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16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 33: Performance to LEO
V∆ Calculations for Launches to Low Earth Orbit (LEO)
Ideal Earth-to-orbit launch
α ∆ = '
1 2EV cos R v R
µ µ ∆− = −
2 '2
1 2
2 2E
V V
R R
µ α
⎛ ⎞= ∆ −⎜ ⎟
⎝ ⎠
2
1
1
2E
RV cos
R R
µ µ
α
∈
∈
−∆ =
⎛ ⎞− ⎜ ⎟
⎝ ⎠
2
1 2
2
2
1
R RV
Rcos
R
µ
α
−∆ =
⎛ ⎞− ⎜ ⎟
⎝ ⎠
1 2
2
12
1
E
E E
R
RVR R
cosR
µ
=cv R
µ µ µ α α
α α
∈
∈
⎡ ⎤⎢ ⎥− −⎢ ⎥
∆ = − = − = −⎢ ⎥⎛ ⎞ ⎛ ⎞⎢ ⎥− −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
2 2 2 2
2 2
1 1' 2 2
1 1
E E
E Ec
EE E
R RR R RR RV v v cos cos
R R R R R RR Rcos cos
R R
1 2V V V∆ = ∆ + ∆
( )η η η η α
η α η α µ ∈
−∆ −= + −
− −2 2 2 2
2 1 12'
1 1 /
Vcos
cos cosR
R
Rη ∈ = ( )
η α η η
η α µ ∈
∆ −= − +
+
12 1
1 /
V cos
cosR α (increasingf.of )
( )η η
η µ
−∆= +
+
21
21 /
MIN
E
V
RFor 0α =
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Note: Max at 0.064178 99,260R kmη = → = (worst altitude)
1.5363 /
MIN
F MAX
V
Rµ
⎛ ⎞∆⎜ ⎟ =⎜ ⎟⎝ ⎠
7910 /m sR
µ
∈
⎛ ⎞=⎜ ⎟⎜ ⎟
⎝ ⎠
ε η
− =1
1 η ε
=+
1
1
ε η
ε − =
+1
1
ε η
ε
++ =
+
21
1
( )ε ε
ε
+
+
2
2 121
ε ε ε ε ε ε ε ε + = − + + − = + −+ +
2 21 3 3 31 ... 1 ...1 /2 1 2 8 4 2 82
ε ε ⎛ ⎞
−⎜ ⎟⎝ ⎠
31 ...
40α = 0.9 1.05128η = →
(approx. 1.05093)
( )
/ /V R
V
µ ∈∆
∆ 0.15095η =
(GEO,
R=42,200Km)
0.23951η =
η ⎛ ⎞
=⎜ ⎟⎝ ⎠
1, 26,580
2
day Km
0.87620η =
( )= 900Z Km
0.91392η =
( )= 600Z Km
0.95502η =
( = 300Z Km
00α = 1.50775(11,918 m/s)
1.45534(11,504 m/s)
1.06387(8,409 m/s)
1.04399(8252 m/s)
1.02275(8,084 m/s)
015α = 1.51366(11,965 m/s)
1.46373(11,570 m/s)
1.07960(8,534 m/s)
1.05952(8375 m/s)
1.03748(8201 m/s)
030α = 1.53109(12,102 m/s)
1.48854(11,766 m/s)
1.12032(8,856 m/s)
1.09755(8676 m/s)
1.06953(8454 m/s)
ε ε
µ
∆+ −
21 5
14 32 /
E
V
R ε ε
µ
∆−
22 7
4 32 / E
V
R NOTE:
So, for LEO, mainly (apogee kick)1V∆
Sticking to 0α = , variation with R
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11.05
R
Rη ∈
= = 1.1 1.05 2 4 6 10 15.6
/
V
Rµ ∈
∆
1.02041 1.04651 1.18164 1.28446 1.44868 1.49934 1.52978 1.53626
Effects of Earth’s Rotation
(a) reductionV∆
∈Ω = 463 / secR m
β is launch azimuth w.r.t
East
( 0α = , near-horizontal
launch)
1V∆ = rocket-imparted V∆
Starting velocity (abs.) is now
( ) ( ) ( )∈ ∈= ∆ + Ω + ∆ Ω2 2
1 1 12v V R cos L V R cos L cos
So, replaces in previous formulation1v 1V∆
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α
µ
µ µ
α
=
−= ⎯⎯⎯⎯→ =
+⎛ ⎞⎛ ⎞ +− ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
1 02
2
21 /2
112
E
E E
E EE E
RR RRv
R R RR Rcos RR
R
( ) ( ) ( )µ
β ∈= ∆ + Ω + ∆ Ω+
2 2
1 1
22E
E
E
RRV R cos L V R cos L cos
R R
( ) ( ) ( )µ
β β ∈∆ = − Ω + Ω + − Ω+
2 2
1
2E
E E
E
R
RV R cos L cos R cos L cos R cos L
R R
( )µ
β β ∈
∈
∆ = − Ω + − Ω+
2
1
2sin
E E
R
RV R cos L cos R cos L
R R
Notice rotation reduces even for . The benefit is low for some larger1V∆ 090 β = β
(Westwards launch). For , need= ∆1v 1V
β ∈Ω= − −
∆
1
0.0562
R co sLcos cos L
V(for 1
8200 /V m s∆ = )
(for )0 028.5 , 92.8L β = =
Example: For , R = 6370+500 = 6870 Km,028.5L =
( )( ) β β ∆ = − + × −
2
27
1
8052.8407 6.48399 10 407V cos sin
( )0 β 0 030± 060± 090± 0120± 0150± 0180±
1V∆ (m/s)’ 7645 m/s 7694 7841 8042 8248 8402 8459
1V∆ reduction 407 m/s 355 m/s 211 m/s 10 m/s -195 m/s -350 m/s -407 m
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(b) Orbit inclination
For 0 β = (launch due East), i = L. For any other azimuth, higher inclination.
∈= ΩRv R cos L
( ) β γ γ
∆=
−
1Rv V
sin sin
γ β γ
γ = ∆
1R
Sinv V sin cos
Cos
γ β
γ
⎛ ⎞−⎜ ⎟
⎝ ⎠cos
sincos
β γ
β
∆=
∆ +1
1
tanR
V sin
V cos v
β γ
β
=⎛ ⎞
+ ⎜ ⎟∆⎝ ⎠1
tantan
1R
v
V cos
Given two angles and the side included, find opposite angle
( ) ( )γ γ γ = − − + − =0 090 90 90 90cos i cos cos sin sin cos L cos cos L
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Example: Continuing from previous example,
= → = = →0
E28.5 407 / , ' 500 RR
L v m s R Km
β
γ
β
=+
∆ 1
tan
tan 4071
V cos
, γ = 0.87882cos i cos
∆ from previous table1V
β 0 030 060 090 0120 0150 0180
α 0 028.54 057.49 087.10 0117.49 0148.55 0180
i 028.5 039.5 061.8 087.5
( )
0
0
113.9
66.1
( )041.4 ( )028.5
retro-orbits (westwards)
Lecture 33Prof. Manuel Martinez-Sanchez Page 6 of 9
030− 060− 090− 028.50− 057.39−
039.4 061.7
slightly different inclination
In reality, we probably require the orbit altitude, the orbit inclination and then launchazimuth β must be calculated
γ =cos i
coscos L
γ − ⎛ = ⎜
⎝ ⎠
1 cos icos
cos L
⎞⎟ γ
γ + = =
22
2 2
11 tan
cos L
cos cos i
β γ
β
∆= − =
∆ +
21
2tan 1
cosR
V sincos L
V vcos i
γ = −2
21
cos isin
cos L
with
µ
β β ∈
∈
∆ = − −+
2 2
1
2
R R
R
RV v sin v cos
R R β γ γ β ∆ + = ∆1 1
tan tanR
V cos v V sin
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or ( )µ
β ∆ + + ∆ =+
2 2
1 1
2
2 ER R
E
R
RV v v V cos
R R γ
β β β γ β
γ tan
∆ = =−
−1
tan
tanR R
v vV
sinsin coscos
µ
β β β β β γ γ
+ + = +⎛ ⎞ −−⎜ ⎟⎝ ⎠
2 2
22
2
2 cos
tantan
R RR
E
R
v vvsin R Rsin coscos
ER
µ β β β
β β β γ γ
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − + − = −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟
+⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
2 2
2
2
1 2tan tan tan
ER
E
R
Rsin sin sinv cos cos cos cos
R R β
γ
β β β β
γ γ
+ + −2
2
2
21
tantan
sin sin coscos
β β
γ
+2
tan
sin cos β − 22cos
β γ
⎛ ⎞+⎜ ⎟
⎝ ⎠
2
2
11
tansin
β
γ
2
2
sin
sin
µ
γ γ β
⎛ ⎞= −⎜ ⎟
+ ⎝ ⎠
2
2 2
21 1
tan tanE
R
E
R
Rv sin
R R
β
γ γ µ
=+−
1tan
1
tan2
R E
E
v R R
Rsin
R
γ β
γ µ
=+
−
tantan
1
2
R E
E
v R R
Rcos
R
Check: ) → 0
0
28.5
61.8
L
i
==
057.47γ = β → = =⎛ ⎞
− ⎜ ⎟×⎝ ⎠
0
1tan 1.7308407
0.63770.8431 8053
( ) β = 0 059.98 60
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β
µ
−=
+−
2
21
tan
1
2
ER
E
cos L
cos i
R RcosLv
Rcos i
R
β
µ
−=
−⎛ ⎞⎜ ⎟
+⎜ ⎟ −⎜ ⎟⎜ ⎟⎝ ⎠
2 2
tan
1
2
R
E
E
cos L cos i
v cos Lcos i
R R
R
R
Directly:( ) β β α
=−
1 RV v
sin sin
γ γ
β = −
1 tanRv sin
cosV
γ γ β
γ γ
−= =
− −
2
1 1
1tan
R R
cossin
v vcos cos
V V
and γ =cos i
coscos L
( )
( )
( ) ( )
β β
−
= = =+ − ⎛ ⎞+
⎜ ⎟ − +⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟− ⎝ ⎠⎜ ⎟⎜ ⎟⎝ ⎠
2
2 2 2 2
2 2
1 1
1 tan1
2
R
R R
R
v Cos LCos i
CosCos L Cos i
v Cos L v cos Lcos i cos L
v Cos LCos i
α β
γ µ
=+
−
tantan
12 /
R E
E
v R R
cos R R
γ
β β
γ µ
∈
∈
= + = +⎛ ⎞+
−⎜ ⎟⎜ ⎟⎝ ⎠
22
2
1 t1 tan 1
12 /
Rcos v R R
cos R R
an
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β β
γ
∆ =
−1
1
tancos1
tan
RvV
γ γ
γ µ
µ
∈
⎛ ⎞+∆ = − +⎜ ⎟
⎜ ⎟+ ⎝ ⎠
2
2
11 t
2 /
2 /
R R E
E
R
E
v cos v R RV
cos R RR Rv
R R
an
=Rv γ
R
cos
vµ γ µ γ γ
µ
⎛ ⎞ ⎛ + ++ −⎜ ⎟ ⎜
+ ⎝ ⎠ ⎝
2
2 2
12
2 / 2 /
2
R E R
E EE
v R R v R R
R R cos R Rcos cosR R
⎞⎟
⎠
E
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