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 $c$
- When arriving in the Alpha Centauri System (Pollution, fallout, etc. are allowed)?
- 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.
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1 answer
Use a solar sail.
Advantages:
- 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.
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