# What is the most effective way to brake from interstellar speeds?

In my story, I have a slower-than-light starship (traveling at 0.6 $c$) going to Alpha Centauri A. There are several planets around the star. The target planet is a terrestrial, habitable world orbiting at 1.2 AU (since Alpha Centauri is slightly bigger than the Sun) and a gas giant 3 times the size of Jupiter at 10-20 AU.

What is the most effective way to decelerate from 0.6 $c$

1. When arriving in the Alpha Centauri System (Pollution, fallout, etc. are allowed)?
2. When returning to the Solar System (Pollution, fallout, etc. are not allowed since we have got colonies as far as in the Oort Cloud; this means radioactive sections of the starship will most likely be detached and dropped onto Jupiter or Sun to be destroyed)?

Old-school "halftime" strategy (acceleration until halfway, then brake)? Quick acceleration and then quick braking, with a coasting period in the middle? Aerobraking? Gravity assist?

Please keep it hard science fiction: no wormholes and stuff; laser propulsion is used for the big part and nuclear propulsion is used for trajectory adjustment inside the Solar or Alpha Centauri system (plane change, escape/capture burns, etc...). The crew is brought to the ship in a shuttle, lands on the planet around Alpha Centauri A in a shuttle and lands on Earth in a shuttle, so the massive ship doesn't need to be aerodynamic unless it is needed for the aerobraking part.

• Do you use Alpha Centauri just as a distance measurement? The planets you talk about don't exist in the real world; for clarity, it might be better to use a fictional star, so people don't confuse this Alpha Centauri and the real one. Also, is "3 times the size of Jupiter" referring to mass or volume? – HDE 226868 Aug 30 '15 at 13:26
• A planet can't be 3 times the volume of Jupiter. Add more mass, it gets denser but only slightly larger. – JDługosz Aug 30 '15 at 21:35
• Citation or link? en.m.wikipedia.org/wiki/Super-Jupiter "Even though they are more massive than Jupiter, they remain about the same size as Jupiter up to 80 Jupiter masses" that's a summary; I learned details from a SETI seminar on brown dwarfs. – JDługosz Aug 30 '15 at 22:06
• @JDługosz I could have sworn I read it somewhere, and I'll keep searching to remember it, but I'll acknowledge that it's exceedingly probable that you're right and I'm wrong. – HDE 226868 Aug 30 '15 at 23:45
• A hot Jupiter would have additional elements for the story. The colony would have to deal with iron rain and ambient temperature with more digits than people are meant for. Though the planet has nothing to do with the breaking problem, unless someone finds a way to use it in the solution. – JDługosz Aug 31 '15 at 1:42

# Use a solar sail.

• Solar sails are lightweight(-ish)
• They can be used for propulsion
• According to Dandouros et al., solar sails built with technology in the near future could easily travel to a nearby star system in 60 years, reaching a top speed of $0.16 c$.
• You might want to use a solar sail merely for braking, so it's good that solar sails are low-mass (see this article, especially the part about the material called CP-1) and can be folded up.

For the calculations, an interesting reference is Solar sail thrust calculation:

In Space Mission Engineering: The new SMAD, page 555, section 18.7.2, the following thrust formula is given for a solar sail: $$F=\frac{2RSA}{c}\sin^2\theta=9.113\times10^{-6}\frac{RA}{D^2}\sin^2\theta$$ Where, $F$ is the thrust; $R$ is the fraction of incident light; $D$ is the distance from the Sun in astronomical units; $S$ the solar flux in $W/m^2$; $c$ the speed of light; $A$ the sail area in $m^2$ and $\theta$ the sail tilt angle.

If we set $\theta=\frac{\pi}{2}\text{ radians}$, then we find $$F=9.113\times10^{-6}\frac{RA}{D^2}\tag{1}$$ If we're being optimistic, and saying that $R\approx 0.5$, then $$W(s)=\int_{s_0}^s F\cdot dD=\int_{s_0}^s 9.113\times10^{-6}\frac{A}{2D^2}dD$$ $$W(s)=-\left[9.113\times10^{-6}\frac{A}{2D}\right]_{s_0}^s$$ Here, $W$ is work. Be careful to also account for changes in potential energy in your calculations. Also, there should actually be a sign flip in there (i.e. the $-$ should by a $+$), but that's superficial.

We can then use $$KE=\frac{1}{2}mv^2$$ to find the speed at any given distance, assuming that $S$ is constant (which it isn't - it's a function of $D$); given that $\frac{dS}{dD}\neq 0$, that must be accounted for for long-distance calculations.

For more fun, use $v=\frac{dD}{dt}$ to find the time it will take the sail to go from one point to another.

Since solar sails don't use fuel, there won't be any pollution or fallout from their use. They're a perfectly clean propulsion and braking technique. Also, you say you're using laser propulsion. Perhaps you could create (hypothetically) a large group of lasers and fire them at the craft, creating additional thrust.

• Great, except . . . every 4-microgram interstellar dust speck will impact your flimsy sail with kinetic energy equivalent to 14.3 kg of TNT at 0.6c relative. Yes. Every dust speck will do the damage of three anti-tank mines. Sorry. – imallett Aug 31 '15 at 14:00
• @imallett Wouldn't that be a problem for pretty much any ship? – HDE 226868 Aug 31 '15 at 14:02
• Absolutely. But most ships have it easier since the way they try to slow down isn't using a mylar bag turned broadside to relativistic bombardment. – imallett Aug 31 '15 at 14:10
• @imallett makes me think of a kind of interstellar parachute.... could you use a thicker, ablative material and a large change in surface area to slow down? – codeMonkey Aug 31 '15 at 16:20
• @Peter Roughly speaking, momentum is what causes penetration, kinetic energy is what causes damage. There's not great data about hypervelocity impacts, certainly none about relativistic impacts, but the general expectation is that (since momentum is linear and KE is quadratic) hypervelocity impacts look less like penetrators and more like point-source explosions.¶ If you somehow survive the dust specks, maybe try the proton flux? At 0.6c, it's about 42.25 petaelectronvolts per square centimeter per second. – imallett Jan 19 '17 at 1:09

If you use laser propulsion to get them to 0.6c, odds are that's the only realistic technology for the job. If you have a better system that works halfway between the sun and Alpha Centauri on whatever power you brought with you, then why not use that to accelerate you as well?

I propose a symmetric solution. Take a small craft (the shuttle) that contains everything your settlers need to build their colony. Figure out what's needed to accelerate that to 0.6c by laser propulsion. Take the whole package (shuttle + laser), and make that your spaceship. Presumably, the majority of mass will be taken up by the laser.

Build one big-ass laser in the solar system (perhaps powered by a partial Dyson sphere) and accelerate the whole thing to 0.6c. Once it gets half way, use the onboard laser to decelerate the shuttle. The rest of the craft travels in to space at 0.6c forever (no doubt bearing a plaque with naked people on it).

To make things a little more economical, the first trip might just contain a load of robots and a second laser to be put into orbit around Alpha Centauri. After that's in place, you can send supplies and people back and forth more economically: using one laser to accelerate, and one to decelerate,

• " Once it gets half way, use the onboard laser to decelerate the shuttle. " Problem with that is, as soon as you have a onboard lazer able to decelerate it would need to be at least as powerful as what accelerated you (or it would need to be turned on a lot longer to decelerate you the same amount). And, if you have a onboard lazer at this point, you can always use a onboard mirror to go whatever way you want anyway. But ideally the "lazer each end" is the best once that's established. Or, at least, "external momentum changing devices" or some sort. – darkflame Aug 30 '15 at 21:00
• @darkflame, No, that's the point. The onboard laser only needs to be powerful enough to decelerate the shuttle. Almost the whole ship stays accelerated at 0.6c: only the shuttle gets decelerated. So you have a huge laser at the sun, and a medium sized laser in the ship. – Peter Aug 30 '15 at 21:06
• Oh, in which case +1. Yes, absolutely. – darkflame Aug 30 '15 at 21:06
• Replace the onboard laser with a big mirror and reflect the laser light from sol back onto the shuttle. – Lex Oct 2 '17 at 5:37
• +2 for the good idea and creative way to solve your decel problem; -1 for leaving behind an enormous piece of space trash that is now hurtling through space at a large fraction of light speed. Whoever lives on the planet this crashes into does not stand a chance. +2-1=1, so I suppose you still get an up-vote. ;) – Loduwijk Feb 13 '18 at 22:57

Silly answer: lithobraking - i.e. just crash whatever piece needs to stop into solid rock.

(This would in fact be very destructive, of both the crew, their vessel, and the destination. The hot Jupiter planet might not notice an aerobraking ship, but the crew sure would.)

• Could you add the important parts from the link into your answer? Answers should be able to stand by themselves, should the link die. – bowlturner Aug 31 '15 at 2:24
• Litho/aerobraking for a relativistic spaceship? That's going to hurt. The planet as well. – Luaan Aug 31 '15 at 11:22
• Just look for the Magic Schoolbus: worldbuilding.stackexchange.com/a/22894/10364 – Green Aug 31 '15 at 14:39
• At a fraction of lightspeed, aerobreaking will not be useful. – JDługosz Aug 31 '15 at 17:53
• No, but it would make for great aurora! – Phil Miller Sep 1 '15 at 1:49

Underused in SF is the idea of magnetic breaking.

I recall reading about the math of a Bussard ramjet that it can't work because the drag is greater than the energy produced.

But, instead of tweeking it as much as possible to reduce drag, turn the problem on its head: optimize it to maximize drag, and you have an excellent brake.

I planned on using that in a story, but lost track of the notes I took at the time. Google shows some suggestive topics, though.

Robert L. Forward wrote about lightsails. in Flight of the Dragonfly/Rocheworld, he used a detachable ring for the main part of the sail. The earth-based laser reflects off the sail and back to the craft with its much smaller piece of the sail, slowing it down.

• Bussard ramjets need slow varying magnetic fields for effective intake capture. Make the magnetic field fast varying and it pushes away the interstellar medium, increasing its braking. I lost interest Bussard ramjets when I found they required one million kilometre of intake diameter per ton of vehicle mass. One of Alan Bond's papers in the mid-1970's had both results. – a4android Jul 30 '16 at 8:20
• That can’t be right, not only because the "need" varies with the acceleration and you can make do with less; the collecting area varies with the square of the diameter and then needed power varies directly with the payload mass. – JDługosz Jul 30 '16 at 10:04

Best answer with currently understood technology:

Only remaining reasonable method not already discussed: use H-bombs (that is, Orion drive). Despite its dry mass, the high efficiency of the Orion drive makes it unbeatable for long flights by current technology. 0.1c is attainable this way, and trying to break with a solar sail is not feasible at this speed due to inadequate force available. We can avoid most fallout problems by using a nuclear-hydrogen rocket for a planeshift burn first.

This speed still makes a 40+ year flight though.

Look, if you wanna get up to 0.6c and back down to 0 again you're gonna need an antimatter drive, and you're still going to need > four times your rest mass in fuel. But right now, we don't have a solution for containing that much antimatter with the fuel fractions to make this remotely reasonable.

A simple light sail isn't going to cut it—you're not going to brake from 60%c with one.

If you want a hard-science answer the only thing I see is to use your launch laser. This uses a light sail but it's powered by the laser beam rather than just sunlight, it can accelerate harder and longer than something simply using the central star for power.

Now, to stop you cut off a ring of your sail—this must comprise most of the sail and most of the mass of your spacecraft. The ring flips over (or more likely simply inverts and reflects off the back side of it) and focuses its energy on the inner piece. Note that this requires incredible accuracy and is almost certainly beyond current technology.

For a more detailed look at this read Robert L Forward's Flight of the Dragonfly. The numbers in that book are scary indeed—the launch laser is a facility around (not in orbit! It would push itself away from the planet in operation if it were) Mercury, the final focusing lens is in the outer solar system and is of planetary scale. Furthermore, his craft was only doing 20% of lightspeed. He ignored the accuracy problem of the ring focusing its energy on the decelerating spacecraft.

In theory the same approach could be used again to return, discarding another ring comprising most of the remaining mass to boost for home. Stopping on the launch laser won't be a problem.

The sails must be gargantuan and incredibly shiny as they will be bathed in a laser beam from hell and they can't have a cooling system, they must simply reflect off enough energy that they can radiate away whatever they absorb.

I believe a much more viable approach is to send out the first craft with only robots on board. It goes considerably slower and brakes on an unboosted sail. It then proceeds to construct a laser that's a duplicate of the launching laser. You still need huge sails but you're not discarding most of the sail so it's nowhere near as big as it would have to be using the separating sail approach and you eliminate the aiming problem.

The more I look at aerobrake the worse it gets.

1) The toughest aerobrake we have ever done was 47 km/sec—and at a cost of ½ of the weight of the probe being devoted to the heat shield, half of which burned away in the process. This is almost 4000× the energy.

2) I'm not competent to figure the actual deceleration involved but I can show that it's well above a million gravities. (Scaling up the Galileo probe gives a million g but it runs out of planet well before it's done. Thus the actual value must be considerably higher.) The toughest electronics we build are artillery shells—something around 1% of this energy.

3) At this kind of velocity air ceases to behave as a fluid, but rather as particles. Heat shields are based on deflecting away most of the energy that hits them but the particles won't be deflected. Your shield is going to absorb energies greater than a matter-antimatter explosion.

There are no good passive braking methods for such speeds that don't smash the mission. The key factor is that maximum mission deceleration is limited to fairly low values, which means a large braking distance.

Solar sails. Using the notation of HDE-226868's answer, the braking force is $F\simeq10^{-5}\frac{N\cdot a.u.^2}{m^2}A/D^2$. It is obvious that mission deceleration will be greatest at the minimum distance: $Mg_{max}\simeq10^{-5}\frac{N\cdot a.u.^2}{m^2}A/D_{min}^2$, where $M$ is the mission mass and $V$ will be the mission velocity. The work of the braking force, braking from infinity, will equal the mission's kinetic energy at infinity:$MV^2\simeq10^{-5}\frac{N\cdot a.u.^2}{m^2}A/D_{min}$ (the term corresponding to maximum distance vanishes). Dividing these two equations, we can obtain the relation $g_{max}D_{min}\simeq V^2$. It holds for any solar sail regardless of efficiency, material, target star &c. For interstellar but not relativistic speeds, let $V=\beta c$, then $D_{min}\simeq\beta^2\frac{c^2}{g}\frac{g}{g_{max}}\simeq10^5\,a.u.\beta^2\frac{g}{g_{max}}$ where $g$ is Earth gravity. Substituting this back into the mission deceleration equation, we see that the area density of the sail is inversely proportional to the fourth power of mission velocity: $M/A\simeq10^{-16}\,kg\,m^{-2}\frac{g_{max}}{g}\beta^{-4}$. If the mission is carrying something squishy like humans, $g_{max}$ must be on the order of $g$. Assuming a mission velocity marginally attainable with known technology $\beta=0.1$, our braking sail must weigh at most $10^{-12}\,kg\,m^{-2}$. This is quite beyond the capacity of any known reflective material. (For comparison, a one-atom-thick sheet of aluminium would be around $10^{-7}\,kg\,m^{-2}$.) A fully solid-state probe could, perhaps, sustain decelerations as high as $1000g$, but the sail would need to sustain them as well. Solar sails aren't likely to be useful for braking an interstellar mission.

Braking on interplanetary medium. Unless one is traveling to a young planetary system full of dust (which, ipso facto, will have no useful planets), IPM seems to be only dense enough to be a nuisance and many orders of magnitude away from being useful for braking a mission traveling at interstellar velocity. If we assume that IPM material simply accretes to the mission, the braking force $Mg_{max}=\rho SV^2$ where $\rho$ is IPM density and $S$ is the cross-section of the mission vehicle. Mean IPM density in the vicinity of Earth is on the order of $10^{-19}\,kg\,m^{-3}$[1], and it falls off with distance from the primary as $r^{-1.3}$[2]. For given $V$ and $g_{max}$, mission vehicle must have $M/S\lesssim\rho V^2/g_{max}$. The latter fraction is just the braking distance: $D_{brake}\sim V^2/g_{max}$, so $M/S\lesssim\rho D_{brake}$. Using again $g_{max}=g$ and $\beta=0.1$, $D_{brake}\sim10^{14}m$ and $M/S\lesssim10^{-6}\,kg\,m^{-2}$. This is much less flimsy than the solar sail, above, needs to be. However, the braking distance ($600\,a.u.$ in our case) is much larger than the effective radius within which IPM is sufficiently dense.

If hypervelocity impacts of IPM particles produce explosions and eject material, the braking force seems to be enhanced by a factor of $\sim\frac{v_e}{L}V$, where $v_e$ is the exhaust velocity of ejected material and $L$ is the specific heat of sublimation of the materials involved. For $\beta=0.1$ and $v_e\sim 3\,km\,s^{-1}$ characteristic of chemical explosions, this factor is on the order $10^2$ and the mission can be heavier, $M/S\lesssim10^{-4}\,kg\,m^{-2}$ (though it must now withstand the explosions, a daunting prospect). Nevertheless, this does not remove the tyranny of braking distance and probably precludes the use of this braking method.

Electrodynamic braking on interplanetary magnetic field does not appear to be effective [3].

• Hats off to your hard work – Garret Gang Oct 2 '17 at 1:45

You mentioned a crew. Crews are squishy.

If your ship has humans in it, it's not a good idea to submit them to accelerations much higher than Earth's gravity.

I have no suggestion on propulsion methots, but the old-school "halftime" strategy as you put it is the only way to go.

To be blunt: the same way it accelerated. At 0.6c, there is nothing that can be used for braking but not for acceleration.

• Okay, so how would it accelerate, then? The laser propulsion idea? – HDE 226868 Aug 31 '15 at 15:19
• @HDE226868 I don't know - the author does. – Agent_L Aug 31 '15 at 15:32
• And who is this author? Can you provide a reference and ideally a better explanation? – Vincent Sep 12 '15 at 14:56
• Actually this is a completely valid answer. The person asking the question has somehow got the ship up to those speeds. We need to know what that method is to know why it won't help for slowing down. Fuel usage? Directional (i.e. laser propulsion?) etc. – Tim B Sep 12 '15 at 21:13
• @Vincent Author of the question, who else? What the heck is wrong with you guys: the author imagined speeding his ship all the way to 0.6c, told us nothing about the details and somehow it's me who's supposed to know what's happening in his mind? Please, read the damn question before you start downvoting answers! – Agent_L Sep 14 '15 at 8:27

I don't have much to add about drive technology -- it's already been covered. (FWIW, I favor the Orion drive at 0.1c max or antimatter tech stack.)

That said, I think it's worth noting that your concerns about radiation are probably overstated. Once you get outside the earth's magnetic field, you are exposed to massive quantities of radiation from the sun. Per space.com:

...the radiation dose received by an astronaut on even the shortest Earth-Mars round trip would be about 0.66 sievert. This amount is like receiving a whole-body CT scan every five or six days.

By implication, any society that can build space habitats is capable of dealing with radiation exposure.

Does the main ship even need to slow down significantly? It might be beneficial to try to keep most of the main ships momentum - instead put it into a big path around the star, using whatever thrust you have to steer inwards over a long period. Meanwhile the smaller "shuttle craft" has detached from it, and its this little craft that has to then slow down. While your still stuck with the prospect of needing to slow down from 0.6C, it should be significantly easier to do with the (presumably much smaller) mass of the little craft. While its never "easy" slowing down from high speeds, the smaller the mass the less energy it will take.

You then either have the big ship slow down over a (much much longer) period using sails or a ion-drive, or try to get that little craft at least briefly upto 0.6C itself to rejoin its main craft for the trip back. (the main craft having never slowed down much at all).

• If you want to put the ship on a big orbit around the star, it needs to have a small momentum. The orbital speed is proportional to the inverse square root of the orbit radius. $0.6c$ is well within the star. To put the ship in orbit at, say, the same distance as Earth from the Sun, you'd need to slow it down to about $10^{-4}c$. – Gilles Aug 30 '15 at 20:00
• unassisted, but I assumed use of the big crafts steering mechanism. Would not the (vastly) bigger distances involved give more time to "steer in" enough? I guess it wouldn't be a orbit any more - but is not the potential of some sweat spot between orbital distance? Where the force you can divert to turning is enough to put you eventually into a orbital path. If not see part of peters answer - dump the big ship completely. Either way getting rid of most of the mass seems the better way to go here. – darkflame Aug 30 '15 at 20:51
• "Steering" doesn't really work in space. If you have nothing to push off, you simply need to change your velocity the old-fashioned way - by using thrust. There are ways to "push off" planets/stars, but it only works well at near orbital-velocities - compared to your 0.6c, this is a tiny drop, and isn't going to help any. – Luaan Aug 31 '15 at 11:27
• I aware you need thrust when there is no medium. By the word "steer" I meant "apply thrust to make the ship slowly point more in that direction then the direction it would normally without". Isn't the term steer used in space then? Pretty sure I have heard it used with regard to firing thrusters. The idea was simply rather then fire them halfway to slow down to a stop, you instead use them to loop around the star on a very big oribit, with the people already left on a shuttle. – darkflame Aug 31 '15 at 22:05
• If you can get the small craft up to 0.6c "briefly", then you can get it up to 0.6c "indefinitely" (ignoring the effects of, for example, bombardment of microscopic interstellar dust). In spaceflight, once you have reach escape velocity, there is very little to nothing that will slow you down, so you can just turn off the engines and coast along your current orbit at whatever velocity you have built up. – a CVn Sep 11 '15 at 20:07