Problem 12.10 Use the method of variation of parameters to solve Equation 1.113,
&&
x2
n&
x
n2xF
0
msin
t
(a)
u1  F
0
m
1
p1p2
cos
tp1sin
t
2p12ep1tC1
A i C1C2
BC1C2
ei
dtisin
dt
Problem 12.11 Use the method of variation of parameters to solve the differential equation
x+2x+2x=tsint
(a)
Problem 12.12 Show that the variation of parameters solution of the differential equation
x+a1t
( )
x+a2t
( )
x=f t
( )
(a)
is
x=u1t
( )
x1t
( )
+u2t
( )
x2t
( )
where
u1=x2f t
( )
x2x1–x2x1
dt
ò+C1u2=x1f t
( )
x1x2–x1x2
dt +C2ò
in which
x1
and
x2
are solutions of the reduced homogeneous equation
x+a1t
( )
x+a2t
( )
x=0
and
C1
and
C2
are constants
&
a
1
a2
x1
x2
3
2L

0
L

0
1,L,6
ta
u
u
u
u
Problem 12.15 If
a
is the two- body relative acceleration vector
u
u
r3
u
u
u
u
Problem 12.17 Combine the results of Problems 12.14 and 12.16 to show that
L
a b
t=0
a
,
b
=1, ,6
(a)
That is, on a given osculating orbit (i.e., for a given set of orbital elements
u
a
), the Lagrangian brackets
are constant. This means that
L
[ ]
may be computed at a point where the orbit formulas have their sim-
plest algebraic form, which usually is at periapsis (as it was for obtaining the specific energy formula in
Section 2.6).
t0
Problem 12.18 For the orbital elements
u1=h u2=e u3=
q
u4=
W
u5=i u6=
w
it can be shown that the Lagrange matrix is
L
[ ]
=
0 0 –1–e
1+e–cos i0–1
0 0 –eh
1+e
( )
20 0 0
1–e
1+e
eh
1+e
( )
20 0 0 0
cos i0 0 0 –hsin i0
0 0 0 hsin i0 0
1 0 0 0 0 0
é
ë
ê
ê
ê
ê
ê
ê
ê
ê
ê
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
ú
ú
ú
ú
ú
ú
ú
ú
ú
Solve Equation 12.47, where
P
{ }
=RhReR
q
R
W
R
w
Ri
ê
ëú
û
T
for the element rates
h
,
e
,
q
,
W
,
w
and
i
.
1
0 0 0 0 1
hsin i0
0 0 0 1
hsin i01
htan i
11e2
eh 0 0 1
htan i0
d
dt 1
hsin i
R
i
di
dt   1
hsin i
R
1
htan i
R
d
dt   R
h1e2
eh
R
e1
htan i
R
i
Problem 12.19 For the orbital elements
u1=
W
u2=i u3=
w
u4=a u5=e u6=tp
L
[ ]
=
0–nabsin i01
2nbcos i–na3e
bcos i0
nabsin i0 0 0 0 0
0 0 0 1
2nb –na3e
b0
–1
2nbcos i0–1
2nb 0 0 1
2n2a
na3e
bcos i0na3e
b0 0 0
0 0 0 –1
2n2a0 0
é
ë
ê
ê
ê
ê
ê
ê
ê
ê
ê
ê
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
ú
ú
ú
ú
ú
ú
ú
ú
ú
ú
12.47 with
P
{ }
=R
W
RiR
w
RaReRtp
ê
ë
êú
û
ú
T
to obtain the Lagrange planetary equations listed in Equations 12.50.
d
dt
di
dt
d
dt
0 0 2
na 0 0 0
0 0 b2
na4e
0 0 b
na3e
2
na b2
0 0 0 0
R
R
i
R
dt   2
n2e
tp
de
dt   b
na3e
R
b2
n2a4e
R
tp
dtp
dt 2
n2a
R
ab2
n2ea4
R
e
Problem 12.20 Show that
dr
dt =0
implies that
dr
dt =0
.
dt 0dr
dt 0