than any launch can supply directly — so that mission launched at a $C_3$ of 154, nearly the record, aimed backwards, and then used seven flybys of Venus to walk its perihelion inward. The most demanding departure in the catalogue and the most flyby-dependent trajectory in it are the same mission, which is the clearest possible statement of where the cost lives.

Where the model stops #

The burn is not impulsive. A departure stage burns for several minutes, during which the vehicle travels a substantial arc and climbs measurably, so the effective escape speed it is working against changes through the burn. The loss relative to an impulsive calculation is a few per cent and it is called a gravity loss; it is the reason a departure is flown from a low parking orbit with a high-thrust stage rather than spiralled out.

There is no infinity. The excess is defined as the speed at infinite distance from the Earth, and the Earth’s domain ends where the Sun’s begins — about a million kilometres out, or 145 Earth radii. The asymptotic speed at that distance is within a fraction of a per cent of the true limit for any interesting excess, which is why the approximation holds, and it is a approximation.

The Sun is missing. Every curve here treats the Earth as the only mass. It is not: a departing vehicle is in the Sun’s field throughout, and the two-body hyperbola is a stand-in for a trajectory that is genuinely three-body. What the patched-conic approximation hides is a discontinuity at the boundary, and real trajectories are integrated rather than patched.

And a low excess is not the same as a cheap mission. The departure is one of several burns, and a trajectory with a small $C_3$ often has a large arrival cost or a long flight time. The quantity minimised is never the departure energy alone.

The arrival is the same triangle, and it is worse #

Everything above prices a departure, and the identical algebra prices the other end with the sign of the inequality reversed.

A spacecraft approaching a planet on a hyperbola arrives with an excess it cannot shed by coasting. Its speed at periapsis is $\sqrt{v_{\rm esc}^2 + v_\infty^2}$ at that radius — the same right triangle — and to be captured into a bound orbit it has to lose whatever takes it below escape speed there. So an arrival burn is

$$\Delta v = \sqrt{v_{\rm esc}^2 + v_\infty^2} - v_{\rm target},$$

and the quadratic combination now works against the mission rather than for it: a large excess, which was cheap to buy, is expensive to give back.

The asymmetry is sharp. At Mars, arriving with an excess of 2.6 kilometres a second and capturing into a loose orbit costs about 1.0; arriving with an excess of 5 costs about 1.9. The saving at the departure end from accepting a faster transfer is smaller than the cost at the arrival end of dealing with it, so the trajectory that is cheapest to leave on is rarely the one that is cheapest to arrive on, and the minimum of the sum sits somewhere neither half would have chosen.

The route out of that is to make the atmosphere do the work. A capture flown through the upper atmosphere rather than on the engine removes the arrival excess for the cost of a heat shield, and the trade between a shield’s mass and a propellant load’s is what decides which missions use it. It is also the reason Venus and Mars orbiters are cheap and Mercury orbiters are not: the planets with air can arrive fast and the ones without cannot.

The generalisation #

The shape to take away is that a threshold is almost never the quantity a problem is about, and the quantity it is about usually behaves differently from it.

Escape speed is a boundary in a one-dimensional parameter, and everything interesting happens on one side of it. The useful variable is the distance past the boundary, and that variable turned out not to be a speed at all — it combines quadratically, so the natural coordinate is its square, and an intuition built on adding speeds gives answers wrong by a factor of seven at the low end.

Find the quantity that adds, and use it. Energies add where speeds do not; magnitudes add where magnitudes and directions do not; optical depths add where transmissions multiply; and in each case the additive variable is the one whose figure is a straight line and whose errors combine simply. The whole of the reason $C_3$ rather than $v_\infty$ appears on a launch vehicle’s performance chart is that a chart against $C_3$ is nearly straight and a chart against $v_\infty$ is not.

There is a second reading, about where value sits. The exchange-rate curve says that effort spent reducing a requirement is worth most when the requirement is already small — which is the opposite of the usual intuition that there is more to gain where there is more to give. A flyby that shaves a kilometre a second off a departure excess of two is worth ten times one that shaves the same kilometre off an excess of fourteen. The marginal value of an improvement is a property of where the system already is, and reading it off the derivative rather than off the size of the change is the habit this is for.

Still open: the arrival, which is the same problem inverted #

What comes next is what the excess does at the other end. A spacecraft arriving at a planet with a given $v_\infty$ has an energy it cannot lose without a burn, and the size of that burn is set by the same right triangle read backwards — which is why arriving is generally more expensive than leaving, and why an orbiter costs so much more than a flyby.

Beside it lies the case where the excess is delivered by the target rather than paid for: an approach at low relative speed, where the two-body hyperbola degenerates and the capture becomes a three-body problem with no impulse in it at all. That is the regime the low-energy transfers work in, and it is where a threshold stops being the wrong variable and starts being the whole of the answer.