Homework EAS4510 #11 Solution #1) Given:
4 + ( 4n 2 - U xx - U yy ) 2 + (U xx U yy - U xx 2 ) = 0
0 0 0 0 0
=
2
2 + b + c = 0 1,2 =
-b b 2 - 4c 2
(i) 1,2 = i
= i
=
0.7071 + 0.7071i 0.7071 - 0.7071i -0.7071 - 0.7071i -0.7071 + 0.7071i (ii) 1,2 = 4 9i
= 4 9i
2.6314 - 1.7101i
=
-2.6314 + 1.7101i
2.6314 + 1.7101i
-2.6314 - 1.7101i
(iii) 1.7101 + 2.6314i 1.7101 - 2.6314i -1.7101 - 2.6314i -1.7101 + 2.6314i (iv) 0 + 2.0000i (v) (vi) 2 2 2 -2 -2 3 -2 -3 3 0 - 2.0000i -3 0 - 2.0000i 0 - 3.0000i 0 + 2.0000i 0 - 2.0000i 0 - 2.0000i
(vii) 0 + 2.0000i (viii) 0 + 2.0000i
0 + 3.0000i
#2) Put in state space form for first order differential equations. x1 = x 1 0 0 x1 x1 0 x -3n 2 0 0 0 x x =x 2 = 2 = 0 a) 2 x3 0 x3 = y 0 0 1 x3 0 0 0 x4 x4 = y x4 0 0 0 0 0 1 0 0 2 -3n 2 0 0 0 0 0 0 - = det 3n det ( I - A ) = 0 = det 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
0 = 4 - 3n 2 2
-1 0 0 0 0 0 -1 0 0
= 0, 0, n 3
x1 0 x -3n 2 b) 2 = x3 0 x4 0 0 det ( I - A ) = 0 = det 0 0
1 0 0
0 0 0
2n 0
0
0 0
0 0 0
0 x1 -2n x2 =0 1 x3 0 x4 0 0 1 0 -3n 2 0 - 0 0 0 2n 0
0 0 2 0 -2n = det 3n 0 0 1 0 0 0
0 0 0 2n 0 -1 -2 n 0
-1
0 = 4 + n2 2
= 0, 0, ni
#3) Same system as #2b) above, of the form: x = Ax . Now take the Laplacian of this:
d x = ( Ax ) dt s X ( s ) - x(0) = AX ( s )
( sI - A) X (s) = x(0)
X ( s ) = ( sI - A ) x(0)
-1
Notice that in problem 2 I called the eigenvalues lambda, but for this problem we are dealing specifically with the s-domain, so just replace lambda with s. The inverse of a matrix the is adjoint of it divided by its determinant. The process for this problem involves symbolically finding ( sI - A )
-1
in a form where every
entry in the matrix is a separate function of s. Then, take the inverse Laplacian of that matrix to find a solution for x(t ) .
-1 -1 X ( s ) = -1 ( sI - A ) x(0) -1 x(t ) = -1 ( sI - A ) x(0)
When you take this inverse Laplacian (term by term), the terms should come out to look like the matrix given in #4. M N -1 x(0) -1 ( sI - A ) = S T #4) a) atransfer =
rchaser + rtar 6700 + 6800 = km 2 2 ntransfer atransfer 1.5 = 0.0346rad = 1 - atar atar = 6800km
( - lead ) =
ntar
lead
n = 1 - tar ntransfer
In the target coordinate system, the vectors r 0 and v 0 can be found now that lead is known.
rchaser cos lead - rtar r 0 = -rchaser sin lead 0 sin lead rchaser 0 -ntar 0 - - - ntar cos lead - 0 0 r 0 v0 = rtar rchaser 0 0 0 0 The last thing to evaluate is t f . This is the time at which rendezvous occurs.
-
-
-
TOF = t f - t0 =
( - lead ) =
ntar
ntransfer
M N Just say that t0 = 0 . Now, M and N in can be evaluated at t f , where S T n is the mean motion of the reference orbit, previously called ntar .
v(t0 ) = - N
(
-1 f
M f r 0 + v0
-
-
)
****(original problem statement had error)***
-0.0018 v(t0 ) = 0.0355
v(t0 ) = 0.0355 km/s
Now for Hohmann, we know that the delta-v magnitude is:
vHohmann =
rtar
atransfer rchaser
-
rchaser
=
(6800)
atransfer (6700)
-
6700km
= 0.0285km/s
We can easily see that the estimate from the CW equations is not the same as the actual delta-v needed for Hohmann. Why is this?