Chapter 4 - Part 1

D. Mortari, AERO-423

Interplanetary trajectories

D. Mortari, AERO-423 2

Problem: Determine the minimum fuel (Δvtot=min)trajectory from Earth to Mars under the assumptions:

a) Earth and Mars are in the same plane, andb) Earth and Mars are in circular orbits.

Three Phases:Phase I: Escape from Earth (Earth gravity is dominant),

Phase II: Hohmann transfer from Earth to Mars (Sun gravity is dominant), and

Phase III: Mars Capture (Mars gravity is dominant).

Trip to Mars

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Phase 2: Earth to MarsHohmann Transfer to Mars

Initial Earth-Mars angular separation is θ12 so that at apogee of the transfer orbit rendezvous with Mars occurs. Return trajectory is just the reverse. This means that there is a wait time at Mars for the planets to get aligned.

31 2

2 2 12

2 2Mars has moved by the angle

tr tt tr

s

tr

T aR Ra t

n t

+= ⇒ = = π

μ

θ = = π−θ

R1 R2θ12

Earth(arrival)

Mars (arrival)

Mars (departure)

Earth(departure)

θ1

1 1

12 1 1

During the transfer time the Earth arrives at Return trip geometry is identical:

tr

ret tr

n tn t

θ =θ = π−θ = π−

12 2Initial Earth-to-Mars displacement trn tθ = π−

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1 2

The is the time for Earth-Mars configuration to repeat.2It is evaluated from the relative angular velocity: 0syn

Synodic Period

Tn n

π= ≥

−

Phase 2: Mars to Earth (wait)

wait 1 1Earth-to-Mars relative angle increases to 2 2( ) 4 2θ = π− θ − π = π− θ

R1 R2θ12

Earth(arrival)

Mars (arrival)

Mars (departure)

Earth(departure)

θ1

Mars (departure)

Mars(arrival)

Earth (arrival)

Earth (departure)

2R

1R

1θ

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1

1 2

The waiting time to accomplish the return trip will be:4 2 (Eq. 11.13 in Wiesel is wrong)

2syn

wait wait

Tt

n nπ− θ

= θ =π −

Phase 2: Mars to Earth (return)Mars (departure)

Mars(arrival)

Earth (arrival)

Earth (departure)

2R

1R1θ

Overal mission time is 2total tr waitt t t= +

R1 R2θ12

Earth(arrival)

Mars (arrival)

Mars (departure)

Earth(departure)

θ1

1 2Therefore, we have: wait waitn n tθ = −

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Planetary Trip Times

Planet 1 Planet 2 Ttransfer

Twait Ttrip Δv (km/s)

Saturn Neptune 44.16 36.3 124.6 15.94 Jupiter Uranus 21.33 5.586 48.25 17.42 Venus Uranus 15.74 0.51 31.99 11.80

Mercury Venus 75.55 days 40.8 days 191.9 days 6.262 Saturn Pluto 44.89 34.38 124.2 15.94

Neptune Pluto 83.5 7695 7862 4.01 Mars Jupiter 3.08 1.61 7.78 5.211

Planetary Transfer Times

Planet Tsyn Ttransfer Twait Ttrip Mercury 115.8 105.4 66.9 277.9 Venus 583.9 146.1 467.0 2.08 y Mars 779.9 258.8 454.3 2.66 y

Jupiter 398.8 2.73 y 214.6 6.05 y Saturn 378.1 6.05 y 363.2 12.2 y

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Sphere of Influence/ActivityEarth to Mars trip has three regimes:

- Earth gravity field is dominant- Sun gravity field is dominant- Mars gravity field is dominant

Sphere of influence - the sphere about a body in which its gravityfield is dominant. S/C is in orbit about m with radius r, and m is in

orbit about M with radius R. Find point where forces balance.

2 2Generally , 1 2 therefore,rr R R r RR

− − ⎛ ⎞⇒ − + +⎜ ⎟⎝ ⎠

( ) ( )2 22 22 2 3 30 sinceGM Gm GM Gm GM GMR r R r

r r R RR r R r− + +ω − = − + + − = ω =

− −

1/3

2 2 2 3 21 2 1 0 33

M r m M r M m r mrR R r R R R r R M

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + + + − = ⇒ = ⇒ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

rmM

Rω

2 /3 2/32 1/10(Lagrange) and (Tisserand) (1 3cos )r m r m

R M R M−⎛ ⎞ ⎛ ⎞= = + α⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

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Trajectory DesignTo escape Earth gravity field, trajectory must be hyperbolic in Earth sphere of influence.It then becomes elliptic in Sun sphere of influence. When it arrives at Mars it becomes hyperbolic in Mars sphere of influence. When arriving at Mars engines fire to slow downand put into circular orbit. Preliminary orbit design makes these assumptions and matches the orbits at the boundary of the sphere of influence. Called patched conics.

Assume initially in circular orbit of radius r0 about Earth.Δv is applied to transfer from circular to hyperbolic escape.

, - velocity and radius of Earth orbit (in HCI),, - position and velocity vectors relative to the ECI,

, - position and velocity with respect to the HCI,

- velocity at infinityTerms:

E EV Rr v

R V

v∞0

0

0 0

of escape hyperbolic trajectory, - radius of circular Earth parking orbit, - speed in Earth circular orbit of radius , - speed at perigee of Earth hyperbolic escape trajectory (at ), and,

C

A

rv rv rV V - speed at aphelion and perihelion of heliocentric transfer ellipse.P

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

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Trajectory design (departure)

At sphere of influence

(Patch conditions)E

P E

R R r

V V v∞

⎧ = +⎪⎨

= +⎪⎩

Process: First determine the transfer ellipse. This defines the velocity at perigee.The radius of the sphere of influence is small compared to the Earth orbit radius,so it is usually assumed that perigee of the transfer ellipse is at the Earth orbit radius.

( )

2

2

The Earth to Mars transfer orbit has

2The equation of energy applied to Perihelion

2 2allows us to evaluate the velocity there

2

E Mtr

P H H H

E tr E M

H MP

E E M

R Ra

VR a R R

RVR R R

+=

μ μ μ− = − = −

+

μ=

+

(Earth orbit)EV

v ∞1 Cv v v= + Δ

departure hyperbola

sphere of influence

d

ψ0r

0

is the distancewhen then

is the parameterplanet

d approachr r d b

b impact

= =

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Trajectory design (departure)

22 21

0

21

0

1

1 1

On the escape trajectory In fact, the equation of energy implies

02 2 2

which allow us to evaluate the velocity

2

Therefore, the required is

P E

SOIE E

SOI

E

v V V v

vv vr r

v vr

v

v v v

∞

∞

∞

= −

μ μ= − = − >

μ= +

Δ

Δ = −

E

2

0 0

2 E Ec v

r r∞

μ μ= + −

20

0

The semi-major axis of the transfer hyperbola is (negative)2

(1 )(1 ) 1 1 therefore1 cos 1 cos

1At , we will have therefore cos

Ea

r p a er a e e ra e e

re

μ= −

−= − → = − > = =

+ ϕ + ϕ

= ∞ ϕ ψ ψ = −

E

(Earth orbit)EV

v ∞1 Cv v v= + Δ

departure hyperbola

sphere of influence

d

ψ0r

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Impact parameter sensitivityb is called the impact parameter. Pure Hohmann transfer gives b=0.

Small changes in velocity at perihelion create large changes in b.Assume a pure Hohmann transfer which is directed at center of Mars.

( )

22

2 2 2

3

( )2 2

2 (differentiating) 2

( )2 2 2

42 4

Km 35,000 (m/s)

E Mtr tr

H H HP P P tr

E tr tr

tr tr trH M H Mtr P P P P

S S tr E H tr E

tr M Mtr P P

H E tr E

P

R b R ba a

V V V aR a a

a a ab R Ra V V V Va R a R

a R Rb a V VR n R

b V

+ + δ= ⇒ δ =

μ μ μ= − δ = δ

μ + μδ = δ = δ δ

μ μ μ

δ = δ = δ = δμ

δ δ

D. Mortari, AERO-423 12

approach distance d

and impact parameter b

dcosr d∞ α =

2 2 2

22 2

cos

2 2 2

221

p p

p

p

pp p

p

h r v v d r v

v v vr r

rd r r

r v v

∞ ∞ ∞

∞ ∞

∞

∞ ∞

= α = =

μ μ= − = − ≈

μμ= + = +

E

D. Mortari, AERO-423 13

Planetary Flyby/Gravity AssistHohmann transfers are the minimum energy transfer between two planets but often taketoo long. We can use the gravitational attraction of other planets to increase or decrease our velocity and thus require less fuel and take less time.

The concept is simple: As the satellite approaches a planet it is on a hyperbolic trajectorywith respect to the planet. It will leave the Sphere Of Influence (SOI) with the same speedthat it entered the SOI but at a different direction. This velocity when added to the velocityof the planet is the new velocity which can be less than or greater than its entering velocity.

Trailing side flyby increases velocity

Leading side flyby decreases velocity

1 1 2 2

1 2

P Pv V V V V vv v

∞ ∞

∞ ∞

= − = +

=

V P

V 1

PV

1v ∞

2v ∞

V P

V 2

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Planetary Flyby/Gravity Assist

Consider the energy gain

( )

( )

2 21 1 2 2

1 2

1 2

2 2 2 2 2 22 1 2 2 1 1

2 21 2 1 2

2 1

1 1(before) and (after) 2 2

1 1 [( 2 ) ( 2 )]2 2

2 cos

true anomaly at sphere or at

H H

p p p p

p p

V VR R

R R

V V V V v v V V v v

v v v v

V v v V v

∞ ∞ ∞ ∞

∞ ∞ ∞ ∞

∞ ∞ ∞

μ μ= − = −

≈

Δ = − = + ⋅ + − + ⋅ +

= → =

Δ = ⋅ − = ψ

ψ = ∞

E E

E

E

1 1 2 2P Pv V V V V v∞ ∞= − = +

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Planetary Flyby/Gravity Assist

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Possible maneuver points

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Timeline: Earth to Mars

• Wait for the launch window (Mars 44.3°ahead of Earth)

• Escape from Earth (3.2 days),• Flying from Earth to Mars (258.8 days),• Wait for coming back (454.3 days),• Flying from Mars to Earth (258.8 days),• Return to Earth (3.2 days).

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Gravity assist trajectories

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Optimal planetary capturing• Given v∞, find the parking orbit so that

the Δv is minimized.

d = approach distance

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Optimal Planetary CaptureThe objective here is to go into a circular orbit, when arrivingat the planet, that minimizes the Δv needed for capture.

2 2 2 2

sph

1/ 2

22

At the boundary of the activity sphere:

1 1 1 22 2 2

1 2 2 1 1 02 2

p p cp p p

p p p p p

v v v v v v vr r r r

d v vdr r r r r

∞ ∞ ∞

−

∞

μ μ μ μ= − = − ⇒ Δ = − = + −

⎛ ⎞ ⎛ ⎞ ⎛ ⎞Δ μ μ μ= + − + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

E

2 2 2

min2

2 2Rearranging you obtain 2 4

2Optimum radius , is the one for which 2 2 , therefore 2

p pp p

p esc cp

r v r vr r

vr v v v vv r

∞ ∞

∞∞

∞

⎛ ⎞μ μμ = μ + ⇒ μ = μ +⎜ ⎟⎜ ⎟

⎝ ⎠

μ μ= = = = Δ =

D. Mortari, AERO-423 21

Capturing at Mars

The transfer orbit atr = 1.8877×108 KmAt aphelion, Va = 21.48 Km/s

v∞ =Vm–Va= 24.13–21.48 = 2.65 Km/sΔv = 1.87 Km/s

Mars Gravitational parameterμmars = 4.269×104 Km3/s2

rp = 24,316 Kmparking orbit alt. = 24,316–3,395 = 20,921 Km