Givens Operational Procedure

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```					               Interplanetary Mission Design
ASEN 5519
Spring 2008
Lecture #2

1. Application of Patch Conic to Deep Impact
2. Orbit Change maneuvers
a. Energy Change
b. Inclination Change
c. Node and Inclination Change
3. Sphere Of Influence

1
Temple 1 Ephemeris
•   ClassificationJupiter-family Comet
Osculating Orbital Elements
•   Epoch Julian day 2453440.5, 2005 March 11 00:00:00.0 UT, Friday
•   Semi major axis 3.1219 AU
•   Period [Julian yrs] 5.516 Julian years
•   Inclination 10.530 °
•   Eccentricity 0.5176
•   Lon. of ascending node 68.941 °
•   Argument of perihelion 178.838 °
•   Passage of perihelion 2005-Jul-05.316301
•   Aphelion distance 4.7378 AU
•   Mean anomaly 339.217 °
Physical Data
Discovery Circumstances
•   Discoverer Ernst Wilhelm Leberecht Tempel
•   Discovered1867-04-03

Orbit Configuratoin

Application of Hohmann Transfer to the orbit of
Deep Impact and its encounter of Temple 1
Assume that the Deep Impact spacecraft is on a Hohmann Transfer.

.   Launch 1/12/05            1) Energy of Temple orbit
        1.327 x1011
E     
2a   2  3.1231.5 x108
 141.637 km 2
s2
2) Compute the velocity of
Deep Impact at perihelion
Impact 7/4/05                                         12
 
            
V p  2  E       
1AU  1.5 x108 km                                   
 
rp p  

 
                      1.327 x1011    
  1.327 x1011 km3 s 2            Vp  2  141.637                        
 
                1.506  1.5x108   

V p  29.86 km s

Orbit of Deep Impact spacecraft
1.5 x108  2.26 x108
at                        1.88 x108 km
2
Energy             
E
2at
1.327 x1011

2 1.88 x108 
 352.93 km 2 s 2
Velocity at aphelion of Deep Impact spacecraft
12
 
            

VA  2  E       
 
       rp   

12
 
                1.327 x1011     
 2  352.93                          21.66 km s
             1.506 1.5 x108   
                               
Orbit of Deep Impact spacecraft
If we assume all orbits are in the ecliptic the impact velocity of the spacecraft
will be (neglecting the mass of Temple 1)

V  29.86  21.66  8.20 km s

The mass of Temple 1 is 7.2 x 10
13
kg , G  6.67 x 1020 kg 3 s 2 kg

Hence,
GM  4.802 x 106 km3 s 2

and the V due to the mass of Temple 1 is insignificant.

Orbit of Deep Impact spacecraft
However, we must account for the inclination of Temple 1’s orbit relative to the
ecliptic.
29.86
V   10.53
21.66

From the law of cosines:
V  29.862  21.662  2  29.86 21.66 cos10.53
2

V  9.44 km s

Actual value of V was 10.20 km s

Hence, the Hohmann Transfer is a good approximation to the actual
value of V

Maneuver to change semimajor axis (Energy)

In general, if we are trying to increase or decrease the energy (semimajor axis)
of an elliptical orbit, the most efficient place to do this is periapsis.

V 2  
From the energy equation:          E     
2 r 2a

E  V V            2
a
2a
2a 2V
a              V

a 
V  2
2a V
Hence, for a given a , V will be minimized when V has its maximum value

Inclination Change
If it is only desired to change the inclination, the most efficient place to do this
is at the nodal crossing (the only points where the orbits intersect).

V2          V
i
V1

V2 i V                     From the law of cosines
1

V 2  V12  V22  2V1V2 cos i               (1)

If there is to be no Energy change, V1  V2
and                      i
V  2V1 sin                      (2)
2

Inclination Change
From Eq. (2) it is seen that the most efficient place to change inclination
is at the point of lowest velocity, i.e. apoapsis.
Hence the bielliptic transfer which changes the orbital energy is also an
ideal opportunity for an inclination plane change. Note that the Vs are
applied at a nodal crossing and transfers are 180° Hohmanns.

rf
Low energy plane
V3                ri
change opportunity.
V2
 V1
rf
If         15.58 , then
ri
the bielliptic transfer is
more efficient than a
single Hohmann Transfer.
General Plane Change – Both Ω & i
Final Orbit
Initial Orbit

V



u
Equatorial
Plane
if
ii
i               f

From the law of cosines for spherical triangles
cos   cos ii cos i f  sin ii sin i f cos                  0     (3)
  i   f                    Assuming no energy change the equation for

V
is given by Eq. 2 (V  2V1 sin )
2
General Plane Change (cont.)
The argument of latitude on the initial orbit at which the maneuver is
performed is given by the law of cosines for spherical triangles.

cos i f sin ii  sin i f cos ii cos 
cos u                                               0u         (4)
sin 

Note that   u   is independent of the sign of  since cos    cos  
A reference for Eqns (3 )and (4) is C. D. Brown, Spacecraft Mission Design,
2nd Edition. Note that Eq (3.10) of Brown, which is based on the law of
sines, should be replaced by Eq (4) which yields the proper quadrant.

Example
Consider an initial circular orbit:

r  6878 km
i  30
  100 E

It is desired to make a plane change to:

r  6878 km
i  10
  60 E

Example (cont.)
  40
cos   cos 30 cos 10  sin 30 sin 10 cos  40
  23.164
The argument of latitude on the original orbit is,

cos10 sin 30  sin10 cos 30 cos 40        thus u  16.45
cos u 
sin 23.164

398600
Vc          7.61 km s
6878

 23.164 
V  2  7.61 sin           3.06 km s
 2 
Example (cont.)
Note that if ii  10   and i f  30

  23.164

But
cos 30 sin10  sin 30 cos10 cos 40
cos u 
sin 23.164

u  125.21

Now the argument of latitude is > 90°.
Draw a sketch to show that u  90 in this case.

Example (cont.)
If   0 , then

cos   cos ii cos i f  sin ii sin i f
 cos  ii  i f 
  i
cos i f sin ii  sin i f cos ii
cos u 
sin i
sin  ii  i f 
                      1
sin i
u0
i.e. The plane change occurs at a nodal crossing.

Sphere of Influence of the Earth
Ref-Curtis, H.D., Orbital Mechanics for Engineering Students (p354)
v

rsv                    r                            sun,  s
rse
 Earth,  e
rv
rs                          re                               v    spacecraft,    v
Inertial Frame

We wish to find      r to describe the motion of the satellite relative to the Earth.

r  rv  re
or

r  rv  re                     (1)

Sphere of Influence of the Earth
From Newton’s 2nd law
mv rv   Fv
so
rsv          r
M v rv  GM v M 3  GM v M  3     (2)
rsv         r
rse GM  M v r
M  re  GM  M 3                 (3)
rse   r3
Subst (2) + (3) into (1)

r       rsv rse 
r  G  M   M v  3  GM  3  3      (4)
r        rsv rse 

Sphere of Influence of the Earth
We may ignore     Mv       Mv  M            and M  

r      rsv rse                         but
3
rsv     3
rse
r    3    3  3 
r       rsv rse                         and   rsv  rse  r
Hence,                         r      r
r          3   3
r      rse
or
r  a2 B  a p ,              a p  Perturbation acceleration
The ratio of the magnitude of the perturbing acceleration to the primary
acceleration is
 r                         3
ap                r3
  r 
          se
     
a2 B        r             rse 
r3
Sphere of Influence of the Earth
Likewise if we consider the spacecraft motion relative to the sun,

0 compared to r
rsv    r             rse                r3
rsv    3    3  3 
rsv     r rse 

rsv  A2 B  Ap
And the ratio of the magnitudes is,
r     1
 2
Ap    rsv
2
r 3
r
                , using
A2 B  r 2                     rsv   1
 2
3
rsv rsv

Sphere of Influence of the Earth
The sphere of influence (SOI) is the sphere about the primary body where the
perturbations of the sun on an orbiter of the primary are equal to the
perturbations of the primary on an orbiter of the sun.
Hence,        ap           Ap                   setting r  rSOI
                                                 25
a2 B          A2 B                                   
rSOI    rse     
  
 r 3  rsv
2

 rse  r 2
3
  1.327 x 1011 km3 s 2
  3.986 x 105 km3 s 2
But rse       rsv , so                     rse  1.5 x 108 km
5              2       The sphere of influence for the Earth is,
 r    
                                            3.986 x10 
25
rse    
5
                               rSOI  1.5 x108                         9.24 x 105 km
 1.327 x1011 

Sphere of Influence of the Earth
Sphere of Influence
Planet          x 106 km
Mercury           0.111
Venus             0.616
The SOI, Patch Conic Technique              Earth             0.924
does not work well for Earth-Moon
system because of the perturbing            Mars              0.577
effects of the sun.                         Jupiter          48.157
Saturn           54.796
Uranus           51.954
Neptune          80.196
Pluto             3.400
Moon             0.0662