# Tag Info

36

No, because there's nothing like water for a keel to work against. In water sailing there are two force vectors, the vector from the reaction of the wind against the sail, and the vector from the keel and rudder against the water. These vectors add together to propel the sailboat. This works for almost any direction on the compass except where the wind ...

33

You cannot directly propel the solar sail towards the sun. A solar "sail" is basically a mirror. The analogy of wind and sails on ships is not useful for understanding how solar sails work. Each photon from the sun which strikes the sail is reflected. Each photon imparts a small amount of momentum. If the sail is pointed directly at the sun then you get ...

25

This is actually somewhat easier than you would think. In the world of Orbital Dynamics, you only have to accelerate or decelerate your orbit to move closer/further away from the object you are orbiting. So, all you have to do is create a net momentum that pushes to slow down your orbital velocity. However, a big part of what makes tacking work is the fact ...

15

If you're asking about interstellar travel, then the answer is pretty simple; For diffuse, unfocused sources of light like the one emitted by stars, photon flux density decreases with the inverse square of the distance to its source, and with it photon pressure (imparted momentum of absorbed or, better yet, reflected photons) on the sail. So you accelerate ...

13

First, there are several forces acting on a sailcraft. Aerodynamic drag Solar radiation pressure Gravity field non-sphericity Electrodynamic drag If you go anywhere lower than 770 km, atmospheric drag is by far the largest force that pulls your apoapsis down. Quoting Vulpetti (2008): When unfurled in LEO, the drag on the sail produced by its flight ...

13

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 ...

13

If you're already in a solar orbit, then yes. You can use a sail at an angle and send the reflections prograde. The result is to reduce your orbital energy and you spiral in. I recall it was a standard physics problem to find the angle that maximized the energy transfer (it's not 45 degrees). Time of journey depends on the mass of your item and the ...

12

The answer is a clear no because Hubble Space Telescope (HST) uses a fixed focal length of 57.6 m. The only thing you'd achieve by re-purposing it as a solar concentrator would be to melt its focal plane assembly and likely everything around it. HST is a Cassegrain reflector and its mirror assembly is utterly unsuitable to applications where adjustable focal ...

11

You could expect to get a reasonable propulsion only if you plan ahead. Specifically, if you were to shine a very high powered laser on said solar sail from somewhere, you might be able to get propulsion even further out. Without a laser, you really can't do much once you get far away. As this table shows, here's a few solar constants at the orbits of ...

11

A few years ago the Japanese spacecraft Ikaros traveled from Earth to Venus using a solar sail. See http://en.wikipedia.org/wiki/IKAROS. From the article: The IKAROS probe is the world's first spacecraft to use solar sailing as the main propulsion. Other spacecraft (noteably Hayabusa) have used solar pressure for attitude control. Again, quoting ...

11

As far as I know, no solar sail has been used for any purpose, other than demonstration. Well, except for MESSENGER. It was not designed as a solar sail, but since it flies so close to the Sun, they ended up using it like one. They were able to use solar sailing instead of firing rockets for trajectory correction maneuvers for Mercury flybys. Sunjammer ...

10

As long as other variables are held constant (such as the orientation with respect to the direction to the sun, reflectivity of the sail surface, etc.) the force produced by a solar sail is always proportional to its area—at least, until the sail size gets rediculously large, a significant fraction of the distance to the sun!—so bigger always produces more ...

10

Ignoring the ISS, the question is simply whether light pressure on the sail can counteract the drag on the sail from atmosphere. Light pressure near Earth is about $10 \mu Pa$ (with optimum geometry and no eclipse). So the question is what is the drag force per square meter of sail. We know that the ISS experiences about 0.2N of drag, and its area is ...

8

The problem is that you have to go from low-speed regime in the atmosphere (the balloon), to a high-speed regime out of the atmosphere (~7000m/s). The low forces available from a solar sail don't allow this to happen quickly enough. At an intermediate speed it is either destroyed by forces from interacting with the atmosphere or it is insufficiently ...

7

How large would the sail need to be for a five day transfer? Too large. Solar sails are typically not very useful to increase your semimajor axis when orbiting a body other than the Sun (eg Earth). This is because when you're travelling away from the sun, the solar pressure will accelerate you, and when you're travelling towards the sun, it will slow you ...

7

Between atmospheric drag out through almost 1000km, there's the issue that solar light and solar wind are both cut by the passing into the shadow of a planet. Since this puts more of the energy on the solar side, even for high orbits, the sail is, if passively deployed, making the orbit more ellipsoid and thrusting it planetward on the sunward lobe. Let's ...

7

No. They degrade gracefully with small punctures, with the degradation approximately proportional to the area ratio of the pinholes to the total sail. So almost not at all.

7

While you can't do tricks common for normal sails due to lack of water to keep your keel from drifting sideways and normally lets the ship travel upwind, you are still able to extract force diagonal to the sun radius (the angle of incidence equals the angle of reflection; resulting force is perpendicular to the surface), aimed outside the Solar system. Now, ...

7

You need 1000km of altitude to break even between air drag and light pressure of sunlight. Below that the solar sail acts as a parachute, dragging against the remains of the atmosphere. You must reach 1000km altitude by other means. Definitely not a balloon, and not an orbital gun. Also, solar wind (protons and ions) accounts for about 0.5% of "thrust" ...

7

Intro This is a tricky question, primarily because solar sails benefit from a low mass spacecraft and with the addition of an electric propulsion system, you are increasing the mass decreasing the effective of acceleration and total delta V capabilities. System Architecture and Supporting Systems Solar Sails – Solar sails require photons to gain ...

6

There is a paper on what orbital drag looks like as a function of orbit. In order to determine if the solar sail would work to raise the orbit, let's make a few assumptions. Approximately 1/4th of the orbit, the sail will be providing useful energy. That is probably a high assumption, but I'll put it out there. Drag is a constant effect. The orbit is ...

6

Yes, it is entirely possible for lasers to give a propulsive power. In fact, there is a fairly commonly discussed idea of pointing a large laser at a solar sail to accelerate it faster than would be possible by just the sail itself. The key is, the lasers cannot be on the spaceship, they have to be away from the spaceship. And yes, given these laws, lasers ...

6

Rotate the sail. Centrifugal force will help maintain flatness. Unless there's some engineering requirement that dictates otherwise, it would probably work better to spin both craft and sail. Getting the rotation on the craft up to speed before deploying the sail should do the trick. JAXA's IKAROS solar sail demonstrator used rotational deployment. However, ...

6

Thrust is a reaction force, so yes this formula provides instantaneous thrust. I think what you're asking is how does thrust (force) translate into velocity. Force is a rate of change of momentum (i.e. F=ma). To get velocity you need to integrate that over time. If you think about it, you're given F and m, you solve for acceleration and then integrate over ...

6

tl;dr: Can we get close to the speed of light with that? No, at least not very easily. The terminal velocity $v_{\infty}$ is only about 0.2% the speed of light if you start at 1 AU using a 10 nanometer thick sail, and scales only as the inverse square root of the distance to the Sun where you start accelerating (as well as sail thickness), so you'd melt ...

5

This is not a complete answer (there really isn't one, since what's considered "reasonable" will vary with the person answering), but: With current technology, assuming that we didn't do fabrication in space (which would allow much thinner sails that don't require folding and packing in a rocket), and without relying on anything that hasn't been developed ...

5

In sailing, tacking is used to sail as close to the wind as you can (obviously not straight into it, though the hard wing Americas Cup boat might have been able to) while still generating lift in the sail. For normal boats this means at best a 25-30 degree angle to the wind. Thus you sail a zig zag pattern and need to tack, else you get way off course. ...

5

In his text Solar sailing: Technology, Dynamics and Mission Applications (see here for some of his open access titles), Colin R. McInnes derives the pressure on a perfect sail face on to Sol as $$P(r) = (L/(3 \pi c R^2)) * (1 -[1 - (R/r)^2]^{1.5})$$ $P$ is photon pressure $c$ is speed of light $r$ is distance from center of Sol $L$ is luminosity of Sol $R$ ...

5

This response addresses the last part of the question: "what is the proportion between "propulsive power" between the two accelerating factors of solar sail?" Solar Wind is between 1-6x10-9 N/m2 "The wind exerts a pressure at 1 AU typically in the range of 1–6 nPa (1–6×10−9 N/m2), although it can readily vary outside that range." Solar Radiation ...

5

Simply white-painting an asteroid probably isn't enough to avoid a predicted collision in typical cases according to my back-of-the-envelope math. A 10km diameter asteroid has a mass of around 1e15kg. According to Wikipedia, at 1 AU from the sun (i.e. in an orbit near that of Earth) you get about 9 µN/m^2 of radiation pressure via perfect perpendicular ...

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