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16.346 Astrodynamics
Fall 2008
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Lecture 25
The Restricted Problem of Three Bodies
#8.2
Jacobi’s Equations
If the third mass m3 ≡ m is infinitesimal, then
d2 r
dr dω
Gmm1
Gmm2
m
+
2ω
+
×
×
r
+
ω
×
(ω
×
r)
=−
ρ1 −
ρ2
3
2
dt
ρ1
dt
dt
ρ32
where
ρ 1 = r − r1
ρ 2 = r − r2
r = r3 −
with ω 2 =
ω = ω iζ
m1 r1 + m2 r2
= ξ iξ + η iη + ζ iζ
m1 + m2
G(m1 + m2 + m)
G(m1 + m2 )
≈
3
3
r12
r12
With m1 and m2 on ξ -axis, then
r1 = ξ1 iξ
ρ21 = (ξ − ξ1 )2 + η 2 + ζ 2
r2 = ξ2 iξ
ρ22 = (ξ − ξ2 )2 + η 2 + ζ 2
Now
ω × ( ω × r) = −ω 2 (ξ i ξ + η i η )
so that
d2 r
dr
Gm1
Gm2
2
+
2ω
×
=
ω
(ξ
i
+
η
i
)
−
ρ
−
ρ2
ξ
η
1
3
dt2
dt
ρ1
ρ32
Define
J (ξ, η, ζ) =
ω2 2
Gm1
Gm2
(ξ + η 2 ) +
+
2
ρ1
ρ2
Then
T
d2 r
dr
∂J
+ 2ω ×
=
dt2
dt
∂r
or
d2 ξ
dη
∂J
− 2ω
=
2
dt
∂ξ
dt
2
d η
dξ
∂J
+ 2ω
=
2
dt
dt
∂η
2
d ζ
∂J
=
2
dt
∂ζ
Jacobi’s Integral
Take scalar product with dr/dt
d2 r dr
dr dr
∂ J dr
+ 2ω ×
·
·
=
2
dt dt ∂rdt
dt dt =0
1 d dr dr
dJ
·
2 dt dt dt
dt
Integrate to obtain
16.346 Astrodynamics
Lecture 25
1
2
dr dr
+C
·
dt dt
=J
or
2
vrel
= 2J (ξ, η, ζ) − C
Hence
2
vrel
= ω 2 (ξ 2 + η 2 ) +
2Gm1
2Gm2
+
−C
ρ1
ρ2
Surfaces of Zero Relative Velocity
In ξ, η, ζ space, surfaces of zero relative velocity are
ω 2 (ξ 2 + η 2 ) +
2Gm1
2Gm2
+
= constant
ρ2
ρ1
In the ξ, η plane (in terms of bipolar coordinates) curves of zero relative velocity are
Gm1
ρ21
2
+
3
ρ
ρ1
+ Gm2
ρ22
2
+
3
ρ
ρ2
= constant
Note: In the two-body problem, curves of zero relative velocity are
2 1
−
=0
v =µ
r
a
2
or
r = 2a
Fig. 8.1 from An Introduction to the
Mathematics and Methods of
Astrodynamics. Courtesy of AIAA.
Used with permission.
16.346 Astrodynamics
Lecture 25
Rectilinear Oscillation of an Infinitesimal Mass
Jacobi’s integral
dζ 2
dt
=
4Gm
−C
ρ
where
Page 373
ρ2 = D 2 + ζ 2
and
ω2 =
Gm
4D3
4Gm
− v02 .
D
dζ 2
D
= v02 − 16ω 2 D2 1 −
dt
ρ
With velocity v0 when ζ = 0, then
Hence
C=
v02
Define ζ = D tan θ , ρ = D sec θ and B =
.
16ω 2 D2
dθ 2
Then
= 16ω 2 cos4 θ[B − (1 − cos θ)]
dt
dθ
dθ
Now
= 0 if and only if B ≤ 1 . Define θ = θm when
= 0 . Then B = 1 − cos θm .
dt
dt
dθ 2
Therefore,
= 16ω 2 cos4 θ(cos θ − cos θm )
dt
Define x = cos θ and xm = cos θm .
dx 2
= 16ω 2 x4 (1 − x2 )(x − xm )
Then
dt
1
dx
Let T be the quarter period. Then 4ωT =
with P (x) = (1 − x2 )(x − xm )
2
P (x)
xm x
1
P (x) 1 1 x dx
1 1
dx
+
Integrate by parts
4ωxm T =
+
x 2 xm P (x) 2 xm x P (x)
xm
Convert P (x) to fourth degree with the substitution x = 1 − z 2
α
α
(1 − z 2 ) dz
dz
4ωxm T =
+
2
Q(z)
0
0 (1 − z ) Q(z)
z2
z2
2
1 − 2 ≡ R(y) = (1 − y 2 )(1 − k 2 y 2 )
Q(z) = 2α 1 −
2
α
where we have defined α = 1 − xm , z = αy and k 2 = 12 α2 and obtain
1
1
√
(1 − 2k 2 y 2 ) dy
dy
4 2 ωxm T =
+
2 2
R(y)
0
0 (1 − 2k y ) R(y)
After some algebra, convert to Legendre form with y = sin φ to obtain
√
4 2 ωxm T = 2E(k) − K(k) + Π(2k 2 , k,
16.346 Astrodynamics
Lecture 25
1
2
π)