lecture 33-rocket theory

8/9/2019 Lecture 33-rocket theory

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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 =

16.512, Rocket Propulsion Lecture 33Prof. Manuel Martinez-Sanchez of 9



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




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∆




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




(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




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 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

16.512, Rocket Propulsion



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




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




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


lecture 33-rocket theory

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