How to "set" the lightsail?

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: $$\begin{array}{|cc|}\hline \text{Value of }\beta & \text{Type of trajectory}\\ \beta =0& \text{circular Keplerian}\\ 0<\beta <\frac{1}{2}& \text{elliptical}\\ \beta =\frac{1}{2}& \text{parabolic}\\ \frac{1}{2}<\beta <1& \text{hyperbolic}\\ \beta =1& \text{rectilinear}\\ 1<\beta & \text{flipped hyperbolic}\\ \hline\end{array}$$ 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 $\beta \approx 2$.

There are two equations of motion for polar coordinates $(r,\theta )$: $$\begin{array}{}\text{(4.37a)}& \frac{{\mathrm{d}}^{2}r}{\mathrm{d}{t}^2}-r{\left(\frac{\mathrm{d}\theta }{\mathrm{d}t}\right)}^2=-\stackrel{\text{gravitational}}{\overbrace{\frac{\mu }{r^2}}}+\stackrel{\text{radiation}}{\overbrace{\beta \frac{\mu }{r^2}{\cos }^3\alpha }}\end{array}$$ $$\begin{array}{}\text{(4.37b)}& r\frac{{\mathrm{d}}^2\theta }{\mathrm{d}{t}^2}+2\left(\frac{\mathrm{d}r}{\mathrm{d}t}\right)\left(\frac{\mathrm{d}\theta }{\mathrm{d}t}\right)=\beta \frac{\mu }{r^2}\cos {\alpha }^2\sin \alpha \end{array}$$ 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 $\alpha \approx 0$. 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

• $\beta \approx 0.75$ (I've chosen a value for a normal hyperbolic trajectory)
• $\alpha \ne 0^{\circ }$. It could, but that might not be optimal.

McInnes goes through several substitutions, leading to $$\begin{array}{}\text{(4.41)}& r^3{\left(\frac{\mathrm{d}\theta }{\mathrm{d}t}\right)}^2=\mu [1-\beta {\cos }^2\alpha (\cos \alpha -\tan \gamma \sin \alpha )]{\cos }^2\gamma \end{array}$$ 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 $$\begin{array}{}\text{(4.44)}& v(r)=\sqrt{\frac{\mu }{r}}{[1-\beta {\cos }^2\alpha (\cos \alpha -\sin \alpha \tan \gamma )]}^{1/2}\end{array}$$ There's a fairly complicated relationship between $\gamma$ and $\alpha$, but it can be simplified for small $\gamma$: $$\begin{array}{}\text{(4.45,4.48)}& \frac{\beta {\cos }^2\alpha \sin \alpha }{1-\beta {\cos }^3\alpha }=\frac{\sin \gamma \cos \gamma }{2-{\sin }^2\gamma }\approx \frac{1}{2}\tan \gamma \end{array}$$ This integration is important when we try to find a relationship between $r$ and $t$. We integrate $(\text{4.44})$: $$\begin{array}{}\text{(4.46)}& {\int }_{r_0}^r\sqrt{r}\mathrm{d}r={\int }_{t_0}^t{(2\beta \mu \sin \alpha {\cos }^2\alpha \tan \gamma )}^{1/2}\mathrm{d}t\end{array}$$ Integrating this and substituting in $(\text{4.48})$ yields $$\begin{array}{}\text{(4.49)}& t-t_0=\frac{1}{3}(r^{3/2}-r_0^{3/2}){\left(\frac{1-\beta {\cos }^3\alpha }{{\beta }^2\mu {\cos }^4\alpha {\sin }^2\alpha }\right)}^{1/2}\end{array}$$ 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. $\beta <0.5$), ${\alpha }_{\text{max}}\approx {35.26}^{\circ }$. 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 $\tan \gamma \approx 1.362$, which gives us $\gamma \approx {53.7}^{\circ }$. 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 $$t=r^{3/2}\times 1.23\times {10}^{-10}$$ and for a final radius of three light-years ($2.838\times {10}^{16}$ meters), we find that $t\approx 5.88\times {10}^{14}$ 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 $$\frac{{\mathrm{d}}^{2}r}{\mathrm{d}{t}^2}-\frac{h^2}{r^3}=\frac{F(r)}{m}$$ 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 $$\begin{array}{}\text{(355)}& r(\theta )=\frac{l}{1+\epsilon \cos \theta }\end{array}$$ where $k=-GM$ (in general, $k=(\beta -1)GM$), and $$\begin{array}{}\text{(356)}& l\equiv \frac{m{h}^2}{k},\epsilon \equiv \frac{l}{a}-1\end{array}$$ 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 $$v_0=\sqrt{\frac{\mu }{r}}=2.97\times {10}^4\text{ m/s}$$ $$h=\frac{|L|}{m}=\frac{rmv}{m}=rv=4.46\times {10}^{15}{\text{ m}}^2\text{/s}$$ $$k=(\beta -1)GM=1.327\times {10}^{20}{\text{ m}}^3{\text{/s}}^2$$ $$l\equiv \frac{m{h}^2}{k}=3\times {10}^{14}$$ $$\epsilon \equiv \frac{l}{a}-1=2000$$ From this, I get $$r=\frac{3\times {10}^{14}}{1+2000\cos \theta }$$ $\epsilon >1$ (as was expected, given that $\beta >1$), and in fact $\epsilon \gg 1$.

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.

posted over 8 years ago

HDE 226868‭

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