Decelerating: laser vs gun vs rocket

post by Stuart_Armstrong · 2019-02-18T23:21:46.294Z · LW · GW · 16 comments

Contents

  Decelerating with guns
  In practice: laser and solar sail
    Extra, theoretical, efficiency
None
16 comments

In our paper on exploring the universe, some of our probes required huge quantities of reaction mass in order to decelerate on arrival.

This is due to the tyranny of the rocket equation: to decelerate our final mass $m_f$ by $\Delta v$, we need an initial mass $m_0$ so that:

• $\Delta v=v_e\ln \left(\frac{m_0}{m_f}\right)$, where $v_e$ is the exhaust velocity.

Given this equation, the initial mass $m_0$ needed grows as exponential in the required deceleration $\Delta v$.

The relativistic rocket equation is similar but even worse, with an extra $\tanh$ term in it.

So the most effective way of allowing high speed exploration of the universe is to somehow get around the rocket equation. I was vaguely thinking about ways of using the target galaxy or solar system to do that - maybe the probe could scoop up interstellar dust or use gravity slingshots. But I realised that we can get around the rocket equation more directly.

Decelerating with guns

Suppose you are shooting through space, and you want to decelerate by pointing a gun in the direction of travel. Because of relativity, we can consider that you are at rest, and that you are accelerating by shooting a gun in an opposite direction.

You have two bullets, and you can shoot the bullets sequentially or simultaneously; imagine that you had two guns strapped together. If you shoot the bullets one after another, the first will start moving at velocity of $v$ while you recoil at some $-v^{\prime }$. Then when you shoot the second bullet, it will start moving at $v-v^{\prime }$ (if we stay in the classical model for the moment). So the total backwards momentum is $m(2v-v^{\prime })$, where $m$ is the mass of a bullet.

If you fire the bullets simultaneously, the total backwards momentum is $m\left(2v\right)$, however, and $m\left(2v\right)>m(2v-v^{\prime })$. By conservation of momentum, you will therefore be recoiling faster than if you shot the bullets sequentially.

What's happened? When you shot the bullets sequentially, part of recoil of the first bullet went into moving the second bullet at the same speed as you, which you actually didn't want. When you shot both bullets together, the recoil of both went purely to moving you. Therefore simultaneous fire is more effective at accelerating/decelerating. The real tyranny of the rocket equation comes from the fact that the early fuel needs to move the later fuel that needs to move the even later fuel. And most of that momentum gain is completely wasted: we don't actually care that exhaust fuel has gained momentum through the process. We'd like that extra momentum to be applied to the payload or probe, not to the fuel.

In practice: laser and solar sail

So there's a theoretical way around the rocket equation; can we do this in practice? Expending all fuel simultaneously would help (the equivalent of shooting all your bullets at once), but that extreme discharge might tear the probe and the rocket to pieces.

In space, there's no difference between the gun and the bullet - they're both just pieces of mass that fly off in opposite directions due to an explosion. So now imagine that there are ten thousand guns, floating independently in space, pointing at you. Everything is at rest with each other, and all the guns will fire in some sequence, and you will catch all the bullets (completely inelastic collision). Assume each gun, of mass $m$, will recoil with velocity $v$. Then the guns will have a total momentum of $10,000mv$, and, by conservation of momentum, you and the bullets will have the same momentum in the opposite direction. If the mass of the bullets (and you) is small compared to $m$, this will be an effective way of accelerating you. And note that your total final momentum depends on your mass, the mass of the bullets, the number of guns, $m$, and $v$. So it does not depend on the guns being fired at the same time, or any details of when they were fired. As long as you can catch every bullet, your final acceleration/deceleration will be the same. So you don't need to burn all your energy at once.

Catching bullets is hard, and we want to minimise their mass. So it's even better if we do this with lasers! Unfurl a solar sail around yourself, and have ten thousand free-floating lasers shoot at you in some sequence. This will gain you all the momentum of the lasers, independently of the sequence of firing.

The only real practical consideration is that you can cool down fast enough that each laser can fire before your sail moves out of their focus range; but a bigger sail can make both cooling and long distance firing easier.

Extra, theoretical, efficiency

What if your sail doesn't perfectly absorb all the laser light, but reflects some of it back? That's even better! In terms of bullets, that's the equivalent of elastic collisions, and you'll accelerate/decelerate even faster, losing less energy. Think in terms of conservation of momentum again: some light is now moving backwards, away from you. This can only happen if you've yourself gained some forward momentum.

In fact, the perfectly efficient way of decelerating would be for you to deploy a giant mirror, and for a single giant laser to do the same, then for the laser to blast you. The laser beam would bounce back between your mirror and the laser's mirror, gradually getting redshifted as you and the laser move faster and faster apart. This setup preserves both momentum and energy, and is the most perfectly efficient way of decelerating - and it doesn't depend on how fast the laser fires, a slow burn reaches the same conclusion as a swift burst. Why? Because conserving energy and momentum dictates the speeds at which you and the laser will end up.

Of course, in practice, the mirrors would not be perfectly reflective, the beam would lose focus, there would be some cosmic dust, and so on. Still, it's interesting to note that, in theory, we can completely do away with the rocket equation and accelerate/decelerate in the most efficient way possible, while using up energy arbitrarily slowly to do so. This hints that there may be practical methods that could get very efficient as well.

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