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9 Orbital Maneuvers
Philosophy is such an impertinently litigious lady that a man had as good be engaged
in lawsuits as to have to do with her.
Isaac Newton in a letter to his friend Edmund Halley, June 20, 1687
10.1Introduction
10.1.1 Orbital Energy
Spacecraft is not inserted in an orbit to stay forever! A spacecraft may need
to change its orbit once or more during its life time due to many reasons. A
launch vehicle may insert a geostationary (GEO) satellite into an initial low
Earth orbit (LEO) which is much lower than the final operational orbit. Then,
the satellite should transfer from the initial orbit to its final orbit. Another
need may arise if a surveillance satellite has to change its orbit in order to
track a new target. Interplanetary missions usually require many orbit
transfers until the spacecraft is inserted into the operational orbit or to use
the same spacecraft to accomplish more than one mission. At the satellite
end of life (EOF), the satellite may be kicked out of its orbit whether to
reenter the Earth’s atmosphere or to rest in a graveyard orbit.
Any analysis of orbital maneuvers, i.e., the transfer of a satellite from one
orbit to another by means of a change in velocity, logically begins with the
energy as
( 10-1)
Sir Isaac Newton (1643-1727). English
Physicist, Astronomer and Mathematician
who described universal gravitation and
the three laws of motion, laying the
groundwork for classical mechanics,
which dominated the scientific view of the
physical Universe for the next three
centuries and is the basis for modern
engineering.
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Where V is the magnitude of the orbital velocity at some point, r the
magnitude of the radius from the focus to that point, a the semimajor axis
of the orbit, and μ the gravitational constant of the attracting body. Fig.
illustrates r, V, and a .
Equation can be rearranged as
( 10-2 )
Where it is evident that
Kinetic Energy
Potential Energy
Total Energy
+
=
Satellite Mass
Satellite Mass
Satellite Mass
Note that total energy/satellite mass is dependent only on a. as a increases,
energy increases.
v
Apogee
Earth
r
Perigee
2a
Figure 10-1. Conservation of energy relates r, V and a.
10.2Basic Orbital Maneuvers
Orbital maneuvers are based on the principle that an orbit is uniquely
determined by the position and vector at any point. Conversely, changing
the velocity vector at any point instantly transforms the trajectory to a new
one corresponding to the new velocity vector. Any conic orbit can be
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transformed into another conic orbit by changing the spacecraft velocity
vector.
∆V
V1 α
V2
Figure 10-2. Basic orbital maneuver.
10.2.1 Delta–V Budget
Orbital transfers are usually achieved using the propulsion system onboard
the spacecraft. Since the propellant mass on board is limited, it is very
crucial for mission planning to estimate the propellant required for every
transfer. The overall need for propulsion is usually expressed in terms of
spacecraft total velocity change, or DV (Delta-V) budget. We assume the
propulsion is applied impulsively, i.e. the velocity change will be acquired
instantaneously. This assumption is reasonably valid for high-thrust
propulsion.
V+DV
V
V
(3)
(2)
(1)
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Figure 10-3. Delta-V Budget.
From rocket theory,
Figure 10-4. Delta-V Budget.
( 10-3 )
)
Where
=specific impulse = thrust/rate of fuel consumption
= spacecraft initial mass
= spacecraft final mass
=propellant mass used
=9.81m/s²
( 10-4)
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10.3 Satellite Launch
High-altitudes (above 200 km) may be achieved through two burns separated
by coasting phase. The first burn is nearly vertical and places the satellite into an
elliptic orbit with apogee at the final orbit radius. The satellite then coast (no
burn) until it reaches the apogee. A second burn can be used to insert the
satellite into its final LEO orbit.
Figure 10-5.Satellite launch.
10.4 Coplanar Maneuvers
When a satellite is launched, it can be placed into desired orbit through:
1. Directly from launch,
2. A booster at particular point to transfer into another orbit.
In section 10.3, the method (2) has been introduced, which is known as orbit
maneuver. Orbit maneuver had its roots in the classical formulas and dynamics
of Astrodynamics from several centuries ago. However, the application of orbit
maneuver did not occur until after the launch of Sputnik in 1957.
Orbit maneuver is based on the fundamental principle that an orbit is uniquely
determined by the position and velocity at any point. Therefore, changing the
velocity vector at any point instantly transforms the trajectory to correspond to
the new velocity vector. Thus, the orbit of a satellite is changed.
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Coplanar maneuver only involves the change of semimajor axis and eccentricity
of the orbit without changing the orbit plane. In this section, four kind of
coplanar maneuvers are introduced:
i.
Tangential-Orbit Maneuver
ii.
Non-tangential Orbit Maneuver
iii.
Hohmann Transfer
iv.
Bielliptic Orbit Transfer
10.4.1 Tangential-Orbit Maneuver
Tangential-orbit maneuver occurs at the point where the velocity vector of
spacecraft is tangent to its position vector, typically at perigee point.
EXAMPLE 10-1
Determine the ∆V required to transfer from a circular orbit into elliptic
orbit.
SOLUTION
The ∆V between two orbit can be shown as follow:
( 10-5)
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Figure 10-6.Single coplanar maneuver.
Figure 10-6 shows a typical tangential orbit maneuver at perigee point.
Using the equation 10-5, the ∆V required is,
DV 
2 

 
R a
R
10.4.2 Non-Tangential Coplanar Maneuver
The orbit maneuver does not limited only at apogee and perigee point. If
condition is allowed, the satellite able to perform the orbit maneuvers at
any point.
Figure 10-7.Non-Tangential coplanar maneuver.
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Figure 10-7 shows the ∆V vector required for a non-tangential orbit
maneuver, where α is the difference angle between the flight path angle of
V1 and V2.
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Error! Objects cannot be created from editing field codes.
( 10-6 )
10.4.3 Hohmann Transfer
The Hohmann’s transfer is the minimum two-impulse transfer between
coplanar circular orbits. It can be used to transfer a satellite between two
nonintersecting orbits (Walters Hohmann 1925).
The fundamental of the Hohmann’s transfer is a simple maneuver. This
maneuver employs an intermediate elliptic orbit which is tangent to both
initial and final orbits at their apsides. To accomplish the transfer, two burns
are needed. The first burn will insert the spacecraft into the transfer orbit,
where it will coast from periapsis to apoapsis. At apoapsis, the second burn
is applied to insert spacecraft into final orbit.
Figure 10-8 represents a Hohmann’s transfer from a circular orbit into
another circular orbit. A tangential ΔV1 is applied to the circular orbit
velocity. The magnitude of ΔV1 is determined by the requirement that the
apogee radius of the resulting transfer ellipse must equal the radius of the
final circular orbit. When the satellite reaches apogee of the transfer orbit,
another ΔV must be added or the satellite will remain in the transfer ellipse.
This ΔV is the difference between the apogee velocity in the transfer orbit
and the circular orbit velocity in the final orbit. After ΔV2 has been applied,
the satellite is in the final orbit, and the transfer has been completed.
(10-7)
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(10-8)
(10-9)
(10-10)
Figure 10-8.Hohmann transfer.
EXAMPLE 10-1
Determine the total ∆V required and time of flight for Hohmann transfer
to transfer from a circular orbit with hinitial  191.344 km into another
circular orbit with hinitial  35781.348 km
SOLUTION
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The initial and final radius is,
rinitial  191.344  6378.145  6569.489 km
r final  35781.348  6378.145  42159.493 km
At first impulse, the delta-v required is,
DV1 
2


 
rinitial a
rinitial
where,
rinitial  r final
a
 24364.491 km
2
Thus, DV1  2.457 km/sec
For the second impulse, the delta-v required is,
DV2 

r final

2 
  1.478 km/sec
r final a
The total delta-v require is,
DVTOTAL  2.457  1.478  3.935 km/sec
The time of flight for the Hohmann transfer is,
TOF   
a3

 18924.187 sec  5.257 hr
10.4.4 Bi-elliptic Transfer
Another type of orbit transfer that based on Hohmann transfer, which is
called Bi-elliptic Transfer involves series of two Hohmann transfer.
The bi-elliptic transfer requires total of three impulse burn with two transfer
orbit. The first burn is injected to insert the spacecraft into first transfer
orbit at periapsis. When the spacecraft coasts to the apoapsis of the first
transfer orbit, second impulse is injected to insert the spacecraft into
second transfer orbit. Then, the spacecraft orbits along the transfer orbit to
new apoapsis point. Finally, another impulse is injected to insert the
spacecraft into the destination orbit. Figure 10-9 illustrates a bi-elliptic
transfer between two circular orbits.
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Figure 10-9.Bi-elliptic Transfer.
The bi-elliptic transfer requires much longer transfer time compared to the
Hohmann Transfer. However, bi-elliptic is more efficient for long distance
orbit transfer. Fig. 10-10 shows the cost comparison between Hohmann and
Bi-elliptic Transfer. R is the ratio offinal to inital radius for both orbits,
where R* is the ratio of apogee radius of transfer orbit to initial orbit in bielliptic orbit. For R < 11.94, Hohmann transfer requires less cost than bielliptic transfer. For R > 15.58, bi-elliptic transfer performs better.
Figure 10-10.Delta-v Cost Comparison between Hohmann and Bi-elliptic Transfer.
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EXAMPLE 10-2
Determine the total ∆V required and time of flight for a bi-elliptic
transfer with given orbit properties:
Initial orbit, hinitial  191.344 km
Apogee altitude of transfer orbit, hapog  503873 km
Final orbit, hinitial  376310 km
SOLUTION
The initial, transfer orbit apogee and final radius are,
rinitial  191.344  6378.145  6569.489 km
rtrans  503873  6378.145  510251.145 km
r final  376310  6378.145  382688.145 km
And the semimajor axis for both transfer orbits are,
r
r
a1  initial trans  258410.317 km
2
r final  rtrans
a2 
 446469.645 km
2
At first impulse, the delta-v required is,
DV1 
2


 
 3.156 km/sec
rinitial a1
rinitial
At the second impulse, the delta-v required is,
DV2 
2 
2 
 
  0.677 km/sec
rtrans a2
rtrans a1
At the third impulse, the delta-v required is,
DV3 

r final

2


 0.0705 km/sec
r final a2
The total delta-v require is,
DVTOTAL  3.156  0.677  0.0705  3.9035 km/sec
The time of flight is,
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 a3
a23 
TOF     1 
 2138113.26 sec  593.92 hr
 




10.4.5 General Coplanar Transfer between Circular Orbits
Transfer between circular coplanar orbits only requires that the transfer
orbit intersect or at least be tangent to both of the circular orbits. It is
obvious that the periapsis radius of the transfer orbit must be equal to or
less than the radius of the inner orbit and the apoapsis radius must be equal
to or exceed the radius of the outer orbit if the transfer orbit is to touch
both circular orbits. This condition can be expressed mathematically in
Figure 10-11.
Figure 10-11.General coplanar transfer between circular orbits.
10.4.6 Phasing Maneuver
Most coplanar maneuver involves change of orbit size and shape. However,
in some situation, the spacecraft required to change its position at a given
time. Especially for the spacecraft rendezous case where the interceptor
spacecraft required to intercept (or meet) the target spacecraft when it is
behind or ahead of the target spacecraft in the orbit.
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Original Orbit
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Target
Phasing Orbit
∆θ
∆V
Interceptor
Phasing Orbit
Figure 10-12. Phasing Maneuver.
Figure 10-12 shows an illustration of phasing maneuver. If the interceptor is
behind the target spacecraft, then the phasing orbit required to be smaller
than the original orbit, and vise versa.
Given that the phase angle (or difference of two true anomaly) between
two spacecraft is ∆θ. Then, the one orbit period required by the phasing
orbit is:
 phase 
2  D
ntgt
(10-11)
where ntgt is the mean motion of target spacecraft (or the original orbit).
Then, we can determine the semimajor axis for the phasing orbit, that is,
a 3phase
2
 phase 
 2
n phase

  phase  

a phase  
 2 


2/3
(10-12)
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10.5 Out-of-Plane Orbit Maneuvers
A velocity change which lies in the plane of the orbit can change its size or
shape, or rotate the line of apsides. To change the orientation of the orbit
plane in space, the ∆V impulse-vector inserted to the spacecraft should not
parallel to the spacecraft velocity vector.
10.5.1 Simple Plane Change
Orbital maneuvers are characterized by a change in orbital velocity. If a
velocity vector increment, ΔV that is perpendicular to a satellite velocity
vector, V1 is added, then its results a new satellite velocity vector, V2. The
perpendicular ΔV does not change the speed and flight-path angle of the
satellite, but only the inclination of the orbit. The maneuver is called simple
plane change (see Figure 10-13).
Figure 10-13.Simple plane change.
For a circular orbit spacecraft that performs the simple plane change
through an angle θ, the semimajor axis, a and eccentricity, e are remain the
same. Thus, the velocity of spacecraft at before and after the plane change
are equal, V1  V2 . Using the velocity vector triangle illustration in Figure
10-14, the delta-v required is,
DV
V
θ
V
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Figure 10-14.Velocity vector triangle for circular orbit plane change.
(10-13)
EXAMPLE 10-3
Determine the ∆V required for a satellite to change its orbit plane from
inclination 10° to inclination 25° at altitude 600km.
SOLUTION
The radius of the orbit is,
r  600  6378.145  6978.145 km
The delta-v required for the plane change is,
DV  2 

 25  10 
 sin 
  1.973 km/sec
r
 2 
10.5.2 General Plane Change Maneuver
In general, plane change maneuver involves the inclination and RAAN
change while the size and shape of orbit remain the same. The change of
the RAAN in plane change maneuver results that both orbit do not intersect
at the original RAAN location. Figure 10-15 shows the example of general
plane change maneuver.
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Z
Orbit 2
Orbit 1
X
 initial
node 1

 initial
node 2
ê
Figure 10-15.General Plane Change Maneuver.
The delta-v required for the general plane change maneuver is shown in
equation 10-17. The α angle is determined using equation 10-14, and ALa is
the argument of latitude of intersection point, that is shown in Figure 10-16.
cos   cos iinitial cos i final  sin iinitial sin i final cosD
sin ALa 
sin i final sin D
sin 
(10-14)
(10-15)
  ALa  
(10-16)
 
DV  2V sin  
2
(10-17)
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node 2
ALa
Π-ifinal
iinitial
D
node 2
Figure 10-16.Argument of latitude of intersection point.
EXAMPLE
Exam 2 Example insert here.
SOLUTION
10.5.3 Combined Maneuver
Frequently, the spacecraft orbit needs to be raised as well as titled. Two
orbital transfers may then be applied:
-A coplanar maneuver to raise the orbit (change radius), then
-A plane change to tilt the orbit.
However, performing two separates orbit maneuvers is fuel inefficient
because number of burns increased. Also, the time required for spacecraft
to arrive at final orbit is much longer. Therefore, as an alternative, these
two maneuvers can be combined in one maneuver to perform both tasks in
one burn which is more economic (require less fuel) and faster.
There are a few type of combined maneuver available in study. In this
section, we will introduce the minimum inclination maneuver.
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M
K̂
Final Orbit
Initial Orbit
Transfer Orbit
∆Vb
∆Va
Jˆ
Iˆ
Figure 10-17.Orbit transfer of a spacecraft using combined maneuver.
Figure 10-17 shows the minimum inclination maneuver for a spacecraft.
Both initial and final velocity of plane change maneuver contains the
Hohmann transfer’s contribution. The change of inclination between initial,
transfer and final orbit is chosen in the way such that the required cost is
minimum. Here, a scaling term, s is introduced to determine change of
inclination required between orbits.
Diinital  sDi
(10-18)
Di final  1  s Di
The total delta-v that required for the combined maneuver is,
2
2
DV  Vtrans
_ a  Vinitial  2VinitialVtrans_ a cossDi 
2
2
 Vtrans
_ b  V final  2V finalVtrans_ b cos1  s Di 
Now, the optimum scaling, s is required to determine to produce minimum
cost. Then, we take dDV  0 :
ds
(10-19)
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



1
sin Di 
1 

s
tan
 VinitialVtrans_ a

Di
 cosDi 

 V finalVtrans_ b

M
(10-18)
For circular initial and final orbits:
VinitialVtrans_ a
V finalVtrans_ b
 R3 
r final
rinitial
EXAMPLE 10-5
Calculate the total delta-v required for a spacecraft to transfer from a
orbit, r1 = 1.02 DU to r2 = 2.33 DU with the change of inclination ∆i = 10°.
SOLUTION
We have the initial and final radius, r1 and r2. Then semimajor axis for the
transfer orbit is,
r1  r2
 1.675 DU
2
The velocities at each location are:
a
Vinitial 

r1
 0.9901 DU/TU
Vtrans_ a 
2 
  1.1678 DU/TU
r1 a
Vtrans_ b 
2 
  0.5112 DU/TU
r2 a
V final 

r2
 0.6551 DU/TU
Then, we need to determine the scaling, s, that is:
s
 sin Di  
1
tan 1  3 / 2
  0.224105
Di
 R  cosDi 
Therefore, the total delta-v is,
(10-19)
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2
2
DV  Vtrans
_ a  Vinitial  2VinitialVtrans_ a cossDi 
2
2
 Vtrans
_ b  V final  2V finalVtrans_ b cos1  s Di 
DV  0.3464 DU/TU
Problems
1. Given two circular orbits:
Initial
r1=6660km(h1=282km)
i=30 deg
Final
rf=133200km
i=0 (equatorial)
Calculate the component and total ΔVs for the following transfer techniques
from the initial orbit to the final orbit:
a) Plane change and then Hohmann transfer:
Vc1
Descending
node
ΔV2
VATR
Equator
30˚
Vc1
Apogee
ΔV1
b) Hohmann transfer and then plane change:
υ
ΔV2
VATR
Equator
30˚
Vc1
ΔV3
30˚
Ω
ΔV1
Vcf
c) Hohmann transfer with plane change at apogee in a vectorial combination
(two impulses):
υ
Vc1
30˚
ΔV1
ΔV3
ΔV2
VATR
Ω
Vcf
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d) Bi-elliptic transfer with vectorial plane change at rt=266400km (three
impulses):
υ
30˚
Vc1
ΔV1
ΔV2
VATR1
30˚
Ω
VATR2
ΔV3
Vcf
υ
VPTR2
e) Hohmann transfer with optimally split-plane changes (two impulses)
2. The sketch illustrates three circular orbits about the Earth. The radii, as show,
are 9, 16 and 25 Earth radii. Determine the characteristic velocity in meters per
second. (ΔVT=sum of ΔV) for a double Hohmann transfer from the inner orbit
to the outer orbit (A--B--C--D). Calculate ΔVT in meters per second for a
single Hohmann transfer (A--D). Finally, determine ΔVT for an
intermediate bi-elliptic transfer (A--B--E).
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3. Given an elliptical orbit whose apogee radius rA=9r0 and perigee radius
rP=3r0 (where r0 is the radius of the assumed spherical Earth), compute
the velocity requirements for two modes of transfer from the surface of
the Earth to the ellipse. The first mode is an impulsive launch into a
bitangential transfer ellipse that is tangent to the Earth’s surface and to
the target ellipse at perigee of the target ellipse. At this point, the
vehicle impulsively achieves the target orbit. The second mode is via a
bitangential transfer ellipse that is tangent to the Earth’s surface and the
target ellipse at its apogee.
a) Calculate four velocity increments in meters per second.
b) Determine the most economical mode.
4.
A satellite is in a circular polar orbit. If, at the ascending node, the
velocity vector is rotated counterclockwise 90 deg, what is the new orbit
inclination? If the rotation is clockwise 90 deg, what is the new i? if the
same rotations occur after the satellite has moved 60 deg and 90 deg
from the ascending node, what are the new inclinations?
5. Given a set of injection conditions corresponding to the sketch, determine the
true anomaly of the injection point as a function only of γ (and perhaps
constants), and determine the eccentricity of the resulting orbit as a function
only of γ (and perhaps constants).
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6. A space vehicle at the South Pole is instantaneously launched, ΔV1, in a
horizontal direction into a parabolic orbit. When the vehicle crosses the
equator, point 2, a velocity increment ΔV2 is applied that instantaneously
places the vehicle into a polar, circular orbit. Assuming a spherical Earth radius
r0=6371km, determine the magnitudes of ΔV1, V2, Vc2, and ΔV2 in meters per
second, and determine the values of γ2 and α in degrees.
7. An astronaut is heading east in a circular equatorial orbit about the Earth at an
altitude h=3r0. At 0˚ longitude, he applies a velocity increment ΔV1, which
places him in a polar orbit whose perigee grazes the Earth’s surface 180 deg
away in central angle on the equator.
a) What is the magnitude of ΔV1?
b) What is the angle between ΔV1 and the original circular orbit velocity?
c) What is the retro velocity increment ΔV2 at perigee that will reduce his
total velocity to zero (soft-land)?
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8. A satellite is in a polar orbit (orbit 1 on the sketch) about a spherical Earth with
no atmosphere. Its perigee and apogee are in the equatorial plane. The perigee
altitude is 400 n.mi.; the apogee altitude is 2000 n.mi. Transfer from orbit 2 to
orbit 1 can occur in several ways. Determine the total ΔV for transfer via
circular orbit 3 from apogee to apogee. Determine the total ΔV for transfer via
circular orbit 4 from perigee to perigee. Determine the single ΔV at point X to
accomplish the transfer. Would the ΔV at point Y be identical in magnitude? In
direction?
At an arbitrary point, 1, in an initial orbit i, a velocity increment ΔV is added
in the radial direction. A final orbit f is thus achieved. Compare the angular
moment h and the semilatus recta p in the two orbits. Determine the radius
in the final orbit at the point that is 180 deg in central angle away from
point 1.
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R B I T A L
Boris, a Russian cosmonaut, is in a circular equatorial orbit of radius
r=1.44r0 about the moon (see sketch). He decides to pay a surprise visit to
his American friends camped at the North Pole by transferring with ΔV1 into
a polar elliptical orbit whose pericenter is at the camp. When Bpris reaches
the camp, he retrofires with ΔV2 to reduce his total velocity to 0. Determine
ΔV1 and ΔV2 in meters per second. For the moon, Vc0 =
=1679 m/s.
8.4 References
Chobotov, V. (2002). Orbital Mechanics. Reston, Virginia, American Institute
of Aeronautics and Astronautics, Inc.
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