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:
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}];
Show[orbit]
This is the orbit:
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.
