Can I derive a combined equation for velocity of solar sail?

Question

In this article on The Planetary Society website, I read this about solar sails:

At an acceleration rate of 1 millimeter per second per second (20 times greater than the expected acceleration for Cosmos 1), a solar sail would increase its speed by approximately 310 kilometers per hour (195 mph) after one day, moving 7500 kilometers (4700 miles) in the process. After 12 days it will have increased its speed 3700 kilometers per hour (2300 mph).

I find these numbers quite staggering, as they open up a variety of avenues in interstellar travel.

Now I know that this acceleration won't quite be constant, becuase as the craft travels away from the Sun, the force exerted by light decreases. This article highlights this by saying:

The force on a sail and the actual acceleration of the craft vary by the inverse square of distance from the sun (unless close to the sun), and by the square of the cosine of the angle between the sail force vector and the radial from the sun, so

$$F = F_0 \frac{\cos^2 θ}{R^2}$$

(perfect sail) and

$$F = F_0 \frac{(0.349 + 0.662 \cos{2θ} − 0.011 \cos{4θ})}{R^2}$$

(realistic sail)

I have asked for a detailed explanation of this equation in my question, but their general meaning is that the light force on the sail is inversely proportional to the distance between the sail and the Sun.

Finally, this same article calculates the light force on a solar sail at 1 AU from the Sun thusly:

The momentum of a photon or an entire flux is given by $p = \frac{E}{c}$, where $E$ is the photon or flux energy, $p$ is the momentum, and $c$ is the speed of light. Solar radiation pressure is calculated on an irradiance (solar constant) value of $1361\ W/m^2$ at 1 AU (earth-sun distance), as revised in 2011: An actual sail will have an overall efficiency of about 90%, about $8.25\ μN/m^2$

How can I combine these three factors: Force imparted, distance to the sun, and the resulting acceleration, and derive an equation giving $v$ = velocity of craft, with all the factors accounted for?

UPDATE for clarity: I am NOT looking for a differential/integrated equation which actually calculates the final velocity as the craft moves. I am looking for an equation with velocity on one side, and the force, distance and initial velocity etc. on the other, an equation into which I can plug the values in manually to get the velocity at a given time. So no calculus please, I want a simple velocity vector formula.

Answer 1

First, we need an equation for the acceleration:

$$F = \frac{F_0}{R^2}$$

One thing to note is that this equation is missing a factor (the units don't work out). Technically, $F_0$ needs to be multiplied by $1\ AU^2$, that way the units cancel out properly.

Assuming an optimal angle, so:

$$A = \frac{F_0 \cdot 1AU^2}{M\cdot R^2}$$

Where $M$ is the mass of our probe. Your cited $F_0$ is 8.25 µN per square meter of sail. so:

$$A = \frac{8.25\cdot 10^{-6} N/m^2 \cdot \text{Sail Area} \cdot 1 AU^2}{\text{Mass} \cdot R^2}$$

We will need to convert our $\text{Sail Area}$ to $AU^2$ in order to work with that force equation and have units properly cancel.

However, we can rewrite this as:

$$A = \frac{K}{R^2}$$

Where $K$ is just a constant based on the $\text{Sail Area}$ and $\text{Mass}$ (I would have used $C$, but worried people might mix it up with the speed of light).

$$K = \frac{8.25*10^{-6} N/m^2 \cdot \text{Sail-Area} \cdot 1 Au^2}{\text{Mass}}$$

This means we are going to have to deal with differential equations.

$$A = \frac{d^2R}{dt^2} = \frac{K}{R^2}$$

$$A = \frac{dV}{dT}$$

$$\frac{d^2R}{dt^2} = \frac{dV}{dT} = \frac{dV}{dR} \cdot \frac{dR}{dT} = \frac{dV}{dR} \cdot V$$

$$\frac{dV}{dR} \cdot V = \frac{K}{R^2}$$

so

$$dv*V = \frac{K}{R^2} * dR$$

Indefinitely integrate

$$\frac{1}{2} V^2 = \frac{-k}{R} + P$$

$$V = \sqrt{\frac{-2k}{R} + 2P}$$

$P$ is an additive constant, once again, trying to avoid using $C$. Our additive constant $P$ will depend on our starting velocity at $1 AU$. If our initial velocity is $0$ for example,

$$\frac{1}{2} 0^2 = \frac{-k}{1 AU} + P$$

$$P = \frac{k}{1 AU}$$

But it will vary based on that initial condition.

So we have an expression for $V$ based on distance from the Sun. However, I suspect that you want an answer in terms of time traveled, not distance traveled (if not we are done).

so $V=\frac{dR}{dt}$ so

$$\frac{dR}{dt} = \sqrt{\frac{-2k}{R} + 2P}$$ $$\frac{dR}{\sqrt{\frac{-2k}{R} + 2P}} = dT$$

Integrate

At this point it is horrible. I tried doing an arctan substitution but, nothing seemed to work. So I dumped it into Mathematica.

T = (2*Sqrt[p]*Sqrt[p + k/r]*r - k*Log[k + 2*p*r + 2*Sqrt[p]*Sqrt[p + k/r]*r])/ (2*Sqrt[2]*p^(3/2))

At this point you would need to solve for $R$ in terms of $P$, $K$, and $T$. Something that does not seem like a fun proposition. You plug that back into your velocity equation to get the velocity as a function of time.

If anyone sees an easier way to end up that math, please let me know. Maybe someone else can take it from here.

Actually it basically boils down to $T = R-\ln{(R)}$ as a horrible approximation.

This actually makes a lot of sense. Initially we are going to have something that looks somewhat quadratic, however as we get beyond $10 AU$ or so (will really depend on constants), it really becomes more or less linear as the acceleration drops to nearly zero and we are just going at a cruising speed.

Google $x-\ln(x)$ and zoom the graph out a bit to see.

So distance is more or less linear with time, and $V$ hits some constant, which we can find looking at the velocity equation.

We can take the limit of $V = \sqrt{\frac{-2k}{R} + 2P}$ as $R$ goes to infinity. It becomes

$$V = \sqrt{2P}$$

is what our velocity asymptotically approaches. Using the value I came up with for $P$ earlier using the starting conditions.

$$\sqrt{\frac{-2k}{R} + \frac{k}{1AU}} = \sqrt{k \left(\frac{1}{1AU} - \frac{2}{R}\right)}$$

Once $\frac{2}{R}$ is less than a hundredth of $\frac{1}{AU}$ we have more or less reached our final velocity (we are only 1/10th away from it).

So

$$\frac{2}{R} = \frac{1}{100 AU}$$

$$R = 50 AU$$

Which happens to be just beyond Pluto's Orbital distance.

Kind of disappointing. However, this is in line with literature that I have read that shows outside of our solar system solar sails are not an efficient means of propulsion. This of course changes if you have some sort of laser system shooting at it.