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Artists depiction of the Japanese solar sail IKAROS

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.

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  • $\begingroup$ You need to consider two more factors: total effective area of the sail, and total mass of the craft (including sail and rigging). $\endgroup$ – Jerard Puckett Dec 2 '13 at 17:47
  • $\begingroup$ @JerardPuckett Yes, but these are already accounted for in the given equations, so I didn't feel the need to name them separately. You may edit the question to fit your request for clarity. $\endgroup$ – Vedant Chandra Dec 3 '13 at 10:19
  • $\begingroup$ You should definitely switch from km/h to km/s. The numbers become far less staggering, especially when compared to requirements for various space travel destinations. $\endgroup$ – SF. Dec 18 '15 at 11:40
  • $\begingroup$ No calculus is a weird requirement for a physical situation that by definition needs integration. $\endgroup$ – Jan Doggen Dec 18 '15 at 13:03
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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.

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    $\begingroup$ Shouldn't you also consider the gravitational force of the sun (assuming that the prob does not encounter any other celestial bodies closely). Since both forces are proportional to $1/r^2$, it would have the same effect if the gravitational parameter of the sun would be lowered, which would increase the eccentricity, possibly turning an elliptical orbit (before extending the solar sail) into an parabolic or even hyperbolic trajectory. $\endgroup$ – fibonatic Dec 4 '13 at 3:00
  • $\begingroup$ Doh! You are absolutely right about that, though it will not change the overall form of the equation, just some of the coefficients. Actually thinking about it that way, its no different than finding the velocity caused by gravity and I am sure there is probably a much cleaner approach using Lagrangian. I will revisit when I have had a chance to get some sleep. $\endgroup$ – Richard Hansen Dec 4 '13 at 3:14
  • $\begingroup$ I really appreciate your well thought-out answer, however as my clarified question shows, it is not what I am looking for. Could you please read the clarification and attempt another answer? $\endgroup$ – Vedant Chandra Dec 8 '13 at 9:05
  • $\begingroup$ The usual term for the ratio of a solar sail's thrust to its weight at any given distance is lightness, so you may wish to use $L$ instead of $K$. $\endgroup$ – Nathan Tuggy Dec 18 '15 at 6:18
  • $\begingroup$ @RichardHansen I have a series of questions related to this one could you help me with them? Nice answers by the way. space.stackexchange.com/questions/27906/… $\endgroup$ – Muze the good Troll. Jun 16 '18 at 20:53

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