16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 32: Orbital Mechanics: Review, Staging
Mission Planning, Staging
The remaining lectures are devoted to Mission Planning and Vehicle Design, which in
reality occurs even before the rocket engines are fully specified (although iterations
continuously proceed throughout the process, and engine characteristics do affect
the mission plan).
Very roughly, the iteration steps in planning a launch mission are:
(a) Estimate the required ∆VTOTAL using impulsive thrusting formulae, plus addons for gravity losses, drag losses, turning losses, etc.
(b) Distribute this ∆VTOT optimally among vehicle stages (since all orbit launches
so far require multiple stages in order to avoid carrying empty tankage in the
later stages).
(c) Using the mass fractions from (b), perform more detailed flight simulations
and refine the partial and total ∆V for the mission.
During stage (b), the total ∆V is assumed to be unchanged when the mass
distribution for the stages is varied. This is not strictly true, because often the
mission optimization leads to changes in the altitude and velocity at which the
various firings are executed and, as we will see, this may alter the various ∆V ’s.
This is the role of stage (c) above.
Another point to be made is that “stages” and “firings” may not map one-toone. A given stage may be turned off, allowed to coast, and then re-ignited. Or the
firing of two consecutive stages may occur with no interruption (or minimal
interruption), so that both can be idealized as occurring in the same place. As long
as the ∆V ’s are still regarded as insensitive to mission profile details (as per the
comment above), these distinctions do not impact the stage mass calculations, but
they can be of great practical importance nonetheless.
Impulsive Thrusting-Gravity Losses. Because of the large accelerations imparted by
rocket engines, their firings are usually short, from under one minute to about 10
minutes. In fact, there is a performance incentive in minimizing the firing time, as
long as the accelerations remain below structural or other limits. This can be most
easily seen in the context of a vertical ascent against gravity. The vehicle’s equation
of motion is then (ignoring drag)
m
and
dv
= F − mg
dt
•
F = m c = −c
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
dm
dt
(1)
(2)
Lecture 32
Page 1 of 16
dv
d ln m
= −c
−g
dt
dt
(3)
and integrating,
∆V = V − V0 = c ln
m0
− gt
m
(4)
The “ideal”, or gravity-free velocity increment is the familiar ∆Videal = c ln
m0
m
But the presence of gravity reduces the velocity increment by ∆VGravity = gt
(5)
(6)
which can be made insignificant if t is short, but can be very important otherwise. In
the limit when the thrust is barely enough to cancel weight, the vehicle just hovers
indefinitely with no velocity gain.
In practice, the significant item is the fuel used in the firing, which is
contained in the mass ratio m0/m.
The common procedure is then to first ignore gravity, as if the firing was
impulsive (t=0), and calculate the ∆V required for the mission under this
assumption. In our simple ascent example, the “mission” is to reach a velocity V,
starting at V0, and so the impulsive ∆V is simply V-V0. From (4) then
c ln
mo
= ∆Vimp. + gt
m
(6)
and so the extra ∆VGrav. = gt is added on as a correction, with the implication of
additional fuel being used for a given V-V0.
In a more general ascent trajectory (but still over a “flat Earth”, since gravity
losses occur only near the beginning of flight, before the path becomes nearly
horizontal) we would have
dv
F
=
− g sin γ
dt
m
(7)
V
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
dγ
= −g cos γ
dt
(8)
Lecture 32
Page 2 of 16
Here we assumed thrust to be aligned with velocity. This is called a gravity
turn, and is not the most general maneuver. It is, however, the most economical
strategy for turning, since any lateral component of thrust uses propellant without
adding flight energy.
Formal integration of (7) now gives
∆V = c ln
m0 t
− g sin γ dt
m ∫0
(9)
and so the gravity loss is
y
∆VGrav. =
∫ g sin γ dt
(10)
0
Of course, the particular γ ( t ) to be used here must come from simultaneously
solving (8) with (7). This solution cannot be done in simple analytical terms when
thrust is constant, since a nonlinear 2nd order differential equation is involved. But,
F
interestingly, there is a relatively simple solution when he thrust acceleration a =
m
is assumed constant (i.e., throttling down as mass is consumed). Although this is not
a very realistic option, it still is useful in giving information about the initial rotation
of the trajectory near the ground, which happens before the mass has time to
change much.
Eliminate time by dividing Eqs. (8) and (7) by each other, which separates the
variables V and γ :
dV
a − g sin γ
=−
dγ
V
g cos γ
We introduce
a
= n and also change angle variable to
g
⎛π γ⎞
Γ = tan ⎜ − ⎟ ;
⎝4 2⎠
dγ = −
(11)
sin γ =
1 − Γ2
;
1 + Γ2
cos γ =
2Γ
1 + Γ2
2dΓ
1 + Γ2
(12)
The variable Γ varies between 0 when γ = 900 (initial configuration) to 1 when
γ = 00 (orbit insertion). Substituting in (11) and simplifying,
dV
dΓ 2Γ dΓ
= (n − 1)
+
V
Γ 1 + Γ2
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 32
Page 3 of 16
which can be integrated to
(
V = C Γn−1 1 + Γ2
)
(13)
Here C is a constant of integration. The solution (13) satisfies V = 0 when Γ
= 0 (vertical start) for all C (n>1), so C must be calculated by imposing a particular
trajectory angle γ (or Γ ) at some specified velocity V (or, from later results, at
some time or altitude).
The time t is calculated from Eq. (8):
dt = −
V dγ
C
= Γn−2 1 + Γ2 dΓ
g cos γ g
(
)
or, imposing t = 0 at Γ = 0.
t=
Γn+1 ⎞
C ⎛ Γn−1
+
⎜
⎟
g ⎝n − 1 n + 1⎠
dz
= V sin γ :
dt
Similarly, the altitude z follows from
C2 2n−3 2n+1
Γ
Γ
dΓ
g
(
dz = V sin γ dt =
or, with z = 0 at Γ = 0 z =
C2
g
(14)
)
⎛ Γ2n−2
Γ2n+2 ⎞
−
⎜
⎟
⎝ 2n − 2 2n + 2 ⎠
(15)
We can use this model to calculate gravity losses. Starting from (10), and using the
relationships (12),
Γ
∆VG = g
∫
0
or
∆VG =
1 − Γ2 C
1+ Γ
2
g
(
)
Γn−2 1 + Γ2 dΓ
C ⎛ Γn−1
Γn+1 ⎞
−
⎜
⎟
g ⎝n − 1 n + 1⎠
(16)
We could now use (14) to calculate the constant C by specifying the time to turn to a
given angle ( Γ ). Alternatively, we can eliminate C by division of (16) and (14):
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 32
Page 4 of 16
n−1 2
Γ
+1 F
n
∆VG = gt
n −1 2
Γ
1+
n+1 F
1−
(17)
where ΓF = Γ ( γF ) , and γF is the angle reached at t, starting from γ =
π
at t = 0.
2
As an example, say n = 3, γF = 200 ( ΓF = 0.7002). We find from (17)
∆VG
= 5.94 m/s2
t
and if t = 60 sec., ∆VG = 357 m / s , which is a substantial loss.
An alternative procedure would be to set the velocity VF reached when γ = γF .
Eliminating C now between (13) and (16) gives
1
Γ2
−
∆VG = VF n − 1 n2 + 1
1+Γ
(18)
Say n = 3, VF = 1,500 m/s, γF = 200 . We calculate ∆VG = 380 m / s in this case.
Maximum Dynamic Head (“Max-q”) During Ascent
1
ρ V2 . Initially, V 0 and ρ is high.
2
Later, V increases, but ρ decrease. There is a point of “max-q” in between, which is
important for design.
Aerodynamic forces are proportional to q =
Assume Vertical flight . Neglect drag:
m
dv
= F − mg
dt
dz
=v
dt
dv
F
=
− g = (n − 1) g
dt m
or
v
⎛
F ⎞
⎜n ≡
⎟
mg
⎝
⎠
dv
= (n − 1 ) g
dz
Assume n = const.
(F ∼ m)
v2
= (n − 1) gz
2
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 32
Page 5 of 16
Atmospheric “Lapse Rate”
“Adiabatic Lapse Rate”
Also,
T = T0 − Γz
and
dp = −ρ g dz = −
Γ < Γa ≡
p
g dz
RgT
γ −1 g
g
=
∼ 10 K / km
γ Rg
cp
dp
g dz
=−
p
R g ( T0 − Γz )
g
dp
g d ( T0 − Γ z )
=+
p
Γ R g T0 − Γ z
⎛
Γ z ⎞ Γ Rg
p
= ⎜1 −
⎟
p0 ⎝
T0 ⎠
g
⎛
ρ
Γz ⎞ Γ R g
= ⎜1 −
⎟
T0 ⎠
ρ0 ⎝
g
⎛
ρv 2
Γ z ⎞ Γ Rg
q=
= ρ0 ⎜1 −
⎟
2
T0 ⎠
⎝
(n − 1) gz
(19)
⎛ Γ ⎞
⎜−
⎟
⎛ g
⎞ ⎝ T0 ⎠ 1
− 1⎟
+ =0
⎜⎜
⎟
⎝ Γ Rg
⎠1− Γz z
T0
dln q
=0
dz
For qMAX
−1
−1
Some altitude, regardless of
acceleration or lapse rate.
−
g
Γ
1
Γ
+
+ −
=0
R g T0
T0
z T0
Air: R g = 287 J / Kg / K ,
T0
zqMAX =
R g T0
g
(20)
g = 9.8 m/s2
290 K ,
zqMAX = 8, 490 m
Then
qMAX
⎛
Γ R g T0
= ρ0 ⎜ 1 −
⎜
g
T0
⎝
g
⎞ Γ Rg
⎟
⎟
⎠
g
qMAX
ΓR g ⎞ Γ R g
⎛
= ⎜1 −
⎟
g ⎠
⎝
−1
(n − 1) g
−1
R g T0
(n − 1) P0
g
(21)
(proportional to acceleration)
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 32
Page 6 of 16
Say Γ = 6 K / km
Γ Rg
g
=
0.006 x 287
= 0.176
9.8
and n = 3
qMAX = (1 − 0.176 )
1
−1
0.176
(3 − 1) P0
(
2
Based on local T, MMax
q =
'
2
MMax
=
q
)
= 0.808 x 14.7 x 122 = 1710 psf
1 atm
Also, then v2Max q = 2 (n − 1) g
= 0.808 atm
R g T0
'
2
MMax
=
q
g
2 (n − 1)
T
2
(n − 1) 0 =
γ
γ
T
2 (n − 1)
M2 =
ΓRg ⎞
⎛
γ ⎜1 −
⎟
g ⎠
⎝
2
(n − 1)
γ
(based on C0, at ground)
1
1−
Γ R g T0
g
T0
2x2
1.4 (1 − 0.176 )
MMax q = 1.862
Drag Losses: Like gravity losses, drag losses are important only near the ground,
peaking somewhat above z ( q MAX ) . Therefore, they should be estimated and added
to the 1st stage ∆V budget alone. The “drag loss” is defined by analogy to ∆VG as the
decrease in velocity due to the accumulated drag deceleration:
"∞"
∆VD =
∫
0
D
dt
m
(22)
Drag is D = q CD A, where A is the frontal area, and CD varies with vehicle
shape and Mach number (from about 0.02 at low M to a peak of perhaps 0.15 in
transonic flow, then decreasing again). For estimation purposes only, we will use a
mean CD = CD and write (22) as
∆VD =
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
A CD
M0
m0 dz
v
∫q m
(23)
Lecture 32
Page 7 of 16
Our estimate will be based on quantities evaluated at qMAX , and an effective
∆z ∼ 3 z ( qMAX ) :
∆VD
The “ballistic coefficient”
3z ( qMAX )
A CD
⎛m ⎞
qMAX ⎜ 0 ⎟
m0
⎝ m ⎠qMAX v ( qMAX )
A CD
can be related to the vehicle length L and its mean
M0
density ρ . Assuming an given shape with (Volume) =
2
AL, we find
3
A CD 3 CD
=
M0
2 ρL
The mass ratio
(24)
(25)
v
−
m0
= e c can be estimated using v = 2 (n − 1) g and so
m
−
⎛ m0 ⎞
=e
⎜
⎟
⎝ m ⎠qMAX
2(n−1)R g T0
(26)
c
Using as well the values found previously for qMAX and z ( q MAX ) , and
simplifying, our approximate expression is
g
∆VD
4.5 CD
Γ R g ⎞ Γ Rg
P ⎛
n −1
R g T0 0 ⎜ 1 −
⎟
2
g ⎠
ρ gL ⎝
−1
e
−
2(n−1) R g T0
c
(27)
For an example, take CD = 0.1 , n = 3, T0 = 290 K, ρ = 500 Kg/m3 (half the
water density), Γ = 6 K/km = 0.006 K/m, and c = 3,000 m/s. We calculate
∆VD
1060
(m / s )
L (m)
(28)
For a large vehicle (say, L = 30 m) this is small ( ∆VD = 35 m / s ). But for a 3
m. vehicle this amounts to ∆VD = 353 m / s , a substantial loss. The difference can be
traced to the larger Area/Volume of the smaller vehicle.
To conclude, note the dependence ∆VD ∼ n − 1 , which shows that fastaccelerating vehicles, like interception missiles, suffer more drag losses than slowly
accelerating ones. There is here a tradeoff with gravity losses, which vary in the
opposite manner.
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 32
Page 8 of 16
Optimum Staging
Msi = εi Mi
Mi+1 = MLi = λi Mi
MLi + MSi = Mfi = e
−
∆Vi
Li
Mi
Mi+1
∆V
Mi+1
− i
= e Ci − εi
Mi
ML Mn−1
M
M
.... 2 = L
Mn−1 Mn−2
M1 M1
∆V
⎞
n ⎛ − i
ML
= π ⎜ e Ci − εi ⎟
=
i
1
⎜
⎟
M1
⎝
⎠
Maximize subject to ∑ ∆Vi = ∆V (assume εi is independent of Mi . In reality it may
i
depend on absolute mass.)
φ = ln
ML
⎛
⎞
− α ⎜ ∑ Vi ⎟ =
M0
⎝ i ⎠
⎡ ⎛ − ∆Vi
⎤
⎞
∑i ⎢⎢ln ⎜⎜ e Ci − εi ⎟⎟ − α ∆Vi ⎥⎥
⎠
⎣ ⎝
⎦
∆Vi
For each i,
∂φ
=
∂∆Vi
−
e
1 − ci
e
ci
∆V
− i
ci
−α
− εi
∆Vi
1
+ 1 = εie ci
αci
Then, find α from
1
⎛
⎜ 1 + αc
i
ci ln ⎜
∑
⎜ εi
i=1
⎜
⎝
n
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
∆Vi
⎛
1
= −ci ⎜1 − εi e ci
⎜
α
⎝
⎞
⎟
⎟
⎠
1
⎛
⎜1 + α c
i
∆Vi = Ci ln ⎜
⎜
εi
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
⎞
⎟
⎟ = ∆V , then find ∆Vi from
⎟
⎟
⎠
Lecture 32
Page 9 of 16
Assuming ci = c (same all stages), then
n
⎡⎛
1 ⎞ ⎤
1 ⎞
⎛
⎢ ⎜1 +
⎟ ⎥
⎜1 + α c ⎟
n
αc⎠ ⎥
∆V
⎟ = ln ⎢ ⎝
= ∑ ln ⎜
⎢
⎥
πεi
c
⎜ εi ⎟
i=1
⎢
⎥
⎜
⎟
⎝
⎠
⎢⎣
⎥⎦
n
( )
1 ⎞
⎛
⎜ 1 + αc ⎟ = πi εi
⎝
⎠
1
n
e
1
∆V
+
c
e nc
α=
1− < εi >G
−
∆V
nc
( )
< ε >G = π εi
1
n
i
⎡ ⎛
⎤
1 ⎞
∆V
⎡
⎤
∆Vi = c ⎢ln ⎜1 +
− ln εi ⎥ = c ⎢ln < ε > +
− ln εi ⎥
⎟
nc
αc ⎠
⎣
⎦
⎣ ⎝
⎦
∆Vi
ε
∆V
=
− ln i
c
nc
<ε>
n
⎛ − ∆ncV
⎞
ε ⎞
⎛ ∆V
∆V
⎤ n ⎛ ε
n ⎡ −⎜
n
−ln i ⎟
−
⎞
⎛ ML ⎞
e
⎛
⎞
⎜
<ε> ⎠
i
⎝ nc
nc
So, ⎜
= π ⎢e
− εi ⎥ = π ⎜
− εi ⎟ = ⎜ π ε i ⎟ ⎜
− 1 ⎟⎟
e
⎟
i=1
i=1 < ε >
i=1
<
ε
>
M
⎝
⎠
⎢
⎥
⎝ I ⎠OPT
⎝
⎠
⎜
⎟
⎣
⎦
⎝
⎠
So, less ∆Vi when stage is less structurally efficient.
n
⎛ − ∆V
⎞
⎛ ML ⎞
= ⎜ e nc − < ε > ⎟
⎜
⎟
⎝ MI ⎠OPT ⎝
⎠
Note:
⎛ M ⎞
∂ ⎜ ln L ⎟
M0 ⎠
Meaning of α = ⎝
<0
∂∆V
Generally: Max f(xi )
Sensitivity of payload ratio to overall
∆V changes (after re-optimizing)
dGj =
∂f =
∑
i
∂ Gj
∂xi
∂f
∑ ∂x
i
i
dxi =
dxi =
∂ gi
∂gj
∂f
= ∑ λj
j
∂xi
∂xi
∑ ∂x
dxi
and
⎛
∂ gi ⎞
⎟ dxi =
∂xi ⎠
∑ λ ∑ ∂x
i
⎝
j
j = 1 to m < n
i
∑ ⎜∑ λ
i
φ = f − ∑ λ j gj
gj (xi ) = Gj
given
i = 1 to n
j
j
j
j
∂ gi
i
i
dxi =
∑ λ dG
j
j
j
⎛ ∂f ⎞
So, λ j = ⎜
⎜ ∂ G ⎟⎟
j ⎠at optimum
⎝
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 32
Page 10 of 16
Review of Orbital Dynamics
r=
p
1 + e cos θ
(Single center)
( θ from perigee)
“true anomaly”
2
c
⎛b⎞
e = = 1−⎜ ⎟
a
⎝ a⎠
;
Apoapse (apogee, aphelion): θ = π → ra =
p
1−e
Periapse (perigee, perihelion): θ = 0 → rp =
p
1+e
1 ⎞
⎛ 1
ra + rp = p ⎜
+
= 2a
1
e
1
−
+ e ⎟⎠
⎝
2
p
= 2a
1 − e2
→ p = a (1 − e2 )
→ rp = a (1 − e ) ,ra = a (1 + e)
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 32
Page 11 of 16
Energy Conservation:
1 2 µ
v − =E
2
r
At perigee
1 2
µ
vp −
=E
2
a (1 − e )
At apogee
1 2
µ
va −
=E
2
a (1 + e )
( µ = GM)
µ
E+
2
a (1 − e )
⎛ vp ⎞
⎜ ⎟ =
µ
⎝ va ⎠
E+
a (1 + e )
*
Angular momentum conservation: r v θ = h (or r 2 θ = h )
at perigee:
h = a (1 − e ) vp
at apogee:
h = a (1 + e ) v a
equate (*) = (**)
vp
va
=
µ
E+
2
a (1 − e )
⎛1 + e ⎞
⎜
⎟ =
µ
⎝1 − e ⎠
E+
a (1 + e )
1+ e
1−e
**
2
µ 1+e
µ
⎛1 + e ⎞
=E+
⎜1 − e ⎟ E + a
2
a
1
− e)
(
⎝
⎠
(1 − e )
⎡ (1 + e )2
⎤
1+ e⎞
µ
⎛
⎥ =
E⎢
1
1−
−
⎜
2
1 − e ⎟⎠
⎢⎣ (1 − e )
⎥⎦ a (1 − e ) ⎝
E
and then
and
4e
(1 − e )
2
=
υp2 =
µ
−2 e
E=−
a (1 − e ) (1 − e )
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
⎤
µ a 1 − e2 ⎥
⎦⎥
(
)
indep. of e (given a)
vp =
µ 1+e
=
a 1−e
µ 2ra
rp ra + rp
and
va =
µ 1−e
=
a 1+e
µ 2rp
rp ra + rp
or
h=
2µ
µ µ 1+e
− =
a (1 − e ) a a 1 − e
⎡
µ 1+e
=
⎢h = a (1 − e )
a 1−e
⎣⎢
µ
2a
µp
Lecture 32
Page 12 of 16
Period
r2θ = h
dA =
1
r (r dθ )
2
dA h
=
dt
2
A=
h
T
2
T =
A = πab = πa2 1 − e2
(
h=
2
µa 1 − e
T = 2π
)
2A
h
a3/2
indep. of e
µ
Velocity: From energy conservation 1 v2 − µ = − µ
2
v=
vθ =
h
r
Time in orbit:
vθ =
(
2a
2µ µ
−
r
a
µa 1 − e2
r
r
)
=
µrarp / (ra + rp )
r
vr =
=rθ
(
µa 1 − e2
dθ
h
= 2 = 2
dt r
a 1 − e2
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
(
)
)
(1 + e cos θ )
2
(
2
2µ µ µa 1 − e
− −
r
a
r2
→
)
=r
t = t (θ)
not easy – Lambert’s prob.
(except for full orbit)
Lecture 32
Page 13 of 16
Path angle:
tan γ =
Circular orbits
r=a
→
v=
e ( + sin θ )
vr
dr
=
=+
vθ
r dθ
1 + e cos θ
⎡
2πv
r3 ⎤
= 2π
⎢T =
⎥
v
µ ⎥⎦
⎣⎢
µ
r
check,
Time in orbit (elliptic case)
dθ
=
dt
µ
3
(
(1 + e cos θ )
2
2
a 1−e
)
3
θ
1 + cos θ
1
= cos2 =
2
2 1 + t2
µ
(
a3 1 − e2
)
3
dt =
dθ
tan
(1 + e cos θ )
=
2
(
)
2 1 + t2 dt
(1 + t
2
+ e − et2
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
)
2
=
θ
=t
2
cos θ =
2
1 − t2
−
1
=
1 + t2
1 + t2
1 + t2
2
(1 + e )
2
1−e 2⎞
⎛
⎜1 + 1 + e t ⎟
⎝
⎠
2
dθ =
2dt
1 + t2
dt
Lecture 32
Page 14 of 16
Define E by
1−e 2
E
t = tan2
1+e
2
µ
3
(
2
a 1−e
µ
1 − e2
dt =
3
a
1+e
)
3
2
dt =
(1 + e )
2
1+e
E
tan2
2
1−e
2 dE = 1 − e
1+e
⎛
2 E⎞
⎜1 + tan 2 ⎟
⎝
⎠
1+
t=
1+e
E
tan
1−e
2
1+
1+e
E
tan2
1−e
2
⎛
2 E⎞
⎜ 1 + tan 2 ⎟
⎝
⎠
2
dt =
1+e
1−e
1
dE
2
E
cos2
2
1 + e dE / 2
1−e
E
cos2
2
1+e
1 − e2 ⎛ 1 + cos E 1 + e 1 − cos E ⎞
⎛
2 E
2 E⎞
+
⎜ cos 2 + 1 − e sin 2 ⎟ dE = 1 + e ⎜
⎟ dE
2
1−e
2
⎝
⎠
⎝
⎠
(1 − e cos E) dE
µ
1 − e2 ⎛ 1
e
⎞
dt
cos E ⎟ dE =
=
−
⎜
3
a
1 + e ⎝1 − e 1 − e
⎠
1 − e2
µ
dt = E − e sinE
a3
(t from perigee passage)
⎛ 1−e
θ⎞
E = 2 tan−1 ⎜
tan ⎟
⎜ 1+e
⎟
2
⎝
⎠
with
from which
t2 cos θ + cos θ = 1 − t2
cos E =
2e + 2 cos θ
2 + 2e cos θ
cos E =
1−e
E
1−
tan2
+
1
e
2
=
cos E =
E
1−e
1 + tan2
1+
tan2
2
1+e
1 − tan2
⇒
θ
2
θ
2
1−e
(1 − cos θ)
1+e
cos E =
1−e
1 + cos θ +
(1 − cos θ)
1+ e
1 + cos θ −
1 − cos θ
t2 =
1 + cos θ
e + cos θ
1 + e cos θ
θ
1 + e − (1 − e) tan2
2=
θ
1 + e + (1 − e) tan2
2
cos θ =
cos E − e
1 − e cos E
1 + e cos θ =
1 − e2
*
1 − e cos E
So, directly
µ
(1 − e2 )3 2
dt =
dθ (1 − e cos E)2
3
a
(1 − e2 )2
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
sin θ =
(1 − e cos E)2 − (cos E − e)2
1 − e2 sinE
=
1 − e cos E
1 − e cos E
Lecture 32
Page 15 of 16
µ
dt =
a3
1
2
1−e
1 − e2
(1 − e cos E)
2
1−e
dE = (1 − e cos E) dE
µ
t = E − e sinE
a3
From (**)
1 − e2 sinE
1 − e cos E
1 − e2 dθ =
r=
a(1 − e2 )
P
=
(1 − e cos θ)
1 + e cos θ (1 − e2 )
dθ =
sinE (1 − e cos E ) + ( cos E − e sinE)
(1 − e cos E)2
1 − e2 dE
1 − e cos E
r = a(1 − e cos E)
ae − a cos E = r(− cos θ)
a (cos E − e) = a (1 − e cos E) cos θ
cos θ =
cos E − e
1 − e cos E
16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 32
Page 16 of 16