F6 in Cartesian coordinates v1a

F.6 Force analysis of the Dumbbell Satellite in Frame S' in Cartesian Coordinates The only true forces on a dumbbell mass are gravity and stick tension T. From ** we then write F'1 = - (GMEm1/r'13) r'1 - T 1 = - (GMEm1/r'13)( b + r1) - T 1 F'2 = - (GMEm2/r'23) r'2 - T 2 = - (GMEm2/r'23)( b + r2) - T 2 . (F.6.1) As noted earlier, the center of gravity is not quite at the center of mass in Fig (F.1.1), but the above equations are exact despite this fact. The force are primed because they are forces in the inertial Frame S'. The reader is reminded that we are using the "swap notation" where prime↔noprime relative to the non-swap notation. In order to use Newton's Law in rotating Frame S, we must include the fictitious forces. We translate the result of (8.1.8) to swap notation to obtain Ffict,1 ≈ – m1S' – m1ω x (ω x r1) – 2m1 ω x v1 – m1 x r1 Ffict,2 ≈ – m2S' – m2ω x (ω x r2) – 2m2 ω x v2 – m2 x r2 (F.6.2) frame centrifugal Coriolis Euler In these equations, the b acceleration is given by the swap version of (7.13) which then states S' = x b + ω x (ω x b) . // Special Case #1 (F.6.3) and the vectors b and ω are given by ω = ω b = b . (F.6.4) Finally we may state Newton's Law for each mass, Feff,1 = m1 a1 (F.6.5) ≈ - (GMEm1/r'13)r'1 - T 1 – m1S' – m1ω x (ω x r1) – 2m1 ω x v1 – m1 x r1 Feff,2 = m2a2 (F.6.6) ≈ - (GMEm2/r'23)r'2 - T 2 – m2S' – m2ω x (ω x r2) – 2m2 ω x v2 – m2 x r2 where we have now a set of six scalar equations. Using (F.1.9) through (F.1.11), (F.6.6) can be rewritten, Feff,2 = - m1a1 (F.6.7) ≈ - (GMEm2/r'23)r'2 + T 1 – m2S' + m1ω x (ω x r1) + 2m1 ω x v1 – m1 x r1 Adding (F.6.5) and (F.6.7) gives 0 = - (GMEm1/r'13) r'1+ - (GMEm2/r'23)r'2 - (m1+m2)S' (F.6.8) This equation is just F = ma in inertial Frame S' for the total satellite where b is the center of mass. Ignoring the small offset between center of mass and center of gravity, the three equations (F.6.8) describe the circular orbit of the satellite around the Earth. We may then regard the equation (F.6.5) as a set of three scalar equations for the three unknowns θ,φ and T where recall r1 = (r1,θ,φ) in the spherical coordinates of Fig (F.1.1). Our next task is to write vector equation (F.6.5) in Cartesian coordinates to obtain the three equations of motion. After expanding the left side, we then consider the right side of (F.6.5) one term at a time: Left side of (F.6.5): m1a1 = m1(ax + ay + az ) (E.3.6) (F.6.9) Term 1: - (GMEm1/r'13)r'1 = - (GMEm1/r'13)(b + r1) = - (GMEm1/r'13) [ b + x + y + z ] = - (GMEm1/r'13) [ x + y + (b+z) ] Term 2: - T 1 = -(T/r1)r1 = -(T/r1) [ x + y + z] Term 3: – m1S' = – m1 x b - m1 ω x (ω x b) // (F.6.3) = - m1ω x (ω x b) // satellite in circular orbit, = 0 = - m1(ωb)ω + m1ω2b //- A x (A x C) = -(AC)A + A2C = m1ω2b = m1ω2b (F.6.12) Term 4: – m1ω x (ω x r1) = -m1(ωr1)ω + m1ω2r1 // identity shown above = -m1ω2(r1) + m1ω2r1 = -m1ω2(x) + m1ω2 [ x + y + z] = m1ω2 [ y + z] Term 5: -2m1 ω x v1 = -2m1 [ω] x ( vx + vy + vz) = -2m1ω [ vy - vz] (F.6.14) Term 6: – m1 x r1 = 0 because we assume = 0 (F.6.15) Having all the bits and pieces, we now assemble the three component equations of (F.6.5). The numbers show the Term above associated with each piece: Feff,1 = m1 a1 (F.6.5) ≈ - (GMEm1/r'13)r'1 - T 1 – m1S' – m1ω x (ω x r1) – 2m1 ω x v1 – m1 x r1 1 2 3 4 5 6 : m1 a1x = - (GMEm1/r'13) x - (T/r1)x 1 2 : m1 a1y = - (GMEm1/r'13) y - (T/r1)y + m1ω2y + 2m1ωvz 1 2 4 5 : m1 a1z = - (GMEm1/r'13) (b+z) - (T/r1)z + m1ω2b + m1ω2z - 2m1ω vy 1 2 3 4 5 Rewrite these trying to simplify m1 a1x = - [(GMEm1/r'13) + (T/r1)]x m1 a1y = - [(GMEm1/r'13) + (T/r1) - m1ω2]y + 2m1ωvz m1 a1z = - (GMEm1/r'13) (b+z) - (T/r1)z + m1ω2(b+z) - 2m1ω vy We now rewrite the three equations dividing by m1 and using (F.5.1) that GME = ω2b3 : a1x = - [(ω2b3/r'13) + (T/m1r1)]x a1y = - [(ω2b3/r'13) + (T/m1r1) - ω2]y + 2ωvz a1z = - (ω2b3/r'13) (b+z) - (T/m1r1)z + ω2(b+z) - 2ω vy and now switch to dot notation = - [(ω2b3/r'13) + (T/m1r1)]x = - [(ω2b3/r'13) + (T/m1r1) - ω2]y + 2ω = - (ω2b3/r'13) (b+z) - (T/m1r1)z + ω2(b+z) - 2ω x2+y2+z2 = r12 where r'12 = (r1+b2) = r12 + b2 + 2 r1 b = r12 + b2 + 2 r1 [b] = r12 + b2 + 2bz This is a system of 4 equations in 4 unknowns x,y,z,T . Rewrite the last equation = ω2[-(b3/r'13) +1] (b+z) - (T/m1r1)z - 2ω The plan is to eliminate z and then have 3 equations in just x,y,T. I will do this in "satellite in Cartesians v4.mws". Define Q = (T/m1r1) so things are simpler 1 = - [(ω2b3/r'13) + Q]x 2 = - [(ω2b3/r'13) + Q - ω2]y + 2ω 3 3 = ω2[-(b3/r'13) +1] (b+z) - Qz - 2ω 4 x2+y2+z2 = r12 My first shot fails with the usual error! Could it be because Q(t) has no derivatives?? What simple test could I do on this? = x+2 = y-3 Here there is no y derivative, will this complain? Yes it does complain. So I need to eliminate Q(t) from my system of equations. How does this work with the above set of equations? 1 = - (ω2b3/r'13)x - Qx 2 = - (ω2b3/r'13)y - Qy + ω2y + 2ω 3 = ω2[-(b3/r'13) +1] (b+z) - Qz - 2ω You could solve equation equation for Q to get Q = A Q = B Q = C Then you would have three equations which do not involve Q: A = B A = C B = C But subjecting the first two equations gives the third, so only two are independent. So I will do this in effect as follows. Start with 1 and 2 1*y y = - (ω2b3/r'13)xy - Qxy 2*x x = - (ω2b3/r'13)yx - Qyx + ω2yx + 2ωx Subtract to eliminate Q y- x = - ω2yx - 2ωx Now repeat using the first and third equation 1*z z = - (ω2b3/r'13)xz - Qxz 3*x x = ω2[-(b3/r'13) +1] (b+z)x - Qzx - 2ω x or 1*z z = - (ω2b3/r'13)xz - Qxz 3*x x = [-(ω2b3/r'13) +ω2] (b+z)x - Qzx - 2ω x Temp define A ≡ (ω2b3/r'13) so this then says 1*z z = - Axz - Qxz 3*x x = [-A +ω2] (b+z)x - Qzx - 2ω x Subtract to eliminate Q z - x = - Axz - [-A +ω2] (b+z)x + 2ω x or z - x = - Axz + [A - ω2] (b+z)x + 2ω x z - x = - Axz + [A - ω2] (bx+zx) + 2ω x z - x = - Axz + Abx +Azx - ω2(b+z)x + 2ω x z - x = Abx - ω2(b+z)x + 2ω x z - x = (ω2b3/r'13)bx - ω2(b+z)x + 2ω x So we then end up with these two equations of motion: y- x = - ω2yx - 2ωx z - x = (ω2b4/r'13)x - ω2(b+z)x + 2ω x r'12 = r12 + b2 + 2bz Maple Problem: Enter eq1 and eq2, eliminate z as usual, and see if Maple can do a numerical solution. Do this in "satellite in Cartesions v5.mws". I am able to get it to fly, but the odeplot runs out of memory. Here is my code: Are these correctly entered? Yes they are. Next, here is how to eliminate z: Next we have This all looks good. Now we come to the dsolve call: There are only two equations, functions unknown are x(t) and y(t), and there are the inits. I start m1 at location x = 0 and y = 0 and no velocity. But then I ask if for f(0.1) and it locks up! So maybe I can try some other methods now. I started with classical and it hangs on f(0.1). I try rfk45 and get this My error was that I had rp1 and r1p both present. Now it works better. and I get finally some kind of semi-reasonable (perhaps) plot, It starts at the right cusp and goes round the loop and repeat Here is just less than one loop I think as it starts to move, there is Coriolis pushing it to the right. This is I think a big breakthrough in my Cartesian satellite plotting efforts, even though I don't understand the result and I may have lots of math errors still. But it is a reasonable motion! This would be my "out of plane" motion, and I do expect a deviation. Let's try x(0) = 0 and y(0) = 1, then I expect to get the "in plane" motion: Here it stays in the x = 0 plane and swings back and forth. This looks VERY good, I will now take a break, it is 2:30 PM. Now let's try for a circular path?