|Nov25-11, 07:47 AM||#1|
Equations of Motion of a Solar Sail HELP!
1. The problem statement, all variables and given/known data
I am reposting an edited version of this problem from a previous post of mine, due to it not being entirely relevant to that post, and also the question was asked after the thread had been replied to, so looks like an answered question. I also aim to give more detail here.
In the American Rocket Society Journal, number 29 page 422-427, there is an article which details, amongst other things, the equations of motion for a solar sail. I would like to know how these have been arrived at.
"Let us now consider a ship propelled by solar sail in interplanetary space. The forces acting on it are Fg (the sun's gravitational force) and Fs, the force due to radiation pressure on the sail, and is equal to pA (p=p0 cos^2θ (r0/r)^2 where p0 is solar radiation pressure on a normally reflecting surface at Earth orbit, r0 = 1AU; A is the area of the sail)
Space drag is not considered. We also neglect gravitational forces due to other planets.
Under the specified conditions, the equations of motion are:
(-Fg + Fs cos θ)/m = du/dt - v^2/r ; where u is radial velocity and v is tangential velocity (making v^2/r angular velocity?) and r is distance from Sun, and m is total mass of spacecraft
(-Fs sin θ)/m = dv/dt + uv/r"
I can't fathom how these have been arrived at. They appear to be resolving horizontally and vertically, but why is the radial component involved in the vertical resolution and vice versa?
Also, what is the value obtained my multiplying radial and tangential velocity?
Any hints/tips would be great, or links to other resources that explain the same thing.
2. Relevant equations
3. The attempt at a solution
|Nov25-11, 08:33 AM||#2|
After further investigation, I have since discovered that these are just a fairly obscure way of writing tangential and radial accelerations.
However, tangential acceleration is given by Equation 286 on this page http://farside.ph.utexas.edu/teachin...es/node89.html
which upon conversion to the notation used in this article becomes:
dv/dt = 2uv/r
So where has the factor of 2 gone in the article?
|Nov25-11, 09:14 AM||#3|
So, backing up a bit, let me try to answer, in part, your original question. I'm not going to give a complete answer (yet) because this looks a bit too much like homework. Besides, you can probably find a complete answer on the 'net with a little bit of searching. What I will do is provide an outline for the solution.
Even though photons are massless, they still carry energy and momentum. It is the momentum carried by photons that are the key to understanding how solar sails work. When a photon hits an object such as a solar sail, the photon will either be reflected or absorbed by the object. Reflection can be further sub-classified as specular (mirror-like) or diffuse (like a dull, matte finish). A good solar sail will reflect light rather than absorb it, and the reflection will be primarily specular. The equations of motion that you described assume 100% specular reflection.
When light is reflected by some object, the incoming photons will transfer their momentum to the object and the object will transfer some of its momentum back to the outgoing (reflected) photons. When light is reflected specularly, the angle of reflection is equal to the angle of incidence. The net momentum transfer is solely along the normal to the surface.
See if you can take it from here (and see if you can determine why the net momentum transfer is solely along the normal to the surface).
|dynamics, equations of motion, kinematics, solar sail|
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