Stopping from 0.1c

How fast can starlight stop a sail?

A mirror of area $A$ held face-on at distance $r$ from a star of luminosity $L_\star$ feels $2L_\star A/(4\pi r^2 c)$: twice the momentum of the light it intercepts, because it sends the light back. Write that as $k\,m/r^2$ for a ship of mass $m$, with $k = 2L_\star A/(4\pi c\,m)$. Gravity pulls with $GM_\star m/r^2$. Both fall as $1/r^2$, so with the sail open the net force is $(k - GM_\star)\,m/r^2$ outward, and energy is conserved in the form

$$\tfrac12 v^2 + \frac{k - GM_\star}{r} = \tfrac12 v_\infty^2$$

A ship falling straight in stops where the left side has no kinetic energy left, at $r_{\min} = (k - GM_\star)/(\tfrac12 v_\infty^2)$. It must stop outside the closest distance the film survives. Take $r_{\min}$ as 5 Ember radii, where the light is 8.9 × 10⁵ W/m², 653 times sunlight at Earth and 3.3 times what the launch film carried. Solving for the mass per area $\sigma = m/A$:

$$\sigma \le \frac{2L_\star}{4\pi c\,\left(\tfrac12 v_\infty^2 r_{\min} + GM_\star\right)}$$

At $v_\infty = 0.1c$ this demands 0.031 milligrams per square metre for the whole ship. One layer of graphene is 0.77 mg/m², 25 times too heavy with no payload at all. Turned round, a bare graphene sheet can be stopped from 0.020 c, 6,031 km/s; the seed was moving 5.0 times faster. The energy goes into the reflected light, which leaves slightly bluer than it arrived.

Why does a magnetic sail fade?

A loop of radius $R$ carrying current $I$ reflects interstellar protons that enter its field. In a gas as thin as 0.1 per cm³ the ions cannot move together as a fluid around the loop, so each proton meets the field alone. Gros traced protons through the loop's field and fitted the reflecting area:

$$A(v) \approx 0.081\,\pi R^2 \ln^3\!\left(\frac{I}{\beta I_c}\right), \qquad \beta = v/c, \quad I_c \approx 1.55 \times 10^{6}\ \text{A}$$

The drag is the momentum the reflected protons carry off, of order $\rho\,A(v)\,v^2$ for a gas of mass density $\rho$. Since $A$ grows only as a logarithm cubed as the craft slows, the drag falls almost as $v^2$, and the time to reach speed $v$ grows as $1/v$: the brake never reaches zero. From Gros' solution, at the best current, $\beta_0 \times 4.2$ MA (421 kA at 0.1c), a craft of total mass $m$ coasts a stopping length

$$x_{\max} = \frac{ 3.1\,m}{m_p\,n\,\pi R^2}$$

and reaching 100 km/s, 0.3 percent of the start, takes 5.2 times $x_{\max}/v_0$. The catch is that $m$ includes the loop. The conductor weighs $2\pi R I$ times its mass per metre per ampere, and Gros doubles that for tethers and structure. A bigger loop shortens $x_{\max}$ as $1/R^2$ but adds mass in proportion to $R$. Solving for the smallest loop that brakes the one-tonne seed in a given time:

Brake time	2017 coated tape, 0.18 kg per m per MA	Advanced conductor, 0.050 kg per m per MA	Stopping length
30 years	987 t, radius 1,036 km	77 t, radius 291 km	0.57 ly
100 years	297 t, radius 311 km	24 t, radius 90 km	1.9 ly
300 years	100 t, radius 105 km	8.5 t, radius 32 km	5.7 ly

Even the 300-year brake with conductor no one has made weighs 8.5 times the seed, and it still leaves the last 100 km/s for something else. A 10 km loop, 2.6 t with advanced conductor, has a stopping length of 22.6 light-years, 0.81 of the whole crossing.

Why does the rocket close at 0.05c and not at 0.03c?

At a tenth of light speed the rocket equation uses rapidity, $\phi = \operatorname{artanh}\beta$ = 0.1003, in place of the speed:

$$R = \frac{m_0}{m_1} = \exp\!\left(\frac{\phi}{v_e/c}\right)$$

Propellant $P$ does not fly alone. The tanks weigh $f = 0.050$ of what they hold. The engine must turn $P/t$ kilograms a second into a jet of power $\tfrac12 (P/t) v_e^2$; at an assumed 1 × 10⁷ W of jet per kilogram of engine, it weighs $kP$ with $k = v_e^2/(2\,t\,P_{\text{spec}})$. The jet carries a share $\eta$ of the fusion energy of the fuel that burns; the rest, $(1-\eta)/\eta$ times the jet power, is heat, and a two-sided panel at temperature $T$ rejects $2\sigma T^4$ per square metre, so the radiator weighs $k_r P$ with $k_r = \frac{1-\eta}{\eta}\,\frac{v_e^2}{2t}\,\frac{a}{2\sigma T^4}$ for a panel of areal mass $a$. After the stop the stage still has to make the capture burn into Kindling's orbit, a second mass ratio $R_c$ = 1.0053 for 79.2 km/s, from Where you can live around a hole and a star. Its propellant is carried through the stop. With $A$ the mass left after both burns, the stage before the stop is $R\,R_c\,A$ plus the reserve, and

$$A = \frac{m_s}{1 - (k + k_r + f)(R - 1)R_c - f(R_c - 1)}$$

The denominator must stay positive: no stage can have a mass ratio above $1 + (1 - f(R_c - 1))/((k + k_r + f)R_c)$, whatever it carries. With these tanks, radiator and reserve that ceiling is 15.9 at 0.03c, below the 28.3 the stop needs, so a Daedalus-class exhaust cannot stop the seed in one stage. At 0.05 c the ceiling is 11.3 against a need of 7.44, and $P$ comes to 17,192 kg, with 14 kg more for the reserve.

The sail that launches this stage must push its mass too. The array keeps the beam focused only over a fixed distance, so a heavier stack needs a harder push, and the flux on the film rises with it. The design holds the flux at 200 suns by enlarging the sail to 8,524 m (Pushed by light has the beam geometry).

Why is an assist capped at twice the body's speed?

In the frame of the body passed, a ship arriving at speed $u$ leaves at speed $u$; the pass only turns the velocity. Take the pair's centre of mass, where the body moves at $\vec V$. A ship with velocity $\vec w$ has $\vec u = \vec w - \vec V$ relative to it, and after the pass

$$\vec w_{\text{out}} = \vec V + R(\delta)\,(\vec w - \vec V), \qquad w_{\text{out}} \ge u - V \ge w - 2V$$

by the triangle inequality twice. One pass lowers the speed by at most $2V$, reached only by being flung straight back. For Ember, $V$ = 82 km/s; for Vesper, which sits close to the centre of mass because it is 17 times heavier, $V$ = 4.8 km/s.

Can the engine be built?

No fusion device had produced net electricity by 2026, and deuterium with helium-3 needs a hotter plasma than the deuterium and tritium most experiments burn. Its advantage for a rocket is that most of the energy comes out in charged particles, a proton and a helium nucleus, which a magnetic nozzle can aim; deuterium and tritium put most of theirs into neutrons, which nothing can steer. The design asks two things beyond 2026 engineering: 0.32 of the reaction energy in the jet, and 1 × 10⁷ watts of jet per kilogram of engine. frontier The engine's mass scales inversely with the second figure, and the stage's propellant grows with it. For scale, Project Daedalus aimed at 100 megawatts per kilogram and about 40 in its flown profile (Long, 2016), a quarter of which is this design's figure. The jet's 0.76 share of the burning fuel's energy is what Razin and colleagues computed for a one-megawatt aneutronic engine (thrust power 0.91 megawatts of 1.19), and they put neutrons at 0.5 percent of the fusion power; the model assumes the stage absorbs every joule that is not in the jet. The radiator's mass per square metre is the figure the budget is most sensitive to. NASA's tested carbon-carbon heat-pipe radiator weighs 2.1 kg/m² and works at 700 to 850 K (Juhasz, 2008); at that mass and 1,500 K the stage would weigh 77 t. The design takes the 1.0 kg/m² that report projects as the least the technology could reach, at nearly twice the tested temperature; at 0.5 kg/m² the stage would weigh 14.9 t. challenge Helium-3 is scarce on Earth, and gathering 17 tonnes of fuel that is part helium-3 is its own industrial problem. challenge

Would antimatter do better?

An antimatter engine whose exhaust is the charged pions from annihilation might reach an effective exhaust near 0.3 c. Frisbee's 2003 estimate, as Keane and Zhang quote it in 2012, is 0.92c for the pions times 0.36 for the share of the energy that ends up in them, about 0.33c; their own simulation gave about 0.69c, and this page takes the lower figure. frontier The stop would then take 397 kg of propellant, half of it, 199 kg, antimatter. Making antimatter creates a matching particle of ordinary matter, so each kilogram kept costs at least two kilograms of mass-energy: 3.6 × 10¹⁹ J for the seed's supply, 0.17 of the launch's electricity. A factory that kept 1 × 10⁻⁴ of its energy as antimatter, far better than any accelerator by 2026, would need 3.6 × 10²³ J, 3,216 years of the world's 2024 electricity.

Why did magnetic sails look better before 2017?

Robert Zubrin and Dana Andrews treated the magnetic sail in 1991 as a fluid problem: the field pushes back a bubble of plasma that swells as the craft slows. In Zubrin's 2000 report to NASA, that model slows a 100-tonne craft on a 100 km loop from 0.1c to 468 km/s in 55.5 years. Gros' particle treatment in 2017 found that a fast transit to Proxima Centauri would need magnetic braking of the order of 1,000 tonnes, and that a 1.5-tonne craft could brake on a slow mission to Trappist-1, with cruise times of the order of 10,000 years. The gas density matters as much as the model: measured electron densities in the clouds near the Sun run from 0.01 to 1.0 per cubic centimetre (Frisch, Redfield and Slavin, 2011), and the stopping length scales inversely with it. challenge

Do light sails stop at brighter stars?

René Heller and Michael Hippke showed in 2017 that a graphene-thin sail of 7.6 × 10⁻⁴ g/m² arriving at Alpha Centauri A at up to 13,800 km/s could be parked in orbit around Proxima Centauri, using the light and gravity of all three stars. Alpha Centauri A and B are both brighter than Ember, and even there the limit is 0.046 of light speed for a sail with almost no payload.

Could the launch beam brake the seed?

Robert Forward's 1984 plan for a round trip stopped its sail with the launch beam itself: the sail splits, and its outer ring turns into a mirror that sends the beam back onto the inner part. That needs the beam focused across the whole crossing, which Forward provided with a huge lens in space. The Moon's array keeps its beam on the sail for only 658 AU before diffraction spreads it (Pushed by light).

How does the seed reach Kindling?

The burn ends with the seed close to Ember's speed, and the move onto Kindling's orbit is a set of short burns and passes. Their path through the fields of both Ember and Vesper is a three-body problem, still to be computed. design

Further reading

1. Gros (2017), Universal scaling relation for magnetic sails: momentum braking in the limit of dilute interstellar media, Journal of Physics CommunicationsProtons traced through a current loop's field; fast interstellar braking needs of the order of a thousand tonnes of sail.
2. Zubrin and Andrews (1991), Magnetic sails and interplanetary travel, Journal of Spacecraft and RocketsThe magnetic sail in the solar wind, and the fluid model of its drag.
3. Zubrin and Martin (2000), The Magnetic Sail, final report to the NASA Institute for Advanced ConceptsThe fluid model applied to braking from a tenth of light speed.
4. Frisch, Redfield and Slavin (2011), The Interstellar Medium Surrounding the Sun, Annual Review of Astronomy and AstrophysicsThe warm, partly ionized clouds around the Sun, their densities and temperatures.
5. Heller and Hippke (2017), Deceleration of high-velocity interstellar photon sails into bound orbits at α Centauri, The Astrophysical Journal LettersStarlight and gravity assists from three stars park a graphene-thin sail arriving at thousands of kilometres per second.
6. Forward (1984), Roundtrip interstellar travel using laser-pushed lightsails, Journal of Spacecraft and RocketsA staged sail that brakes on the launch beam, reflected from its own outer ring.