1
11
9 Relative Motion
11.1
General Relative Motion
Z
Chaser
k
rrel
j
i
Target
X
Y
Figure 11-1. Relative position of Chaser to Target in RSW frame.
Given two spacecraft that moving in an inertial frame (see Figure 11-1),
where the relative position between two spacecraft is defined as:
⃗ 𝒓𝒆𝒍 = 𝒓
⃗𝑩−𝒓
⃗𝑨
𝒓
(11-1)
Let the relative positions between both spacecraft are measured in
spacecraft A’s RSW-reference frame, such that
⃗𝒓𝑨𝒓𝒆𝒍 = [𝑸]𝒓
⃗𝑵
𝒓𝒆𝒍
(11-2)
where the superscript “N” denotes the representation in Inertial reference
frame, superscript “A” denotes the representation in spacecraft A’s RSWreference frame, and [𝑸] is the attitude matrix of spacecraft A for RSWreference frame.
Because the RSW-reference frame rotates along the orbit, the angular
velocity of spacecraft A is considered. Consider the time derivative of
kinematics with rotation, then the time derivative of Eq. (11-1) is,
⃗⃗ 𝑨 × 𝒓
⃗𝑩=𝒗
⃗ 𝑨+𝜴
⃗ 𝒓𝒆𝒍 + 𝒗
⃗ 𝒓𝒆𝒍
𝒗
(11-3)
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⃗⃗ 𝑨 is the angular velocity of the spacecraft A with respect to the
where 𝜴
Earth. Then, taking a time derivative of Eq. (11-3), the acceleration of
spacecraft B expressed in spacecraft A and relative positions, velocities and
accelerations is,
⃗⃗ 𝑨 × (𝜴
⃗⃗ 𝑨 × 𝒓
⃗⃗ 𝑨 × 𝒗
⃗𝑩=𝒂
⃗ 𝑨 + 𝜴̇𝑨 × 𝒓
⃗ 𝒓𝒆𝒍 + 𝜴
⃗ 𝒓𝒆𝒍 ) + 𝟐𝜴
⃗ 𝒓𝒆𝒍
𝒂
⃗ 𝒓𝒆𝒍
+𝒂
(11-4)
⃗ 𝑨.
𝜴̇𝑨 is the rate of change of the angular velocity, ⃗𝜴
Define that the specific angular momentum of spacecraft is,
⃗𝒉𝑨 = 𝒓
̂
⃗𝑨×𝒗
⃗ 𝑨 = 𝒓𝑨 𝒗𝑨⊥ 𝒌
(11-5)
⃗ 𝑨 . In addition, given
where 𝒗𝑨⊥ is the velocity vector that perpendicular to 𝒓
⃗ 𝑨 and 𝒗𝑨 is,
that the relationship between angular velocity, ⃗𝜴
⊥
𝒗𝑨⊥ = 𝒓𝑨 𝜴𝑨
(11-6)
Substitute Eq. (11-6) into (11-5), we have,
⃗⃗𝜴𝑨 =
⃗𝑨×𝒗
⃗𝑨
𝒓
̂
= 𝜴𝑨 𝒌
𝒓𝟐𝑨
(11-7)
Taking the time derivative for Eq. (11-7), we obtain,
⃗𝑨∙𝒗
⃗ 𝑨)
𝟐
𝟐(𝒓
⃗⃗ ̇ 𝑨 = − 𝒓̇ 𝑨 (𝒓
⃗⃗ 𝑨
⃗𝑨×𝒗
⃗ 𝑨) = −
𝜴
𝜴
𝟐
𝟐
𝒓𝑨
𝒓𝑨
(11-8)
We have the angular velocities and accelerations derived in Eqs. (11-7) and
(11-8), then we can determine the relative velocities and accelerations in
Inertial frame using Eqs. (11-3) and (11-4). In addition, the representation
of relative velocities and accelerations in spacecraft A’s RSW-reference
frame are,
⃗ 𝑨𝒓𝒆𝒍 = [𝑸]𝒗
⃗𝑵
𝒗
𝒓𝒆𝒍
⃗ 𝑨𝒓𝒆𝒍
𝒂
=
[𝑸]𝒂
⃗𝑵
𝒓𝒆𝒍
(11-9)
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3
Observation by a spacecraft on relative motion
Consider two spacecraft in two orbits with same semimajor axis. That is,
both spacecraft have same orbit period. Let the first orbit is circular and the
second orbit is elliptic with small eccentricity that is shown in Figure 11-2.
X
3
X
X
4
2
Y
1
5
X
X
Y
6
8
X
Spacecraft A
Spacecraft B
X
7
X
Figure 11-2. Relative direction observation in Inertial frame.
The spacecraft A’s reference frame is always rotating so that one of the axis
is pointing to the earth center. Because spacecraft B orbits in elliptic orbit,
its velocity is not uniform all the time due to the orbit’s characteristic. This
results a bean-shape observation occurred when spacecraft B is observed
by spacecraft A in an orbit period (see Figure 11-3).
There is no any force occurs between both spacecraft. However, spacecraft
B travels faster when it is closes to perigee (or altitude is below spacecraft
A) and travels slower when it is closer to apogee (or altitude is above
spacecraft A). As a result, the direction of observation done by spacecraft A
on spacecraft B varies in a specific clock direction.
EXAMPLE 11-1
Given the orbital elements of spacecraft A at particular time are,
𝐚 = 𝟖𝟎𝟎𝟎𝐤𝐦, 𝐞 = 𝟎. 𝟎𝟏, 𝐢 = 𝟏𝟓°, 𝛉 = 𝟐𝟓°, 𝛚 = 𝟏𝟎°, 𝛀 = 𝟓°
At the same time, the orbital elements of spacecraft B are,
𝐚 = 𝟖𝟏𝟎𝟎𝐤𝐦, 𝐞 = 𝟎. 𝟎𝟏𝟐, 𝐢 = 𝟐𝟎°, 𝛉 = 𝟐𝟎°, 𝛚 = 𝟏𝟓°, 𝛀 = 𝟓°
Determine the relative position, velocity and acceleration that expressed
in spacecraft A reference frame.
Y
3
2
4
5
1
8
X
6
7
Figure 11-3. As viewed from co-moving
frame by spacecraft A.
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SOLUTION
First, the absolute positions and velocities for both spacecraft are,
𝑟𝐴 = [6086.2086 4941.2610 1176.8355]T km
𝑣𝐴 = [−4.5384 5.2771 1.5146]T km/sec
𝑟𝐵 = [6159.0316 4871.8140 1571.0703]T km
𝑣𝐵 = [ −4.5076 5.1031 1.9933]T km/sec
Next, we determine the angular velocity and rate of angular velocity of
spacecraft A that are expressed in Spacecraft A’s RSW frame, which are:
μ
̂ = 8.82334 × 10−4 𝐤
̂ rad/sec
⃗ω
⃗A=√ 3𝐤
aA
𝟐(𝒓𝑨 ∙ 𝒗𝑨 )
̂ rad/sec
⃗ω
⃗̇ A = −
⃗⃗ 𝑨 = −0.6641 × 10−8 𝐤
ω
𝟐
𝒓𝑨
Next, we determine the Directional Cosine Matrix of spacecraft A at that
point, which is,
[Q]A = C(Ω, i, ω + θ)
0.76775
0.62332 0.14845
[Q]A = [ −0.64035 0.73824 0.21201]
0.02256 −0.25783 0.96593
Then, the angular velocity and rate of angular velocity can be expressed in
Inertial frame, which are:
⃗⃗ N = [Q]TA ω
ω
⃗⃗ A
⃗⃗̇ N = [Q]TA ω
ω
⃗⃗̇ A
Now, the relative position in Inertial frame is,
𝑁
𝑟𝑟𝑒𝑙
= 𝑟𝐵 − 𝑟𝐴 = [72.8230 −69.4471 394.2348]T km
The relative velocity in Inertial frame is,
𝑁
⃗𝑁
⃗ 𝑩−𝒗
⃗ 𝑨−ω
𝒗
⃗⃗ N × 𝑟𝑟𝑒𝑙
𝑟𝑒𝑙 = 𝒗
𝑁
⃗ 𝑟𝑒𝑙 = [0.06122 −0.22816 0.46354]T km/sec
𝒗
The relative acceleration in Inertial frame is,
𝑁
𝑁
𝑁 )
⃗ 𝑩−𝒂
⃗ 𝑨−ω
⃗𝑁
𝑎𝑟𝑒𝑙
=𝒂
⃗⃗̇ N × 𝑟𝑟𝑒𝑙
−ω
⃗⃗ N × (ω
⃗⃗ N × 𝑟𝑟𝑒𝑙
− 𝟐ω
⃗⃗ N × 𝒗
𝑟𝑒𝑙
𝑁
𝑎𝑟𝑒𝑙
= [−0.03850 0.11369 −0.29040]T × 10−3 km/sec 2
Thus, the relative position, velocity and acceleration expressed in
spacecraft A’s RSW frame are,
𝐴
𝑁
𝑟𝑟𝑒𝑙
= [Q]A 𝑟𝑟𝑒𝑙
= [71.1472 −14.3185 400.3501]T km
𝐴
𝑁
𝑣𝑟𝑒𝑙
= [Q]A 𝑣𝑟𝑒𝑙
= [−0.02639 −0.10937 0.50795]T km/sec
𝐴
𝑁
𝑎𝑟𝑒𝑙
= [Q]A 𝑎𝑟𝑒𝑙
= [−0.001802 0.047013 −0.310685]T km/sec 2
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Linearized Equations of Relative
Motion
Z
Target
rB
δr
rA
X
Chaser
Y
Figure 11-4. Relative Position of Target Spacecraft to Chaser Spacecraft.
We consider two spacecraft that their position vectors expressed in Inertial
⃗ 𝑨 is the position vector of
frame, that are 𝑟𝐴 and 𝑟𝐵 in Figure 11-4. Let 𝒓
⃗ 𝑩 is the position vector of target spacecraft. The
chaser spacecraft and 𝒓
relative position between two spacecraft can be expressed as,
⃗ =𝒓
⃗𝑩−𝒓
⃗𝑨
𝜹𝒓
(11-10)
If we assume that the relative distance between two spacecraft is very small
compared to the absolute distance to the Earth’s center,
𝜹𝒓
≪𝟏
𝒓𝑨
(11-11)
The assumption in Eq. (11-11) is reasonable, especially for spacecraft
rendezvous case. Next, let the equation of motion for spacecraft B is,
⃗𝑩
𝒓
⃗ ̈ 𝑩 = −𝝁 𝟑
𝒓
𝒓𝑩
(11-12)
Substitute the Eq. (11-12) into the second order time derivation of Eq. (1110) yields,
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⃗ +𝒓
⃗𝑨
𝜹𝒓
⃗ ̈ = −𝒓
⃗̈𝑨 − 𝝁
𝜹𝒓
𝟑
𝒓𝑩
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(11-13)
Then, let the range of target spacecraft, 𝑟𝐵 to the Earth’s center can be
expressed as,
⃗ +𝒓
⃗ 𝑨 ) ∙ (𝜹𝒓
⃗ +𝒓
⃗ 𝑨) = 𝒓
⃗𝑨∙𝒓
⃗ 𝑨 + 𝟐𝜹𝒓
⃗ ∙𝒓
⃗ 𝑨 + 𝜹𝒓
⃗ ∙ 𝜹𝒓
⃗
𝒓𝟐𝑩 = (𝜹𝒓
(11-14)
If we expend the Eq. (11-14), and considering the assumption in Eq. (11-11).
Then, we have
𝒓𝟐𝑩 = 𝒓𝟐𝑨 [𝟏 +
⃗ ∙𝒓
⃗ 𝑨 𝜹𝒓𝟐
⃗ ∙𝒓
⃗𝑨
𝟐𝜹𝒓
𝟐𝜹𝒓
+ 𝟐 ] ≈ 𝒓𝟐𝑨 [𝟏 +
]
𝟐
𝟐
𝒓𝑨
𝒓𝑨
𝒓𝑨
(11-15)
Next, consider that 𝑟 −3 = (𝑟 2 )−3/2, Eq. (11-15) can be expressed as,
−
𝒓−𝟑
𝑩
⃗ ∙𝒓
⃗𝑨
𝟐𝜹𝒓
= 𝒓−𝟑
]
𝑨 [𝟏 +
𝟐
𝒓𝑨
𝟑
𝟐
(11-16)
To linearize Eq. (11-16), we consider the binomial series, that is,
(1 + 𝑥)𝑛 = 1 + 𝑛𝑥 +
where, 𝑥 =
⃗ ∙𝒓
⃗𝑨
𝟐𝜹𝒓
𝒓𝟐𝑨
𝑛(𝑛 − 1) 2
𝑥 +⋯
2!
(11-17)
. By neglecting the high order terms in binomial series, we
have,
−
⃗ ∙𝒓
⃗𝑨
𝟐𝜹𝒓
(𝟏 +
)
𝟐
𝒓𝑨
𝟑
𝟐
⃗ ∙𝒓
⃗𝑨
𝟑 𝟐𝜹𝒓
≈ 𝟏 + (− ) (
)
𝟐
𝟐
𝒓𝑨
(11-18)
Then, substitute Eq. (11-18) into (11-16), we obtain,
−𝟑
𝒓−𝟑
𝑩 = 𝒓𝑨 (𝟏 −
⃗ ∙𝒓
⃗𝑨
𝟑𝜹𝒓
𝟏
𝟑
⃗ ∙𝒓
⃗𝑨
) = 𝟑 − 𝟓 𝜹𝒓
𝟐
𝒓𝑨
𝒓𝑨 𝒓𝑨
(11-16)
Now, Eq. (11-13) becomes,
𝟏
𝟑
⃗ ̈ = −𝒓
⃗ ̈ 𝑨 − 𝝁 ( 𝟑 − 𝟓 𝜹𝒓
⃗ ∙𝒓
⃗ 𝑨 ) (𝜹𝒓
⃗ +𝒓
⃗ 𝑨)
𝜹𝒓
𝒓𝑨 𝒓𝑨
(11-17)
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⃗
⃗𝒓
𝜹𝒓
𝟑
𝟑
⃗ ̈ 𝑨 − 𝝁 [ 𝟑 + 𝑨𝟑 − 𝟓 (𝜹𝒓
⃗ ∙𝒓
⃗ 𝑨 )𝒓
⃗ 𝑨 − 𝟓 (𝜹𝒓
⃗ ∙𝒓
⃗ 𝑨 )𝜹𝒓
⃗]
= −𝒓
𝒓
𝒓
𝒓
𝒓
𝑨
𝑨
𝑨
𝑨
Again, here we assume that the relative positions between two spacecraft is
𝟑
⃗ ∙𝒓
⃗ 𝑨 )𝜹𝒓
⃗ ≪ 1. Then, the Eq. (11small, therefore we can consider that 𝒓𝟓 (𝜹𝒓
𝑨
17) becomes,
⃗
⃗
𝜹𝒓
𝒓
𝟑
⃗ ̈ = −𝒓
⃗ ̈ 𝑨 − 𝝁 [ 𝟑 + 𝑨𝟑 − 𝟓 (𝜹𝒓
⃗ ∙𝒓
⃗ 𝑨 )𝒓
⃗ 𝑨]
𝜹𝒓
𝒓
𝒓
𝒓
𝑨
𝑨
(11-18)
𝑨
𝑟
Recall that 𝑟𝐴̈ = −𝜇 𝐴3 and substitute it into Eq. (11-18). Finally, the
𝑟𝐴
linearized relative motion of equation is,
⃗
𝜹𝒓
𝟑
⃗ ̈ = −𝝁 [ 𝟑 − 𝟓 (𝜹𝒓
⃗ ∙𝒓
⃗ 𝑨 )𝒓
⃗ 𝑨]
𝜹𝒓
𝒓𝑨 𝒓𝑨
(11-19)
The equation of motion for the relative position in Eq. (11-19) is considered
⃗ is only
to be linear to the relative position vector, because the vector 𝜹𝒓
appear in numerator. In addition, Eq. (11-19) stays true as long as the
assumption in Eq. (11-11) applicable.
EXAMPLE 11-2
Given that the initial positions and velocities of two spacecraft are,
𝐫𝟏 = [𝟐𝟐𝟓𝟖. 𝟎𝟏𝟖𝟑 𝟓𝟗𝟐𝟑. 𝟐𝟓𝟏𝟔 𝟏𝟒𝟓𝟕. 𝟗𝟓𝟓𝟑]𝐓 𝐤𝐦
𝐫𝟐 = [𝟐𝟐𝟑𝟓. 𝟗𝟒𝟑𝟗 𝟓𝟖𝟔𝟓. 𝟔𝟐𝟐𝟔 𝟏𝟒𝟒𝟑. 𝟖𝟕𝟔𝟐]𝐓 𝐤𝐦
𝐯⃗𝟏 = [−𝟕. 𝟒𝟕𝟕𝟓𝟔𝟏 𝟐. 𝟗𝟎𝟖𝟓𝟐𝟏 𝟏. 𝟏𝟏𝟓𝟒𝟏𝟗]𝐓 𝐤𝐦/𝐬𝐞𝐜
𝐯⃗𝟐 = [−𝟕. 𝟓𝟑𝟐𝟑𝟑𝟑 𝟐. 𝟗𝟔𝟔𝟕𝟎𝟓 𝟏. 𝟏𝟑𝟑𝟒𝟎𝟎]𝐓 𝐤𝐦/𝐬𝐞𝐜
Using the linearized method, determine the relative position after
30mins. Then compared with the relative position obtained through
determine the absolute position of each spacecraft, using Eq. (11-12).
SOLUTION
First, the relative positions and velocities are,
𝑟12 = 𝑟2 − 𝑟1 = [−22.0744 −57.6290 −14.0791]T km
𝑣12 = 𝑣2 − 𝑣1 = [−0.054772 0.058184 0.017981]T km/sec
Using the Eq. (11-19), where 𝑟𝐴 = 𝑟1 and 𝛿𝑟 = 𝑟12. The relative position
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after 30mins is,
𝑟12 (𝑡 = 30min) = [−52.2169 −30.0041 −5.4692]T km
The absolute position of each spacecraft after 30mins are:
𝑟1 (t = 30min) = [−7414.0065 −376.3305 245.6600]T km
𝑟2 (t = 30min) = [−7465.1619 −407.2900 239.8874]T km
Then, the relative position is,
𝑟12 = [−51.1554 −30.9595 −5.7726]T km
The relative position obtained using Eq. (11-19) is considered to be close
to the actual relative position. The precision that base on the assumption,
𝛿𝑟
𝑟𝐴
≪ 1 can be improved significantly if both spacecraft have a closer
formation flying.
11.3
Clohessy-Wiltshire (CW) Equation
In this section, the relative motions of equations between two spacecraft
are derived. The derivation’s assumptions are based on the both spacecraft
are in closed near circular orbits, that is the relative distance between
spacecraft is very small compared to the absolute distance to the earth.
The derivation of Hill’s (or CW) equation begins with the consideration of
equation of motion for non-circular orbit. However, the final formulation
shows the restriction of the application.
We define that the equation of motion for the target spacecraft is:
̈ = −𝜇 𝑟𝑡𝑔𝑡
𝑟𝑡𝑔𝑡
3
𝑟𝑡𝑔𝑡
(11-20)
Theoretically, the equation (11-20) also applies to the interceptor spacecraft.
However, for spacecraft rendezvous, the other forces applied on interceptor
spacecraft, such as thrust, aerodynamic drags and etc are required to be
taking into account.
̈ = −𝜇 𝑟𝑡𝑔𝑡 + 𝐹
𝑟𝑡𝑔𝑡
3
𝑟𝑡𝑔𝑡
(11-21)
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Define that the relative position between target and interceptor spacecraft
is:
𝑟𝑟𝑒𝑙 = 𝑟𝑖𝑛𝑡 − 𝑟𝑡𝑔𝑡
(11-22)
Taking the second time derivative of equation (11-22), we obtain the
relative acceleration between two spacecraft, that is:
̈ = 𝑟𝑖𝑛𝑡
̈ − 𝑟𝑡𝑔𝑡
̈
𝑟𝑟𝑒𝑙
(11-23)
Substitute equations (11-20) and (11-21) into (11-23), we obtain,
̈ = − 𝜇 𝑟𝑖𝑛𝑡 + 𝐹 + 𝜇 𝑟𝑡𝑔𝑡
𝑟𝑟𝑒𝑙
3
3
𝑟𝑖𝑛𝑡
𝑟𝑡𝑔𝑡
(11-24)
Recall the relative position equation in equation (11-22), the relative
distance between two spacecraft can be expressed as:
2
2
2
𝑟𝑟𝑒𝑙
= 𝑟𝑖𝑛𝑡
− 𝑟𝑡𝑔𝑡
− 2𝑟𝑖𝑛𝑡 ∙ 𝑟𝑡𝑔𝑡
(11-25)
Substitute the equation (11-25) into (11-24), we obtain:
𝑟𝑖𝑛𝑡
3 =
𝑟𝑖𝑛𝑡
𝑟𝑡𝑔𝑡 + 𝑟𝑟𝑒𝑙
2
2
(𝑟𝑡𝑔𝑡
+ 2𝑟𝑡𝑔𝑡 . 𝑟𝑟𝑒𝑙 + 𝑟𝑟𝑒𝑙
)
3⁄
2
(11-26)
=
𝑟𝑡𝑔𝑡 + 𝑟𝑟𝑒𝑙
𝑟3
1
3⁄
2
2𝑟𝑡𝑔𝑡 . 𝑟𝑟𝑒𝑙
𝑟𝑟𝑒𝑙 2
[1 +
+
(
2
𝑟𝑡𝑔𝑡 ) ]
𝑟𝑡𝑔𝑡
{
}
2
Here, we assume that the relative distance, 𝑟𝑟𝑒𝑙
is much smaller compared
2
to the target spacecraft radius, 𝑟𝑡𝑔𝑡 . Then, we have,
2
𝑟𝑟𝑒𝑙
2 ≈0
𝑟𝑡𝑔𝑡
(11-27)
𝑟𝑖𝑛𝑡 𝑟𝑡𝑔𝑡 + 𝑟𝑟𝑒𝑙
3 =
𝑟3
𝑟𝑖𝑛𝑡
1
3⁄
2
2𝑟𝑡𝑔𝑡 . 𝑟𝑟𝑒𝑙
[1 +
]
2
𝑟𝑡𝑔𝑡
{
}
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Consider the binomial series shown in Eq. (11-17), where 𝑥 represents the
2𝑟𝑡𝑔𝑡 .𝑟𝑟𝑒𝑙
2
𝑟𝑡𝑔𝑡
3
and 𝑛 is − 2. Then, the equations (11-27) becomes,
𝑟𝑖𝑛𝑡 𝑟𝑡𝑔𝑡 + 𝑟𝑟𝑒𝑙
3 2𝑟𝑡𝑔𝑡 . 𝑟𝑟𝑒𝑙
{1 − (
)+⋯}
3 =
3
2
2
𝑟𝑡𝑔𝑡
𝑟𝑖𝑛𝑡
𝑟𝑡𝑔𝑡
(11-28)
Substitute equation (11-28) into (11-24),
̈ = − 𝜇 {(𝑟𝑡𝑔𝑡 − 3 𝑟𝑡𝑔𝑡 (2𝑟𝑡𝑔𝑡 . 𝑟𝑟𝑒𝑙 ) + 𝑟𝑟𝑒𝑙
𝑟𝑟𝑒𝑙
3
2
2
𝑟𝑡𝑔𝑡
𝑟𝑡𝑔𝑡
2𝑟𝑡𝑔𝑡 . 𝑟𝑟𝑒𝑙
3
− 𝑟𝑟𝑒𝑙 (
) + ⋯ ) − 𝑟𝑡𝑔𝑡 } + 𝐹
2
2
𝑟𝑡𝑔𝑡
𝑟𝑡𝑔𝑡
+𝜇 3
𝑟𝑡𝑔𝑡
(11-29)
We assume that the higher order terms in the binomial series are ≈ 0. Then,
removing the higher order terms and keeping only the first order term, we
have,
̈ = − 𝜇 { − 3 𝑟 (2𝑟𝑡𝑔𝑡 . 𝑟𝑟𝑒𝑙 ) + 𝑟
𝑟𝑟𝑒𝑙
𝑟𝑒𝑙
3
2
2 𝑡𝑔𝑡
𝑟𝑡𝑔𝑡
𝑟𝑡𝑔𝑡
2𝑟𝑡𝑔𝑡 . 𝑟𝑟𝑒𝑙
3
− 𝑟𝑟𝑒𝑙 (
)} + 𝐹
2
2
𝑟𝑡𝑔𝑡
(11-30)
It is possible to reduce equation (11-30) into a more simple term. Consider
the assumption that 𝑟𝑟𝑒𝑙 is much smaller than 𝑟𝑡𝑔𝑡 . Then, we can conclude
that,
𝑟𝑟𝑒𝑙 (
2𝑟𝑡𝑔𝑡 . 𝑟𝑟𝑒𝑙
)≈0
2
𝑟𝑡𝑔𝑡
(11-31)
̈
𝑟𝑟𝑒𝑙
2𝑟𝑡𝑔𝑡 . 𝑟𝑟𝑒𝑙
𝜇
3
= − 3 { − 𝑟𝑡𝑔𝑡 (
) + 𝑟𝑟𝑒𝑙 } + 𝐹
2
2
𝑟𝑡𝑔𝑡
𝑟𝑡𝑔𝑡
Next, we define that both position vector 𝑟𝑡𝑔𝑡 and 𝑟𝑟𝑒𝑙 can be expressed in
RSW frame, that is,
𝑟𝑡𝑔𝑡 = 𝑟𝑡𝑔𝑡 𝑅̂
̂
𝑟𝑟𝑒𝑙 = 𝑥𝑅̂ + 𝑦𝑆̂ + 𝑧𝑊
(11-32)
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Then, the dot product between 𝑟𝑡𝑔𝑡 and 𝑟𝑟𝑒𝑙 only consists of 𝑥 component.
The equation (11-32) becomes,
̈ = − 𝜇 {−3𝑥𝑅̂ + 𝑟𝑟𝑒𝑙 } + 𝐹
𝑟𝑟𝑒𝑙
3
𝑟𝑡𝑔𝑡
(11-33)
Equation (11-33) represents the inertial relative acceleration between
interceptor and target spacecraft that expressed in target reference frame.
However, because the target reference frame is moving all the time, a
further analysis required to be done to express the equation in inertial
frame. Consider the equation,
̈ )𝑅 = (𝑟𝑟𝑒𝑙
̈ ) −𝜔
̇ ) −𝜔
(𝑟𝑟𝑒𝑙
⃗ ̇ 𝑅 × (𝑟𝑟𝑒𝑙 )𝑅 − 2𝜔
⃗ 𝑅 × (𝑟𝑟𝑒𝑙
⃗ 𝑅 × (𝜔
⃗𝑅
𝐼
𝑅
× (𝑟𝑟𝑒𝑙 )𝑅 )
(11-34)
where, subscript 𝐼 represents the expression in inertial frame and 𝑅
represents the expression in target frame. The 𝜔
⃗ 𝑅 is the angular velocity of
̂ = √ 𝜇3 𝑊
̂ . Because 𝜔
the target spacecraft which is 𝜔
⃗ 𝑅 = 𝜔𝑊
⃗ 𝑅 is constant,
𝑟
𝑡𝑔𝑡
then, 𝜔
⃗ ̇ 𝑅 = 0.
Substitute the angular velocity into equation (11-34), we get:
̈ )𝑅 = −𝜔2 {𝑥𝑅̂ + 𝑦𝑆̂ + 𝑧𝑊
̂ − 3𝑥𝑅̂ } + 𝐹 + 2𝜔𝑦̇ 𝑅̂ − 2𝜔𝑋̇𝑆̂
(𝑟𝑟𝑒𝑙
+ 𝜔2 𝑥𝑅̂ + 𝜔2 𝑦𝑆̂
(11-35)
̂ , we obtain:
Now, collecting the term for each 𝑅̂ , 𝑆̂ and 𝑊
𝑥̈ − 2𝜔𝑦̇ − 3𝜔2 𝑥 = 𝑓𝑥
𝑦̈ − 2𝜔𝑥̇ = 𝑓𝑦
(11-36)
𝑧̈ + 𝜔2 𝑧 = 𝑓𝑧
Eq.(11-36) also known as Hill’s equation or Clohessy-Whitshire (CW)
equation.
11.3.1
Closed-form solution for near circular orbits
It is possible to obtain a closed form solution for the Hill’s equation derived
in equation (11-36). Assume that both spacecraft are in near circular orbits;
with no external forces exert onto the spacecraft continuously, i.e. 𝐹 = 0.
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The assumption is reasonable that we may consider only an impulse thrust,
∆V exert on the spacecraft.
Considering that 𝐹 = 0, and taking one derivative order of x-component in
equation (11-36), we have,
𝑥̈ = 2𝜔𝑦̇ + 3𝜔2 𝑥
2
(11-37)
𝑥⃛ = 2𝜔𝑦̈ + 3𝜔 𝑥̇
Substitute 𝑦̈ = 2𝜔𝑥̇ into equation above, we obtain:
𝑥⃛ = −𝜔2 𝑥̇
(11-38)
Then, taking the Laplace transform of equation (11-38), we get:
𝑠 3 𝑋(𝑠) − 𝑠 2 𝑥0 − 𝑠𝑥̇ 0 − 𝑥̈ 0 + 𝑠𝑋(𝑠)𝜔2 − 𝑥0 𝜔2 = 0
𝑋(𝑠) =
𝑥0
𝑥̇ 0
𝑥̈ 0
+ 2
+
2
2
𝑠 (𝑠 + 𝜔 ) 𝑠(𝑠 + 𝜔 2 )
(11-39)
where 𝑥0 , 𝑥̇ 0 and 𝑥̈ 0 are the initial relative position, velocity and
acceleration in x-component respectively.
To solve the Laplace transform, let
𝑠(𝑠 2
𝑥̈ 0
𝐴
𝐵𝑠 + 𝐶
= + 2
2
+ 𝜔 ) 𝑠 (𝑠 + 𝜔 2 )
(11-40)
To find the coefficient 𝐴, multiply the numerator of 𝐴 on both side and set
𝑠 = 0. Thus,
𝐴=
𝑥̈ 0
𝜔2
(11-41)
Apply the similar method to obtain 𝐵 and 𝐶, that are
𝐵=−
𝑥̈ 0
and 𝐶 = 0
𝜔2
(11-42)
Substitute both Eqs. (11-41) and (11-42) into (11-40), the Eq. (11-39)
becomes,
𝑋(𝑠) =
𝑥0
𝑥̇ 0
𝑥̈ 0
𝑠𝑥̈ 0
+ 2
+ 2−
2
2
𝑠 (𝑠 + 𝜔 ) 𝑠𝜔
𝑠(𝑠 + 𝜔 2 )𝜔 2
(11-43)
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Then, the inverse Laplace transform is,
𝑥(𝑡) = 𝑥0 +
𝑥̈ 0 𝑥̇ 0
𝑥̈ 0
+
sin
(𝜔𝑡)
−
cos(𝜔𝑡)
𝜔2 𝜔
𝜔2
(11-44)
Consider the expression of 𝑥̈ in Eq. (11-37), and let 𝑥̈ 0 = 2𝜔𝑦̇ 0 + 3𝜔2 𝑥0 .
Then Eq. (11-44) becomes,
𝑥(𝑡) = 4𝑥0 +
2𝑦̇ 0 𝑥̇ 0
2𝑦̇ 0
+ sin(𝜔𝑡) − (3𝑥0 +
) cos(𝜔𝑡)
𝜔
𝜔
𝜔
(11-45)
Eq. (11-45) shows the closed form solution of Hill’s equation for the relative
position in x-component. The relative velocity is obtained through taking an
order of derivative. Consider that the initial condition of relative position
and velocity are constant, we have,
𝑥̇ (𝑡) = 𝑥̇ 0 cos(𝜔𝑡) + (3𝑥0 +
2𝑦̇ 0
) 𝜔sin(𝜔𝑡)
𝜔
(11-46)
Substitute Eq. (11-46) into the 𝑦̈ equation,
𝑦̈ (𝑡) = −2𝜔𝑥̇ 0 cos(𝜔𝑡) − 2𝜔(3𝜔𝑥0 + 2𝑦̇ 0 )sin(𝜔𝑡)
(11-47)
Integrate the Eq. (11-47) twice, we obtain the expression of 𝑦̇ and 𝑦,
𝑦̇ (𝑡) = −2𝑥̇ 0 sin(𝜔𝑡) + 2(3𝜔𝑥0 + 2𝑦̇ 0 )cos(𝜔𝑡)+C
2𝑥̇ 0
2
𝑦(𝑡) =
cos(𝜔𝑡) + (3𝜔𝑥0 + 2𝑦̇ 0 )sin(𝜔𝑡) + 𝐶𝑡 + 𝐷
𝜔
𝜔
(11-48)
The constant 𝐶 and 𝐷 can be determined by taking the time, t = 0, where
𝑦̇ (0) = 𝑦̇ 0 and 𝑦(0) = 𝑦0 . That is,
𝐶 = −6𝜔𝑥0 − 3𝑦̇ 0
2𝑥̇ 0
𝐷=−
+ 𝑦0
𝜔
(11-49)
Therefore,
𝑦(𝑡) =
2𝑥̇ 0
2
cos(𝜔𝑡) + (3𝜔𝑥0 + 2𝑦̇ 0 )sin(𝜔𝑡) − (6𝜔𝑥0 +)𝑡
𝜔
𝜔
2𝑥̇ 0
−
+ 𝑦0
𝜔
(11-50)
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𝑦̇ (𝑡) = −2𝑥̇ 0 sin(𝜔𝑡) + 2(3𝜔𝑥0 + 2𝑦̇ 0 )cos(𝜔𝑡) − (6𝜔𝑥0 +)
The solution of z-component in Eq. (11-36) is straight forward compared to
the x and y component, because it is uncoupled. Recall that the equation of
motion in z-component is,
𝑧̈ = −𝜔2 𝑧
(11-51)
Taking the Laplace transform of the equation, we obtain:
𝑠 2 𝑍(𝑠) − 𝑠𝑧0 − 𝑧̇0 = −𝜔2 𝑍(𝑠)
𝑍(𝑠)(𝑠 2 + 𝜔2 ) = 𝑠𝑧0 + 𝑧̇0
𝑍(𝑠) =
(𝑠 2
(11-52)
𝑠𝑧0
𝑧̇0
+ 2
2
+ 𝜔 ) (𝑠 + 𝜔 2 )
Then, the inverse Laplace transforms’ solution for the z-component and its
respective time derivative are,
𝑧(𝑡) =
𝑧̇0
sin(𝜔𝑡) + 𝑧0 cos(𝜔𝑡)
𝜔
(11-53)
𝑧̇ (𝑡) = 𝑧̇0 cos(𝜔𝑡) − 𝑧0 𝜔sin(𝜔𝑡)
Eqs. (11-45), (11-46), (11-50) and (11-53) show that the closed form solution
are expressed in a linear function with respect to the relative positions and
velocities. Thus, they can be expressed in matrices format, that are,
𝑟 = Φrr r0 + Φrv ṙ 0
𝑟̇ = Φvr r0 + Φvv ṙ 0
where,
(11-54)
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4 − 3cos(𝜔𝑡)
Φrr = [6(sin(𝜔𝑡) − 𝜔𝑡)
0
1
sin(𝜔𝑡)
𝜔
0
1
0
11 R
2
(1 − cos(𝜔𝑡))
𝜔
1
(4 sin(𝜔𝑡) − 3𝜔𝑡)
𝜔
0
0
[
Φvr
Φvv
3𝜔sin(𝜔𝑡)
= [6𝜔(cos(𝜔𝑡) − 1)
0
M
O T I O N
0
0 ]
cos(𝜔𝑡)
2
(cos(𝜔𝑡) − 1)
𝜔
Φrv =
E L A T I V E
0
0
1
sin(𝜔𝑡)]
𝜔
0
0
0
0
]
0 −𝜔sin(𝜔𝑡)
(11-55)
cos(𝜔𝑡)
2sin(𝜔𝑡)
0
0 ]
= [−2sin(𝜔𝑡) 4 cos(𝜔𝑡) − 3
0
0
cos(𝜔𝑡)
EXAMPLE 11-3
A GEO satellite strikes some orbiting debris and is found an hours
⃗ = [−𝟓 𝟓 𝟎]𝐓 𝐤𝐦
afterwards to have drifted to the position 𝛅𝒓
relative to its original location. At that time the only slightly damaged
satellite initiates a two-impulse maneuver to return to its original
location in 4 hours. Find the total Delta-v for this maneuver.
SOLUTION
For GEO satellite, the height and mean motion are,
h = 36000km
  7.292  10 5 rad / sec
The closed form solutions of relative position for GEO satellite are,
𝑟1 = Φrr r0 + Φrv ṙ 0
𝑟1̇ = Φvr r0 + Φvv ṙ 0
with 𝑟0 = [0 0 0]T km and 𝑟1 = [−5 5 0]T km
Given that the satellite is drifted to position 𝑟1 after an hrs, then,
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𝑡1 = 3600 secs.
The relative velocities at initial and t1 are,
ṙ 0 = [−0.001669 0.000999
𝑟1̇ = [−0.001093
0.001728
0]T km/sec
0]T km/sec
Time required to return to designate position is 4hrs. The, 𝑡2 = 14400
secs.
Using the CW closed form solution,
𝑟2 = Φrr r1 + Φrv 𝑟1̇ +
𝑟2̇ = Φvr r1 + Φvv 𝑟1̇ +
where 𝑟0 = [0 0
0]T km and,
The required initial and final relative velocity to return to designate
position are,
𝑟̇ + = [0.0008235 0.0001988 0]T km/sec
1
𝑟2̇ = [−0.0001941
−0.0005303
Thus, the total Delta-v is,
∆V = ‖𝑟1̇ − 𝑟0̇ ‖ + ‖𝑟̇ + − 𝑟2̇ ‖ = [0.001593
1
0]T km/sec
0.001458
0]T km/sec
16