16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 35-36: Impulsive and Low-Thrust Maneuvers in Space
See Lectures 3-4 of 16.522 (Space Propulsion) for coverage of Low Thrust
Maneuvers and Re-positing within an orbit.
We add here material on Hyperbolic Orbits and Interplanetary Transfer.
Hyperbolic trajectories:
r=
p
1 + e cos θ
Asymptotes : e cos θ = -1
h=
µP
e>1
cos θ = -
1
e
⎛1⎞
θ = θ∞ = π − cos−1 ⎜ ⎟
⎝e⎠
still valid
Instead of “semimajor
axis” a>0 distance from
perigee to “center” is (-a),
and we still have
E=−
µ
>0
2a
Also
p = a 1 − e2 = ( −a) e2 − 1
(
)
(
)
Similarly, distance |focus center| is (-a)e (it is ae in
ellipse)
Turning angle:
⎛π
⎛1⎞
⎛ 1 ⎞⎞
δ = π − 2(π − θ∞ ) = π − 2 cos−1 ⎜ ⎟ = 2 ⎜ − cos−1 ⎜ ⎟ ⎟
e
2
⎝ ⎠
⎝ e ⎠⎠
⎝
⎛1⎞
δ = 2 sin−1 ⎜ ⎟
⎝e⎠
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 35-36
Page 1 of 7
Miss distance or Impact Parameter
∆ = (−ae) sin(π − θ∞ ) = (−ae) sin θ∞
or
∆=
Excess hyperbolic velocity
" C3 " = v2∞
∆ = (−ae) 1 −
e2 − 1
P
e
e −1
2
∆=
2
e
V∞ = 2E =
µ
=
( −a)
(
1
e2
P
2
e −1
)
µ e2 − 1
p
used in many books, reports
Orbit characterized by any pair of parameters, like
(p, e) , (vp , θ∞ ) , (v∞ , θ∞ ) , (v∞ , ∆) , (γp , v ∞ ) , (γp , δ)
Example : Given (v∞ , ∆) , -a =
then
δ = 2 sin−1
µ
v2∞
e2 − 1 =
∆v2∞
∆
=
−a
µ
⎛ v2 ∆ ⎞
1
= 2 cot −1 e2 − 1 = 2 cot −1 ⎜ ∞ ⎟
e
⎝ µ ⎠
Planetary “Spheres of Influence” (SOI)
Locus of points for which the ratio of (the Sun’s disturbing acceleration of the
relative vehicle-planet motion) to (the planet-induced vehicle acceleration) equals
the ratio of (the planet-induced acceleration of the relative vehicle-Sun motion) (the
Sun-induced vehicle acceleration).
This attracting centers, one vehicle
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 35-36
Page 2 of 7
JJJG
d2 rabs
G MP m → G Ms m →
=−
r−
m
R
2
dt
r3
R3
MP
JJJJJG
d2 rabs,p
dt
2
=−
G MP m → G Ms MP →
ρ
r−
ρ3
r3
→
→
⎛ →
⎞
G(MP + m) →
d2 r
−R
ρ
⎜
r + GMs
=−
+ 3⎟
2
3
2
⎜R
dt
r
ρ ⎟
⎝
⎠
Subt.
Accel. of vehicle
Relative to planet
Relative accel.
due to planet only
Sun’s effect on d. Notice: difference
between attractions vehicle and on planet
We could reverse the roles of planet and Sun, and get
→
→
⎛ →
⎞
G(Ms + m) →
d2 R
−ρ r ⎟
⎜
R + GMP 3 − 3
=−
2
3
⎜ρ
dt
R
r ⎟
⎝
⎠
So, SOI defined by locus of
→
Ms
→
ρ
R
−
ρ3 R 3
Mp + m
r
2
→
Mp
=
−ρ r
−
ρ3 r3
Ms + m
R2
Since m << Mp << Ms and r<<p, R,
→
→
Ms ρ
R 2 Mp 1 2
−
r R
Mp ρ3 R 3
Ms r2
−
Also,
→
→
→
R = ρ+ r
3
2
2
→
→
−3 2
R = (ρ + r + 2 ρ . r )
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
r
cos ϕ
ρ
→ →
⎛
⎞
2 ρ.r ⎟
−3 ⎜
ρ 1+
⎜
ρ2 ⎟
⎝
⎠
−3 2
→→
⎛
⎞
ρr
⎜
ρ 1−3 2 ⎟
⎜
ρ ⎟
⎝
⎠
−3
Lecture 35-36
Page 3 of 7
→
→
→
→
ρ.r
ρ.r G
→
→
→
→
1−3 2 → →
3 2 ρ
→
ρ
R
ρ
r ⎛ →
ρ ⎛
ρ
⎞ −r
⎞
r
1
−
−
ρ+
+
=
−
r − 3 cos ϕ 1ρ ⎟
⎜
⎟
⎜
3
3
3
3
3
3
3
R
ρ
ρ
ρ
ρ
ρ
ρ ⎝
⎝
⎠
⎠
Magn.:
r
ρ3
3r cos ϕ
ρ3
− cos ϕ
→
→
ρ
R
r
− 3 = 3 1 + 9 cos2 ϕ + 6 cos ϕ (1r − 1ρ )
3
ρ
R
ρ
(
)
12
r
1 + 3 cos2 ϕ
ρ3
2
⎛ MP ⎞ R 2
r3
2
1
+
3
cos
ϕ
=
⎜
⎟ 2
ρ3
⎝ Ms ⎠ r
(Mp Ms )
r
=
R
1 + 3 cos2 ϕ
25
(
25
)
1 10
But (1 + 3 cos2 ϕ )
1 10
→
r ⎛ Mp ⎞
⎜
⎟
R ⎝ Ms ⎠
is between 1 and 1.15
Planet
Mercury
Venus
Earth
Jupiter
SOI (Km)
113,000
617,000
924,000
48.3 × 106
Neptune
86.7 × 106
Moon
66,000 (in E-M system) (17.2% of FEM , not too small)
The Patched Conic Method for Interplanetary Transfers
Since the SOI is typically small compared to the interplanetary distances,
when dealing with trajectories that go from one planet to another we can make the
approximation that only one body is attracting the space craft at each time. This is
initially the first plane then the Sun, and finally the destination planet. Further, the
“hand-over” from one planet to the sun or vice-versa can be assumed to be at the
planet’s location, when viewed on the Solar System scale. Care must be taken to
converse momentum at these hand-over points.
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 35-36
Page 4 of 7
Example: Hohmann transfer from Earth to Venus. Assume no plane changes are
involved (in reality the Elliptical plane of the planetary orbits does not coincide with
the equation plane of either, but we ignore this complication).
As a preliminary, notice that this type of (minimum ∆ v) maneuver requires a
specific Earth–Venus configuration at launch; this configuration occurs once every ∆t
days, given by
∆t =
⎛ 1
⎜
⎝ Tv
1
1
=
= 584 days
1
1
⎞ ⎛1⎞
⎟ − ⎜ ⎟ 224.7 − 365.3
⎠ ⎝ TE ⎠
(1)
The Earth head angle at launch ( d0 in the sketch) is calculated by stating that the
time in the ITO (half the ellipse’s period ) is the same as that taken by Venus
between Vo and Vi :
32
1 2π ⎛ R SE + R SV ⎞
⎜
⎟
2 µs ⎝
2
⎠
= ( π + α0 )
R 3SV2
µs
(2)
or
⎡⎛ R + R ⎞32
⎤
SV
α0 = π ⎢⎜ SE
⎟ − 1⎥
⎢⎝ 2 R SV ⎠
⎥
⎣
⎦
(3)
Using Rsv = 0.7223 a.u , this gives α0 = 54.00 .
In the heliocentric part of the trajectory, the apohelion velocity (which the
craft must have as it leaves Earth) is
va =
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
µs
2 R SV
R SE R SE + R SV
(4)
Lecture 35-36
Page 5 of 7
while the Earth’s circular velocity is vE =
µs
. This is more than va , i.e., the
R SE
spacecraft must leave the Earth’s vicinity traveling backwards in the heliocentric
frame. Later on it will overtake Earth as it accelerates in the transfer free-fall
trajectory. The magnitude of this backwards velocity, which is the hyperbolic excess
velocity when viewed from Earth, is
⎛
2 R SV
v ∞,E = vE − va = vE ⎜ 1 −
⎜
R SV + R SE
⎝
⎞
⎟
⎟
⎠
(5)
The situation for time near launch, and when viewed in the Earth frame is as
shown in the following sketch:
Conversation of
energy in the
Earth’s frame gives
v2∞ ,E
2
=
2
µ
vPE
− E
2
rPE
(6)
where υPE is the velocity after application of the escape firing at rPE . Before this
firing, the space craft was in orbit, at vCE =
vPE =
v2∞,E +
µE
, and so we find
rPE
2µE
rPE
(7)
and therefore the single impulse needed to enter the ITO is
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 35-36
Page 6 of 7
⎞
r v2
µE ⎛
⎜ 2 + PE ∞,E − 1 ⎟
⎟
rPE ⎜
µE
⎝
⎠
∆v1 = vPE − vCE =
(8)
The point within LEO where this firing must occur is given, as shown, by δ , where
2
r
r
P,E
⎛1⎞
= 1 + P,E v2∞,E we
δ = 2 sin−1 ⎜ ⎟ is the total hyperbolic turning angle. Since e = 1 +
−
µE
a
( )
⎝e⎠
have all the elements to calculate this angle.
After traveling in the ITO, the spacecraft will approach Venus with an
overtaking excess velocity
v ∞,ν = vp − v v =
µs
R s,v
⎛
⎞
2 R SE
− 1⎟
⎜
⎜ R SE + R SV
⎟
⎝
⎠
(9)
and, working now in
the Venus frame
2
vp.v
µ
v2∞.v
=
− v
2
2
rρv
v c.v =
µv
rρv
(10)
(11)
So that the circularization ∆ v is
∆v2 = vρ,v − vc,v =
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
⎞
rpv v2∞,v
µ v ⎛⎜
2+
− 1⎟
⎟
rρv ⎜
µv
⎝
⎠
(12)
Lecture 35-36
Page 7 of 7