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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
9.1 Introduction
9.1.1 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
2 1
𝑉 2 = 𝜇( − )
𝑟 𝑎
( 9-1)
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
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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of the orbit, and μ the gravitational constant of the attracting body. Fig.
illustrates r, V, and a .
Equation can be rearranged as
𝑉2 𝜇
𝜇
− =−
2
𝑟
2𝑎
( 9-2 )
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
r
Earth
Perigee
2a
Figure 9-2.Conservation of energy relates r, V and a.
∆V
V1 α
V2
Figure 9- 1. Basic orbital maneuver.
9.2 Basic 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
transformed into another conic orbit by changing the spacecraft velocity
vector.
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9.2.1 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
∆V
+
V
=
Impulse
(1)
(2)
(3)
Figure 9-3. Delta-V Budget.
From rocket theory shown in Figure 9-4, we can express the force produce
by the impulsive thrust as:
𝐹 = 𝑚𝜈̇ 𝑒 = 𝑚̇𝐼𝑠𝑝 𝑔0
ve
𝑑𝑉
𝑑𝑀
𝑀
=−
𝜈
𝑑𝑡
𝑑𝑡 𝑒
𝑑𝑉
𝑑𝑀
=−
𝜈𝑒
𝑀
( 9-3 )
𝑀𝑓
𝑀𝑖
∆𝑉 = −𝜈𝑒 ln ( ) = 𝜈𝑒 ln ( )
𝑀𝑖
𝑀𝑓
where 𝐼𝑠𝑝 =specific impulse = thrust/rate of fuel consumption. The spacecraft
‘sinitial and final mass, and the propellant mass are:
𝑀𝑖 = spacecraft initial mass
𝑀𝑓 = spacecraft final mass
𝑀𝑝 =propellant mass used
V
Figure 9-4. Impulsive thrust produced
based on rocket theory.
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𝑔0 =9.81m/s²
𝑀𝑝
∆𝑉 = 𝐼𝑠𝑝 𝑔0 ln(1 + 𝑀 )
𝑓
𝑀𝑝 = 𝑀𝑓 [𝑒𝑥𝑝 (𝐼
∆𝑉
𝑠𝑝 𝑔0
) − 1]
−∆𝑉
= 𝑀𝑖 [1 − 𝑒𝑥𝑝 (𝐼
𝑠𝑝 𝑔0
( 9-4)
)]
9.3 Satellite Launch
2nd burn
1st burn
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.
9.4 Coplanar Maneuvers
When a satellite is launched, it can be placed into desired orbit through:
Figure 9-5. Satellite launch.
1. Directly from launch,
2. A booster at particular point to transfer into another orbit.
In section 9.3, 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.
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,
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Bielliptic Orbit Transfer.
9.4.1 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 9-1
Determine the ∆V required to transfer from a circular orbit into elliptic
orbit.
SOLUTION
The ∆V between two orbit can be shown as follow:
𝜇
2𝜇 𝜇
𝑉𝑐𝑖𝑟𝑐 = √ , 𝑉𝑝 = √ −
𝑅
𝑅 𝑎
( 9-5)
Figure 9-6. Single coplanar maneuver.
Figure 9-6 shows a typical tangential orbit maneuver at perigee point.
Using the equation 9-5, the ∆V required is,
2𝜇 𝜇
𝜇
∆𝑉 = √ − − √
𝑅 𝑎
𝑅
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9.4.2 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 9- 1 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.
∆𝑉 = √𝑣12 + 𝑣22 − 2𝑣1 𝑣2 𝑐𝑜𝑠 ∝
( 9-6 )
∝= ∅1 − ∅2
9.4.3 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 9-7 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 ΔV 2 has been applied,
the satellite is in the final orbit, and the transfer has been completed.
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2𝜇 𝜇
𝜇
∆𝑉1 = 𝑉𝑝,𝑡 − 𝑉1 = √ − − √
𝑟1 𝑎
𝑟1
(9-7)
𝜇
2𝜇 𝜇
∆𝑉2 = 𝑉2 − 𝑉𝑎,𝑡 = √ − √ −
𝑟2
𝑟2 𝑎
(9-8)
𝑇𝑂𝐹 =
1
𝑎3
𝑃𝑡 = 𝜋√
2
𝜇
𝑟𝑝,𝑡 = 𝑟1 , 𝑟𝑎,𝑡 = 𝑟2
rfinal = 42164.215 km
At first impulse, the delta-v required is,
2μ
μ
μ
∆V1 = √
− −√
rinitial a
rinitial
where,
rinitial + rfinal
= 24366.852 km
2
Thus, ∆V1 = 2.4571 km/sec
For the second impulse, the delta-v required is,
a=
μ
2μ
μ
∆V2 = √
−√
− = 1.4782 km/sec
rfinal
rfinal a
The total delta-v require is,
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∆V1
∆V2
r2
Determine the total ∆V required for Hohmann transfer to transfer from
a LEO with hinitial = 191.344 km into GEO.
The initial and final radius is,
rinitial = 191.344 + 6378.145 = 6569.489 km
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EXAMPLE 9-2
SOLUTION
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Figure 9-7. Hohmann transfer.
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∆VTOTAL = 2.4571 + 1.4782 = 3.9353 km/sec
EXAMPLE 9-3
Two geocentric elliptical orbits have common apse lines and their
perigees are on the same side of the Earth. The first orbit has a perigee
radius of 𝒓𝒑 = 𝟕𝟎𝟎𝟎 km and 𝒆 = 𝟎. 𝟑, whereas for the second orbit 𝒓𝒑 =
𝟑𝟐𝟎𝟎𝟎 km and 𝒆 = 𝟎. 𝟓.
a. Find the minimum total delta-v and the time of flight for a
transfer from the perigee of the inner orbit to the apogee of the
outer orbit.
b. Do part (a) for a transfer from the apogee of the inner orbit to
the perigee of the outer orbit.
SOLUTION
a. For 1st orbit:
𝑃1 = 𝑟𝑝1 (1 + 𝑒1 ) = 9100 km
𝑃
𝑎1 = 1−𝑒1 2 = 10000 km
1
2𝜇
𝜇
𝑝1
1
𝑣𝑖𝑛𝑡𝑝1 = √ − = 8.6038 km/sec
𝑟
𝑎
nd
For 2 orbit:
𝑃2 = 48000 km
𝑎=
𝑟𝑎 +𝑟𝑝
2
&
𝑎2 = 64000 km
⇒ 𝑟𝑎2 = 96000 km
2𝜇
𝜇
𝑎2
2
𝑣𝑓𝑖𝑛𝑎2 = √𝑟 − 𝑎 = 1.4408 km/sec
For transient orbit:
𝑟𝑝𝑡 = 𝑟𝑝1 = 7000 km & 𝑟𝑎𝑡 = 𝑟𝑎2 = 96000 km
∴ 𝑎𝑡 = 51500 km
2𝜇
𝜇
𝑝1
𝑡
2𝜇
𝜇
𝑎2
𝑡
𝑣𝑡𝑟𝑎𝑛𝑠𝑝1 = √ − = 10.3027 km/sec
𝑟
𝑎
𝑣𝑡𝑟𝑎𝑛𝑠𝑎2 = √𝑟 − 𝑎 = 0.7512 km/sec
△ 𝑣1 = 𝑣𝑡𝑟𝑎𝑛𝑠𝑝1 − 𝑣𝑖𝑛𝑡𝑝1 = 1.6989 km/sec
△ 𝑣2 = 𝑣𝑓𝑖𝑛𝑎2 − 𝑣𝑡𝑟𝑎𝑛𝑠𝑎2 = 0.6896 km/sec
∴△ 𝑣𝑡𝑜𝑡𝑎𝑙 = 2.3885 km/sec
𝑛𝑡𝑟𝑎𝑛𝑠 = √𝑎3
𝜇
𝑡𝑟𝑎𝑛𝑠
= 5.402 × 10−5 sec −1
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𝑇𝑡𝑟𝑎𝑛𝑠 = 𝑛
2𝜋
𝑡𝑟𝑎𝑛𝑠
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= 1.1631 × 105 sec
Time of flight, TOF:
𝑇𝑂𝐹 =
𝑇𝑡𝑟𝑎𝑛𝑠
2
= 2.3262 × 105 sec = 16.1544 hr
b. For 1st orbit:
𝑟𝑎1 = 𝑎1 (1 + 𝑒1 ) = 13000 km
2𝜇
𝜇
𝑎1
1
2𝜇
𝜇
𝑝2
2
𝑣𝑖𝑛𝑡𝑎1 = √𝑟 − 𝑎 = 4.6328 km/sec
nd
For 2 orbit:
𝑣𝑓𝑖𝑛𝑝2 = √𝑟 − 𝑎 = 4.3225 km/sec
For transient orbit:
𝑟𝑝𝑡 = 𝑟𝑎1 = 13000 km & 𝑟𝑎𝑡 = 𝑟𝑝2 = 32000 km
∴ 𝑎𝑡 = 22500 km
2𝜇
𝜇
𝑎1
𝑡
2𝜇
𝜇
𝑝2
𝑡
𝑣𝑡𝑟𝑎𝑛𝑠𝑎1 = √𝑟 − 𝑎 = 6.6036 km/sec
𝑣𝑡𝑟𝑎𝑛𝑠𝑝2 = √𝑟 − 𝑎 = 2.6827 km/sec
△ 𝑣1 = 𝑣𝑡𝑟𝑎𝑛𝑠𝑎1 − 𝑣𝑖𝑛𝑡𝑎1 = 1.9708 km/sec
△ 𝑣2 = 𝑣𝑓𝑖𝑛𝑝2 − 𝑣𝑡𝑟𝑎𝑛𝑠𝑝2 = 1.6398 km/sec
∴△ 𝑣𝑡𝑜𝑡𝑎𝑙 = 3.6106 km/sec
𝑇𝑡𝑟𝑎𝑛𝑠 = 3.3588 × 104 sec
Time of flight, TOF:
𝑇𝑂𝐹 = 4.665 hr
EXAMPLE 9-4
A spacecraft is in a 300 km circular earth orbit. Calculate the transfer
orbit time for a Hohmann transfer to a 3000 km coplanar circular Earth
orbit.
SOLUTION
For initial orbit, 1:
𝑟1 = 6678.145 km ⇒ 𝑣1 = 7.7258 km/sec
For final orbit, 3:
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𝑟3 = 9378.145 km ⇒ 𝑣3 = 6.5194 km/sec
For elliptical transient orbit, 2:
𝑟2𝑝 = 𝑟𝑖 = 6678.145 km
𝑟2𝑎 = 𝑟𝑓 = 9378.145 km
𝑎2 =
𝑟2𝑎 +𝑟2𝑝
2
= 8028.1 km
Transfer orbit time 𝑇2
𝜇
𝑛2 = √𝑎3 = 8.7771 × 10−4 sec −1
2
∴ 𝑇2 =
(2𝜋⁄𝑛𝑡 )
2
= 0.9943 hr
9.4.4 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 9-8 illustrates a bi-elliptic
transfer between two circular orbits.
∆V1
∆V2
Figure 9-8. Bielliptic Transfer.
∆V3
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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. 9-10 shows the cost comparison between Hohmann and
Bi-elliptic Transfer. R is the ratio of final to initial 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.
Bi-elliptic
Total Cost of ∆V, (km/sec)
Hohmann
R* Increases
R* ≈ 50
R* ≈ 60
R* ≈ 100
R* ≈ 200
R* = ∞
R* ≈ 11.94
R* ≈ 15.58
Final radius to initial radius ratio, R
Figure 9-9.Delta-v Cost Comparison between Hohmann and Bi-elliptic Transfer.
EXAMPLE 9-5
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, hfinal = 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
rfinal = 376310 + 6378.145 = 382688.145 km
And the semimajor axis for both transfer orbits are,
rinitial + rtrans
a1 =
= 258410.317 km
2
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rfinal + rtrans
= 446469.645 km
2
At first impulse, the delta-v required is,
2μ
μ
μ
∆V1 = √
− −√
= 3.156 km/sec
rinitial a1
rinitial
At the second impulse, the delta-v required is,
2μ
μ
2μ
μ
∆V2 = √
− −√
− = 0.677 km/sec
rtrans a2
rtrans a1
At the third impulse, the delta-v required is,
∆V3 = √
μ
rfinal
−√
2μ
rfinal
−
μ
= −0.0705 km/sec
a2
The total delta-v require is,
∆VTOTAL = 3.156 + 0.677 + 0.0705 = 3.9035 km/sec
The time of flight is,
TOF = π × (√
a31
a3
+ √ 2 ) = 2138113.26 sec = 593.92 hr
μ
μ
EXAMPLE 9-6
A spacecraft is in a 300 km circular Earth orbit. Calculate
a. The total delta-v required for the bi-elliptical transfer to a 3000
km altitude coplanar circular orbit shown, and
b. Compare the total transfer time with the Hohmann’s transfer
time in Example 9-4.
SOLUTION
a. For initial orbit, 1:
r1 = 6678.145 km ⇒ v1 = 7.7258 km/sec
For final orbit, 4:
r4 = 9378.145 km ⇒ v4 = 6.5194 km/sec
For first elliptical transient orbit, 2:
𝑒2 =0.3
r2A = r1 = 6678.145 km
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r2
a2 = 1−𝑒A = 9540.21 km
2
r2B = rB = a2 (1 + 𝑒2 ) = 12402.26 km
2μ
μ
v2A = √r − a = 8.8087 km/sec
2
2
A
2𝜇
v2B = √r
2B
𝜇
− a = 4.74316 km/sec
2
For second elliptical transient orbit, 3:
r3C = r4 = 9378.145 km
r3B = rB = 12402.26 km
a3 =
r3C +r3B
2
2μ
= 10890.2 km
μ
v3C = √r − a = 6.9573 km/sec
3
3
C
2𝜇
v3B = √r
3B
𝜇
− a = 5.26088 km/sec
3
△ vA = v2A − v1 = 1.0829 km/sec
△ vB = v3B − v2B = 0.51772 km/sec
△ vC = v3C − v4 = 0.43793 km/sec
∴△ 𝑣𝑡𝑜𝑡𝑎𝑙 = |△ 𝑣A | + |△ 𝑣B | + |△ 𝑣C | = 2.03855 km/sec
b. Total transfer time, 𝑇𝑡𝑜𝑡𝑎𝑙
𝑇𝑡𝑜𝑡𝑎𝑙 =
𝑇2 𝑇3
+
2
2
where,
2𝜋
= 9.2736 × 103 sec
𝜇
√ ⁄𝑎3
2
2𝜋
𝑇3 = 𝜇 = 1.131 × 104 sec
√ ⁄𝑎3
3
𝑇𝑇𝑂𝑇𝐴𝐿 = 1.02918 × 104 sec = 2.859
𝑇2 =
∴
hr
From Example 9-4, Hohmann Transfer only requires 0.9943 hr to transfer
the spacecraft into another orbit. However, the bi-elliptic requires 3 times
longer of transfer time to transfer the spacecraft into another orbit.
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9.4.5 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 9-10.
Transfer Orbit
Transfer Orbit
r2
r1
Possible because
rp < r1 and ra > r2
r2
r1
Impossible because
rp > r1
Transfer Orbit
r2
r1
Impossible because
ra < r2
Figure 9-10.General coplanar transfer between circular orbits.
9.4.6 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 9-11. Phasing Maneuver.
Figure 9-11 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 vice 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:
τphase =
2π − ∆θ
ntgt
(9-11)
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,
τphase =
aphase
2π
nphase
a3phase
= 2π√
μ
τphase √μ
=(
)
2π
2/3
(9-12)
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EXAMPLE 9-7
Determine the semimajor axis of the phasing orbit, given that the
position of target and interceptor spacecraft are:
𝐫⃗𝐭𝐠𝐭 = 𝟒𝐉̂ 𝐃𝐔
̂ 𝐃𝐔
𝐫⃗𝐢𝐧𝐭 = 𝟐𝐉̂ + 𝟐√𝟑𝐊
SOLUTION
First, the phase angle between spacecraft is,
𝐫⃗𝐭𝐠𝐭 ∙ 𝐫⃗𝐢𝐧𝐭
∆θ = cos −1 (
) = 60°
‖𝐫⃗𝐭𝐠𝐭 ‖‖𝐫⃗𝐢𝐧𝐭 ‖
The mean motion of the original orbit is,
μ
ntgt = √ 3 = 0.125 rad/TU
r
Then, the one orbit period required for the phasing orbit is,
2π − ∆θ
τphase =
= 41.888 TU
ntgt
The semimajor axis for the phasing orbit is,
2/3
aphase
τphase √μ
=(
)
2π
= 3.5422 DU
9.5 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.
9.5.1 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
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satellite, but only the inclination of the orbit. The maneuver is called
simple plane change (see Figure 9-12).
V2
α
∆V
V1
Figure 9-12.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 913, the delta-v required is,
V
V
θ
∆𝑉 = 2𝑉 sin
𝜃
2
(9-13)
EXAMPLE 9-8
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,
μ
25 − 10
∆V = 2 × √ × sin (
) = 1.973 km/sec
r
2
V
Figure 9-13.Velocity vector triangle
for circular orbit plane change.
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9.5.2 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 9-14 shows the example of general
plane change maneuver.
Nodes 1 and 2 are the direction of RAAN for both initial and final orbits
respectively, where else the ê is the eccentricity vector (also known as
argument of perigee) for the initial orbit.
Z
Final Orbit
Z
Initial Orbit
Final Orbit
Initial Orbit
X
α
Ωinitial
ALa
θ
ifinal
Equatorial Plane
node 2
iinitial
ω
node 1
∆Ω
eˆ
node 1, Ωi
Figure 9-14.General Plane Change Maneuver.
node 2, Ωf
Figure 9-15.Argument of latitude of intersection
point.
The delta-v required for the general plane change maneuver is shown in
equation 9-17. The α angle is determined using equation 9-14, and ALa is the
argument of latitude of intersection point, that is shown in
Figure 9-15.
cos𝛼 = cos 𝑖𝑖𝑛𝑖𝑡𝑖𝑎𝑙 cos 𝑖𝑓𝑖𝑛𝑎𝑙 + sin 𝑖𝑖𝑛𝑖𝑡𝑖𝑎𝑙 sin 𝑖𝑓𝑖𝑛𝑎𝑙 cos(ΔΩ)
sin ALa =
sin 𝑖𝑓𝑖𝑛𝑎𝑙 sin(∆Ω)
sinα
(9-14)
(9-15)
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ALa = θ + ω
(9-16)
α
∆V = 2Vsin ( )
2
(9-17)
R B I T A L
EXAMPLE 9-9
Compute ΔV required to change the right ascension of the ascending node
of the following orbit to 100o West:
rp1 = 1.1DU, e1 = 0.1, i = 45˚ , Ω = 40˚ West, ω = 10˚
SOLUTION
The orbit of satellite is transfer to Ω = 100˚ West.
The ΔV required is given by,
α
∆V = 2Vsin ( )
2
where α is,
cos𝛼 = cos 𝑖𝑖𝑛𝑖𝑡𝑖𝑎𝑙 cos 𝑖𝑓𝑖𝑛𝑎𝑙 + sin 𝑖𝑖𝑛𝑖𝑡𝑖𝑎𝑙 sin 𝑖𝑓𝑖𝑛𝑎𝑙 cos(ΔΩ)
ΔΩ = −40° + 100° = 60°
𝛼 = 41.4°
The speed of satellite at that particular point is,
2𝜇 𝜇
V=√ −
𝑟
𝑎
where,
a = 1.22DU
And r can be obtain through,
sin 𝑖𝑓𝑖𝑛𝑎𝑙 sin(∆Ω)
sinα
ALa = 67.8˚
θ = ALa – ω = 57.8˚
rp1 (1 + e)
r=
= 1.14878 DU
1 + ecosθ
sin ALa =
Thus,
V = 0.96 DU/TU
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ΔV = 0.679 DU/TU
9.5.3 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.
Figure 9-16 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.
∆iinitial = s∆i
∆ifinal = (1 − s)∆i
(9-18)
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K̂
Transfer Orbit
Final Orbit
∆Vb
∆Va
Initial Orbit
Jˆ
Iˆ
Figure 9-16.Orbit transfer of a spacecraft using combined maneuver.
The total delta-v that required for the combined maneuver is,
2
2
∆V = √Vtrans
+ Vinitial
− 2Vinitial Vtransa cos(s∆i)
a
2
2
+√Vtrans_b
+ Vfinal
− 2Vfinal Vtrans_b cos((1 − s)∆i)
(9-19)
Now, the optimum scaling, s is required to determine to produce minimum
cost. Then, we taked ∆V⁄ds = 0
s≈
1
sin(∆i)
tan−1 [
]
Vinitial Vtrans_a
∆i
Vfinal Vtrans_b cos(∆i)
(9-18)
For circular initial and final orbits:
Vinitial Vtrans_a
rfinal
= √R3 , R =
Vfinal Vtrans_b
rinitial
(9-19)
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EXAMPLE 9-10
Calculate the total delta-v required for a spacecraft to transfer from an
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
a=
= 1.675DU
2
The velocities at each location are:
Vinitial = √
Vtrans_a = √
μ
= 0.9901 DU/TU
r1
2μ μ
− = 1.1678 DU/TU
r1 a
2μ μ
Vtrans_b = √ − = 0.5512 DU/TU
r2 a
μ
Vfinal = √ = 0.6551 DU/TU
r2
Then, we need to determine the scaling, s, that is:
1
sin(Δ𝑖)
s = tan−1 [ 3/2
] = 0.224105
Δ𝑖
R + cos(Δ𝑖)
Therefore, the total delta-v is,
∆V = 0.3464 DU/TU
9.6 Problems
1.
A space vehicle in a circular orbit at an altitude of 0.0784 DU above the
Earth executes a Hohmann transfer to a 0.1568 DU circular orbit.
Calculate the total delta-v requirement.
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2.
Calculate the total delta-v required for a Hohmann transfer from a
circular orbit of radius 4r to a circular orbit of radius 16r.
3.
Determine the total time of flight for Hohmann and bi-elliptic transfer
with given orbit properties:
i. Initial orbit, hinitial = 200 km,
ii. Apogee altitude of transfer orbit for bi-elliptic transfer, hapog = 35000
km,
iii. Final orbit, hfinal = 30000 km.
4.
Determine the total ∆V required for a bi-elliptic transfer with given
orbit properties:
i. Initial orbit, r1 = 1.013 DU,
ii. Apogee altitude of transfer orbit, r2 = 3.021 DU,
iii. Final orbit, r3 = 2.601 DU.
i.
ii.
Determine the total ∆V required and time-of-flight for a bi-elliptic
transfer to place a spacecraft from r1 = 6578.145 km into GEO. Given
that, the apogee radius of transfer orbit is, r2 = 46378.145 km.
Given that the target and interceptor spacecraft are orbiting around
the Earth in equatorial orbit. Determine the semimajor axis of the
phasing orbit. Both spacecrafts’ positions at that time are:
r⃗tgt = 12546.387Î + 10527.667Ĵ km
r⃗int = 16129.324Î + 2844.035Ĵ km
iii. Given that the target and interceptor spacecraft are in an equatorial
orbit with semimajor axis of 15000 km and eccentricity 0.1. Determine
the semimajor axis of the phasing orbit if the distances of target and
interceptor spacecraft to the Earth center are 13,574.4216 km and
13,615.9737 km respectively at the time. (Assume that the true
anomaly of both spacecraft are in first quadrant.)
iv. Determine the ∆V required for a satellite to change its orbit plane from
equatorial orbit to an orbit with inclination 10° at altitude 400km.
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Determine the total ∆V required for a satellite to change its orbit plane
at inclination 5° to GEO.
vi. Compute ΔV required to change from following orbit to the right
ascension of the ascending node at 35˚ and inclination at 15˚:
rp1 = 1.08 DU, e1 = 0.05, i = 20˚ , Ω = 20˚, ω = 5˚
vii. Compute ΔV required to change from following orbit to the right
ascension of the ascending node at 30˚ and inclination at 25˚:
Altitude, h = 300 km, e1 = 0, i = 10˚ , Ω = 25˚
viii. Calculate the total delta-v required for a spacecraft to transfer from an
orbit with altitude, h1 = 400 km to geosynchronous orbit with the
change of inclination ∆i = 25°.
ix. Calculate the total delta-v required for a spacecraft to transfer from an
orbit, r1 = 1.157 DU to r2 = 4.136 DU with the change of inclination ∆i =
20° using:
a. Hohmann transfer followed by simple plane change.
b. Combined Maneuver.
9.7 References
Chobotov, V. (2002). Orbital Mechanics. Reston, Virginia, American Institute
of Aeronautics and Astronautics, Inc.
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