Suppose a starship is being powered by a lasers beamed first from the departure star, and then from the destination which is a moving base. The ship is moving in the regime of 10–30% of light speed.

Picture a letter “T”. The ship is accelerated up the vertical line from the source at the bottom of the T, into the path of the horizontal cross-line. A mobile base (the destination) moving on this horizontal line at 10% lightspeed will supply a laser too.

That is, the home system will push a lightsail craft into the path of a base moving at relativistic speeds. That base would pass home with a closest approach of several light years.


Let me try another mental picture: you are a mile back on a path that intersects with a main road at a right angle. A mobile home is travelling down the main road at 70 miles per hour. The beam from home pushes you into the main road, and the beam from the mobile home needs to get you going in that direction instead for eventual rendezvous.

The mobile base needs to accelerate the incoming craft to its own oncoming speed and kill the transverse velocity.

What would the maneuver look like? I’m supposing that the angle at which it presents the sail will be significantly affected by relativistic effects and aberration of moving source.

Also, the mobile base (from a more advanced civilization) can do any advanced tricks you can imagine, such as synthetic aperture beamforming to make the wavefronts come from a different direction than the actual source, and impart orbital angular momentum to the photons.

The home beam has just enough power and focus for the maneuver it is designed for. The base beam can be more powerful and amazingly well-focused. But the base won’t aim the beam directly at the home; they will require the ship to be some distance out (like a light year) before offering.

Peregrine Rook’s sketch is nicer ☺.

I think the sail would be tipped the other way when catching the beam from the base, though, to slow the “up” component.


  • $\begingroup$ Must the two beams be (roughly) orthogonal or can either of them be "aimed" off of their primary axes? $\endgroup$
    – Nolo
    Commented Aug 24, 2016 at 5:20
  • $\begingroup$ @nilo the beams can be aimed anywhere. One from “home” that’s a straight laser. One from “base” which is more powerful and fancy in any imaginable way. $\endgroup$
    – JDługosz
    Commented Aug 24, 2016 at 7:16
  • $\begingroup$ @PeregrineRook yes, your sketch is nicer than mine! $\endgroup$
    – JDługosz
    Commented Aug 24, 2016 at 7:17
  • $\begingroup$ «impart orbital angular momentum to the photons» I don't know if that would help with this beam, but it’s something I introduced earlier in the story for communications. I'll try and dig up a link. $\endgroup$
    – JDługosz
    Commented Aug 24, 2016 at 7:20
  • 2
    $\begingroup$ This is all pretty beyond be science wise, but as someone who used to sail a lot, remember that you need a keel (or dagger/centreboard on smaller boats) to go reliably in any direction other than away from the wind. A keel is essentially a surface for the force imparted to the sail by the wind to push against and an underwater wing at the same time. It turns movement imparted by the sail into forward movement. My (admittedly limited) knowledge of solar sails would suggest that you'd be completely at the mercy of the "Pushing" of the solar winds, and find it hard to do much manoeuvring! $\endgroup$
    – Miller86
    Commented Aug 24, 2016 at 8:33

3 Answers 3


First, let’s look at the different types of trajectories a solar sail can take. They differ mainly based on something called the lightness number, $\beta$, which depends on the composition and structure of the sail. $\beta$ can be used to determine the type of trajectory the solar sail will follow: $$\begin{array}{|c|c|} \hline \text{Value of }\beta & \text{Type of trajectory} \\ \hline \beta=0 & \text{circular Keplerian} \\ \hline 0<\beta<\frac{1}{2} & \text{elliptical} \\ \hline \beta=\frac{1}{2} & \text{parabolic} \\ \hline \frac{1}{2}<\beta<1 & \text{hyperbolic} \\ \hline \beta=1 & \text{rectilinear} \\ \hline 1<\beta & \text{flipped hyperbolic} \\ \hline \end{array}$$ This is also evident in Figure 4.8 (page 123) of Colin McInnes’ Solar Sailing: Technology, Dynamics and Mission Applications, which is my primary reference in this answer:

enter image description here

Now, you can see that a hyperbolic trajectory of some sort may be exactly what you’re looking for - and, in fact, it requires no assistance from the base it is rendezvousing with! Parabolic trajectories, too, are escape trajectories, but a hyperbolic trajectory might be more efficient. Plus, having a greater lightness number results in a greater characteristic acceleration (see Seboldt & Dachwald (2003)), because $a_c\propto\beta$. Therefore, I’d prefer to work with a flipped hyperbolic trajectory; I’ll choose $\beta\approx2$.

There are two equations of motion for polar coordinates $(r,\theta)$: $$\frac{\mathrm{d}^2r}{\mathrm{d}t^2}-r\left(\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2=-\overbrace{\frac{\mu}{r^2}}^{\text{gravitational}}+\overbrace{\beta\frac{\mu}{r^2}\cos^3\alpha}^{\text{radiation}}\tag{4.37a}$$ $$r\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2}+2\left(\frac{\mathrm{d}r}{\mathrm{d}t}\right)\left(\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)=\beta\frac{\mu}{r^2}\cos\alpha^2\sin\alpha\tag{4.37b}$$ where $\mu$ is the standard gravitational parameter and $\alpha$ is the angle between a vector normal to the sail and a vector pointing from the star to the sail. Compare McInnis’ $(\text{4.37a})$ to $(\text{346})$ here, with the substitution of $h=r^2\dot{\theta}$. The two are identical, with the addition of the radiation term in the solar sail reformulation. Let’s have $\alpha\approx0$. This means that the right-hand side of $(\text{4.37a})$ becomes $(\beta-1\frac{\mu}{r^2}$, and the right-hand side of $(\text{4.37b})$ becomes $0$.

We can arrive at a simple analytical solution if we assume that the solar sail takes the path of a logarithmic spiral, i.e. a path of the form $$r(\theta)=r_0\exp(\theta\tan\gamma)$$ where $r_0$ is the initial radius and $\gamma$ is the spiral angle, the angle between the velocity vector and the transverse direction of the sail’s path. So let’s step back a little, and let’s assume that

  • $\beta\approx0.75$ (I’ve chosen a value for a normal hyperbolic trajectory)
  • $\alpha\neq0^{\circ}$. It could, but that might not be optimal.

McInnes goes through several substitutions, leading to $$r^3\left(\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2=\mu\left[1-\beta\cos^2\alpha(\cos\alpha-\tan\gamma\sin\alpha)\right]\cos^2\gamma\tag{4.41}$$ From this and earlier substitutions, we can derive expressions for the radial velocity $v_r(r)$ and angular velocity $v_{\theta}(r)$. The equation for the former is $$v(r)=\sqrt{\frac{\mu}{r}}\left[1-\beta\cos^2\alpha(\cos\alpha-\sin\alpha\tan\gamma)\right]^{1/2}\tag{4.44}$$ There’s a fairly complicated relationship between $\gamma$ and $\alpha$, but it can be simplified for small $\gamma$: $$\frac{\beta\cos^2\alpha\sin\alpha}{1-\beta\cos^3\alpha}=\frac{\sin\gamma\cos\gamma}{2-\sin^2\gamma}\approx\frac{1}{2}\tan\gamma\tag{4.45,4.48}$$ This integration is important when we try to find a relationship between $r$ and $t$. We integrate $(\text{4.44})$: $$\int_{r_0}^r\sqrt{r}\mathrm{d}r=\int_{t_0}^t\left(2\beta\mu\sin\alpha\cos^2\alpha\tan\gamma\right)^{1/2}\mathrm{d}t\tag{4.46}$$ Integrating this and substituting in $(\text{4.48})$ yields $$t-t_0=\frac{1}{3}\left(r^{3/2}-r_0^{3/2}\right)\left(\frac{1-\beta\cos^3\alpha}{\beta^2\mu\cos^4\alpha\sin^2\alpha}\right)^{1/2}\tag{4.49}$$ However, we can simplify this by letting $t_0=0$ and focusing on cases where $r_0\ll r$ for most $r$, which is the case here when $r=r_f$. We can then find when the function of $\alpha$ in $(\text{4.49})$ is maximized; it turns out that for small $\beta$ (i.e. $\beta<0.5$), $\alpha_{\text{max}}\approx35.26^{\circ}$. However, I chose $\beta=0.75$, and so it turns out that $\alpha$ is maximized at about $35.26^{\circ}$. Plugging this back into our approximation for $\tan\gamma$, we find that $\tan\gamma\approx1.362$, which gives us $\gamma\approx53.7^{\circ}$. This likely makes our small angle approximation for $\tan\gamma$ less accurate, but it will do for now. Plugging this in, and assuming once again that $t_0=0$ and $r_0\ll r$, $(\text{4.49})$ gives us $$t=r^{3/2}\times1.23\times10^{-10}$$ and for a final radius of three light-years ($2.838\times10^{16}$ meters), we find that $t\approx5.88\times10^{14}$ seconds, or about 19 million years. That might seem like it can’t be correct, but Centauri Dreams cites Matloff et al. that it could take a really good solar sail 30 years just to reach the Oort Cloud, 500 AU away - and one light-year is about 60,000 AU. Clearly, a simple logarithmic spiral quite like this won’t work.

In fact, this means that you absolutely need to give the solar sail a very fast initial boost to make interstellar travel on these scales even remotely feasible. This makes the equations a little harder, and it means that yyou might not see an easy analytical solution pop up.

Let’s go back to our original coupled equations $(\text{4.37a})$ and $\text{4.37b})$, where we’ve set $\beta=2$ and $\alpha=0$. This becomes a simple central force problem, which has one equation of the form $$\frac{\mathrm{d}^2r}{\mathrm{d}t^2}-\frac{h^2}{r^3}=\frac{F(r)}{m}$$ where I’ve defined $h\equiv r^2\dot{\theta}$, which is conserved. $F(r)$ is the central force as a function of $r$; normally, in orbital mechanics, it’s simply $$F(r)=-\frac{GMm}{r^2}$$ as is the case in $(\text{346})$; here, as I noted before, we also have to account for the force from radiation pressure. With $\beta=2$, it just so happens that the two forces add up to $$F(r)=\frac{-GMm}{r^2}+\frac{(2)GMm}{r^2}=\frac{GMm}{r^2}$$ which is repulsive, unlike $(\text{346})$. That pdf shows a good derivation of the orbital equation from the central force law, which I’m not going to go through again, as it’s pretty standard. For a generic central force of the form $$F(r)=-\frac{k}{r^2}$$ we arrive at an orbit of the form $$r(\theta)=\frac{l}{1+\varepsilon\cos\theta}\tag{355}$$ where $k=-GM$ (in general, $k=(\beta-1)GM$), and $$l\equiv\frac{mh^2}{k},\quad\varepsilon\equiv\frac{l}{a}-1\tag{356}$$ I’m no expert when it comes to solar sail construction, so I read through McInnes et al. (2001) and came up with a conservative estimate of 2,000 kg. The authors estimated that you could send a 900 kg solar sail to solar orbit, with much of that mass being payload. My guess could be way off, so I’d appreciate it if an expert has better figures.

I assumed that the solar sail starts out on a circular orbit around a sun-like star at roughly Earth’s semi-major axis. From this, I calculated $$v_0=\sqrt{\frac{\mu}{r}}=2.97\times10^4\text{ m/s}$$ $$h=\frac{|L|}{m}=\frac{rmv}{m}=rv=4.46\times10^{15}\text{ m}^2\text{/s}$$ $$k=(\beta-1)GM=1.327\times10^{20}\text{ m}^3\text{/s}^2$$ $$l\equiv\frac{mh^2}{k}=3\times10^{14}$$ $$\varepsilon\equiv\frac{l}{a}-1=2000$$ From this, I get $$r=\frac{3\times10^{14}}{1+2000\cos\theta}$$ $\varepsilon>1$ (as was expected, given that $\beta>1$), and in fact $\varepsilon\gg1$.

I used modified code from this page to solve $(\text{4.37a})$ in Mathematica and plot the motion of the solar sail over the course of one year:

M = 1.99 10^30 (*mass of Sun*)
G = 6.67 10^-11 (*Newton's constant*)
x0 = 1.50*10^11 (*apsidal distance*)
y0 = 0; vx0 = 0;(*on x axis with velocity in y direction*)
vCirc = Sqrt[G M/x0] (*apsidal speed for circular orbit*)
vy0 = 0.8 vCirc (*smaller speed gives elliptical orbit*)
a = 1/(2/x0 - vy0^2/(G M)) (*semimajor axis from E=T+V*)
T = 2 Pi Sqrt[a^3/(G M)] (*period from Kepler's third law*)
beta = 2 (*accounts for radiation pressure*)

r[t_] := {x[t], y[t]} (*position vector*)

equation = Thread[r''[t] == (beta-1) G M r[t]/Dot[r[t], r[t]]^(3/2)]

initial = Join[Thread[r[0] == {x0, y0}], Thread[r'[0] == {vx0, vy0}]]

solution = NDSolve[Join[equation, initial], r[t], {t, 0, T}]

orbit = ParametricPlot[r[t] /. solution, {t, 0, T}];

This is the orbit:

enter image description here

As you can see, it travels in essentially a straight line, going at a little over 5 Au per year, at first. That’s not bad at all. It’s still going to take a long time to reach the base, but this is likely going to be on the order of thousands of years, not millions of years.

  • $\begingroup$ «figure out if we even need to take special relativistic effects into account in the case of the lightsail.» I seem to have missed that in the end. What's the punchline? $\endgroup$
    – JDługosz
    Commented Sep 5, 2016 at 16:21
  • $\begingroup$ I would think that with distances > 500AU and into interstellar space, and speeds at 10%C, that orbital mechanics is not applicable. Other stars and galatic tides will make larger peturbations than the sun of the launch site, and the accelerations will render g unimportant. $\endgroup$
    – JDługosz
    Commented Sep 5, 2016 at 16:24
  • $\begingroup$ That is, model it in free space without gravity. Incoming beam is a momentum source subject to aberation of motion, and reflection gives large percent of same momentum in a chosen direction. I'm guessing* that the apparent position of the beam source is what matters but with both source and receiver moving at relativistic speeds I don’t know how the wavefront direction will look. $\endgroup$
    – JDługosz
    Commented Sep 5, 2016 at 16:28
  • $\begingroup$ @JDługosz Regarding the first comment: I was going to calculate the speeds of the trajectories; I haven't gotten around to that yet. Regarding the second comment: Orbital mechanics is most definitely valid at distances greater than 500 AU. It's valid in the Oort Cloud, which starts ten time as far away! $\endgroup$
    – HDE 226868
    Commented Sep 7, 2016 at 18:00
  • $\begingroup$ «5 Au per year» that's nowhere near the 10 to 30% of c I need. Voyage time should be decades. Remember, it’s a high-thrust lightsail, not an ambient solar flux sail. $\endgroup$
    – JDługosz
    Commented Sep 7, 2016 at 22:03

I believe the desired outcome can be achieved with relatively simple means. The trick with solar sails is that although the incoming light can only push the sail directly, the reflection of that light can push the sail in a different direction. The resulting net thrust is the combination of the incident light and the reflected light. For a solar sail being pushed by lasers at significant fractions of C it’s safe to assume that the the sail will have near 100% reflectivity (otherwise the non-reflected light would incinerate the ship) so the magnitudes of the incident and reflected light will be approximately equal. Since from basic geometry we know that the angle of incident light and the angle of reflection will be equal, then the angle of thrust, or net acceleration will always be directly orthogonal to the plane of the sail. This means the solar sail will always accelerate straight ahead. If we want to accelerate in a different direction, we simply have to turn our sail to face directly away from that direction, treating the sail as if it were any other conventional source of thrust. The caveat though is that changing the orientation of the sail changes the magnitude of the experienced thrust. The acceleration of the sail can be computed by the following functions:

sail equations

Since all the variables in the above equations are constants with the exception of B, the angle of the sail with respect to the laser, we can deduce the relationship between the acceleration of the ship and the angle B to be Accel = cos(B)^2. That is to say when B is 0 and the sail is facing the laser directly it will experience maximal acceleration and when B is 90 degrees and the sail is sideways to the laser is will experience 0 acceleration.

It follows that if we want our trip to be efficient we need to minimize B. Of course, if you want to make the whole thing much simpler you can simply say the light comes from directly behind the craft at all times due to some fancy technology. In that situation the force experienced by the ship will always be constant. But that also takes all the fun out of solar sails doesn’t it?

So with this system we can accelerate in any direction away from the laser source, but never back towards it. However, the efficiency of our acceleration decreases rapidly the harder we try to turn. One consequence of this is that it will be difficult to ever use both lasers simultaneously with any efficiency since they are ~90 degrees apart. This means for the first leg of our journey we are going to want to accelerate directly away from our home laser towards some far off rendezvous point and ignore the station laser. By accelerating directly away and keeping B equal to 0 we maximize our acceleration. At some point though we will need to reduce our velocity towards the oncoming station in order not to overshoot it. We have to use the station laser to do this. Ideally by this time we are closing with the station and thus the angle we must thrust to match velocities with the station is close to that of the laser making our thrust once again efficient.

The below schematic isn’t exactly what I am describing above, it was made before I realized how inefficient the angled thrusting would be, but it still gets the general concept across.

setting sail

With regards to the effects of relativity. Obviously the lasers from both parties will need to be aimed light-years ahead of the ship’s path with absurdly precise calculations and a predetermined course. Even a slight error would compound and eventually throw the ship out of the path of the lasers which had been fired years in advance. Even with faster than light communications this would be a remarkable feat.

To specifically address your concerns regarding relativity and the angle of the sail:

I’m supposing that the angle at which it presents the sail will be significantly affected by relativistic effects and aberration of moving source.

The aberration of light due to the movement of the station will not change the direction the ship accelerates since we determined above that the ship always accelerates orthogonal to the plane of the sail. The aberration of light will however change the magnitude of that thrust by influencing the projected area of the sail (and therefore the amount of light that hits the sail) and the proportion of the incident and reflected thrust vectors that are productive (how much of those vectors cancel each other out).

This answer doesn’t include a precise plot of an optimal course for the ship to take. This is because there are many possible courses and they differ drastically based on the relative strengths of the lasers and the distances and velocities involved. For instance, if the station laser is significantly more powerful than the home laser we will want a course that lets us flip over to utilize it as soon as possible. But that course is very different from one in which the home laser is preferred. Based on the distances and velocities and maximal accelerations involved the ship might have to begin accelerating on a course nearly parallel to that of the station to ensure it matches velocities before the station passes. Or if the station is very far off maybe it can simply accelerate directly into the path of the station and then be brought up to speed by the maximal effectiveness of the station’s laser pushing the ship directly in front of it. I see no way to simply compute a single optimal course even if those missing constants were defined. I do think that this answer provides insight into the principles of the laser-powered lightsail’s operations and the equations necessary to calculate the time a given course will take.

  • 1
    $\begingroup$ +1 for the illustrations. What did you use to draw? But this elementary primer is not hard-science—you didn’t even mention cosine, and the relativistic effects are not just time delay in aiming but complicating factors to know how to handle. $\endgroup$
    – JDługosz
    Commented Sep 7, 2016 at 21:57
  • $\begingroup$ @JDługosz It seems to me that without specifics on the exact distances involved, the relative velocities of the star and station, and the relative strengths of the each point's lasers one can't actually apply the geometry and calculus to compute an exact course. The fundamental concept here is that regardless of where the lasers are the solar sail will accelerate in the direction it is facing. This means the solar sail craft is can be treated approximately as any other conventionally propelled ship. The differences being direction influencing magnitude of thrust and no change in mass. $\endgroup$ Commented Sep 7, 2016 at 22:36
  • 1
    $\begingroup$ I used google drawings as part of google drive to generate the schematic. $\endgroup$ Commented Sep 7, 2016 at 22:38
  • $\begingroup$ I didn't give too much detail because I will adjust to suit the timeline of the plot, once I know what's possible. That is, I'm working backwards from the narrative. $\endgroup$
    – JDługosz
    Commented Sep 7, 2016 at 22:43

There are some ways to break this problem into simpler ones.

One laser at a time

Since both the thrust sources are lasers, they're coherent. And since managing their relative phases at distances of light-years is basically impossible, you want to have just one active at a time, to prevent interference and loss of thrust.

Look at it as two separate voyages

It's a basic physics trick to break down problems in forces to two separate problems at right angles. Since there is essentially no friction in this system, that's quite accurate for this problem. So we can look at this as two separate problems:

  1. Leaving the starting point, accelerating, making turnover, decelerating and stopping at the right distance from the start.

  2. Accelerating from zero velocity on the course parallel to the station to match its actual velocity.

I'm not suggesting that the first voyage should be completed, and then the second started, just that it's easier to think about the problem in two pieces.

Managing the sail

Start out with the sail pointing backwards towards the start point. Leave it there until you reach turnover. Then angle it at $arctan(y/x)$ to your course, where $y$ is your velocity away from the start, which you have to shed, and $x$ is the velocity of the station, which you have to acquire. Leave it to the cunning laser on the station's ability to come from a different effective direction to its actual direction to supply thrust from a constant direction relative to you.

This is not the most power-efficient way to do the voyage, but it makes the least demands on the laser at the start point and on the ship. Light-sail ships are even more weight-critical than ordinary spacecraft, and swinging your sail around continuously while you're riding the station's beam takes (a) reaction mass and (b) keeping the engines and control system that use that reaction mass working. All of that costs weight. Avoiding the need to do those things makes the voyage more likely to succeed.

  • $\begingroup$ You bring up a good point! For your numbered list I can’t tell if #1 is meant as a way of composing the problem only or you plan to actually do that. $\endgroup$
    – JDługosz
    Commented Sep 3, 2016 at 9:12
  • $\begingroup$ Edited to clarify. $\endgroup$ Commented Sep 3, 2016 at 9:28
  • 1
    $\begingroup$ You can easily avoid interference between the lasers by just using lasers of different wavelength. Not that it matters much, given that the solar sail is going to be much larger than the wavelength of the laser (meters vs. nanometers), so the light arriving some hundred nanometers displaced won't matter. $\endgroup$
    – celtschk
    Commented Sep 7, 2016 at 7:59

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