1
2
2 The Two Body Problem
and Keplerian Motion
I can calculate the motion of heavenly bodies, but not the madness of people!
Sir Isaac Newton (1642-1727)
2.1 Introduction
This section provides a brief review of necessary concepts and definitions
from kinematics and dynamics of particles. The reader is strongly
recommended to read these sections first before tackling the two-body
problem.
2.1.1 Particle Kinematics
V
The motion of any particle P can be tracked in a Euclidian space with the
help of a Cartesian coordinate system and a clock! In the frame of
reference XYZ, we can define the particle position r(t) as
r(t)  x(t)i  y(t) j  z(t)k
a
(2-1 )
Z
where i , j , and k are unit vectors in the X, Y, and Z directions.
r  r  r  r 1 / 2  x 2  y 2  z 2
O
( 2-2 )
v(t) 
and its acceleration is
r
s
Y
o
Fig. 2-1. Particle kinematics.
Then, the particle velocity is
dr
 r  xi  y j  zk
dt
 vx i  v y j  vz k
X
P
r
C
( 2-3 )
C
2
H A P T E R
2
T
H E
T
W O
B
O D Y
P
dv
 v  r
dt
 v x i  v y j  v z k
R O B L E M
a (t ) 
( 2-4 )
 axi  a y j  az k
Particle Trajectory
The trajectory or path of a particle is the locus of points the particle
occupies as it moves through space. The velocity vector is always tangent
to the trajectory.
The velocity vector of a particle is always tangent to its trajectory.
V
ut
a
un
r
C
Z
O
X
P
Let us define ut and u n as the tangent and normal unit vectors to the
particle trajectory at its local position, respectively. Since the velocity is
always tangent to the trajectory, then we can write it as
r
v  vu t
s
Y
o
Fig. 2-2. Particle trajectory
and osculating plane.
( 2-5 )
v  v  vv
The distance traveled by the particle along its trajectory, s is related to the
particle speed (magnitude of velocity) through
ds  v.dt
v  s
( 2-6 )
Note that
s  v  r , or
( 2-7 )
d
r 
dt
You can check this by taking, for example, r  3t i  2t j  5t k and comparing
r 
the two sides of the above inequality.
The acceleration of the particle can be expressed in the osculating plane
(the plane of motion) in terms of the unit vectors ut and u n as follows
a  at u t  an un
at  v  s, an 
v2
r
( 2-8 )
C
H A P T E R
2
T
H E
T
W O
B
O D Y
P
3
R O B L E M
where r is the radius of curvature of the trajectory at the particle position,
which is the distance from the particle position to the center of curvature C
as illustrated in Fig. 2-2.
2.1.2 Particle Dynamics
Angular Momentum
The angular momentum of a particle about a point is the moment of
momentum (or more specifically, linear momentum) of the particle about
that point. For the particle P shown in Fig. 2-3, which has mass (m), the
angular momentum H about O is given by
H O  r  (mv)
The first term on the right hand side will cancel by vector identity (Note
that r  v  v  v  0 ). If the net force acting on the particle is Fnet , then
from Newton’s second law (conservation of linear momentum), we can
write, for constant m,
( 2-11 )
Therefore, equation ( 2-10 ) can be written as
H 
O
 r  Fnet
( 2-12 )
Now, r  Fnet is exactly the moment of the net force Fnet about O, or Mnet O .
Then, we can write
H   M 
O
net O
Fnet
P
( 2-9 )
Then, for constant m, we can find the rate of change of angular momentum
H O  d r  (mv)
dt
( 2-10 )
 r  (mv)  r  (ma)
Fnet  ma
V
( 2-13 )
The above equation is analogues to Newton’s second law for linear motion,
and is called Newton’s second law for angular motion or the conservation
of angular momentum.
Z
O
m
r
Y
X
Fig. 2-3. Particle kinetics.
C
4
H A P T E R
2
T
H E
T
W O
B
O D Y
P
R O B L E M
2.1.3 Time Derivatives of Moving Vectors
Consider a rigid Cartesian coordinate system attached to an inertial frame
of reference that is fixed relative to the fixed stars. See figure
Z
r
Moving frame
O
Y
X
Inertial frame
A rigid body B is moving relative to the inertial from XYZ. The body B is
shown at two times separated by interval dt. At time t + dt, the orientation
of vector A is slightly different from that time at t. The magnitude however
is the same. Considering only the rotational motion of the body, the body
rotates about same axis and so vector A rotates about the same axis. See
figure. Body B rotates about the axis of rotation at an angle dθ.
dA= A sin  d n
Fig. 2-4. Inertial frame and
moving frame.
( 2-14 )
Where,
n : normal to axis of rotation unit vector
φ : angle between A and axis of rotation
Fig. 2-5. Rotational Motion.
Let ω be the angular velocity vector of body B; this vector points along the
axis of rotation and its direction is given by the right hand rule. By
definition:
‖𝝎‖ =
𝑑𝜃
𝑑𝑡
( 2-156 )
And the angular acceleration is given by,
𝜶=
𝑑𝝎
𝑑𝑡
( 2-167 )
Therefore,
𝑑𝑨 = (‖𝑨‖ sin 𝜑 ‖𝝎‖ 𝑑𝑡)𝒏 = (‖𝝎‖‖𝑨‖ sin 𝜑 )𝒏 𝑑𝑡
( 2-178 )
Recall that,
𝝎 × 𝑨 = ‖𝝎‖‖𝑨‖ sin 𝜑 𝒏
( 2-189 )
C
H A P T E R
2
T
H E
T
∴ 𝑑𝑨 = 𝝎 × 𝑨 𝑑𝑡
∴
𝑑𝑨
=𝝎𝑥𝑨
𝑑𝑡
W O
B
O D Y
P
R O B L E M
( 2-20 )
( 2-21 )
The derivative of a vector of constant magnitude
Example 2.1
The second derivative of a vector of constant magnitude:
𝑑 2 𝑨 𝑑 𝑑𝑨 𝑑
=
= (𝝎 𝑥 𝑨 )
𝑑𝑡 2 𝑑𝑡 𝑑𝑡 𝑑𝑡
=
∴
𝑑𝝎
𝑑𝑨
𝑥𝑨+
𝑥𝝎
𝑑𝑡
𝑑𝑡
𝑑2𝑨
= 𝜶 𝑥 𝑨 + 𝝎 𝑥 (𝝎 𝑥 𝑨 )
𝑑𝑡 2
Fig. 2-6. Reference Frame.
Vector Q has components in the moving frame:
𝑸 = 𝑄𝑥 𝒊 + 𝑄𝑦 𝒋 + 𝑄𝑧 𝒌
( 2-22 )
and components in the inertial frame:
𝑸 = 𝑄𝑋 𝑰 + 𝑄𝑌 𝑱 + 𝑄𝑍 𝑲
( 2-23 )
𝑑𝑸
= 𝑄̇𝑋 𝑰 + 𝑄̇𝑌 𝑱 + 𝑄̇𝑍 𝑲
𝑑𝑡
( 2-24 )
𝑑𝑸
𝑑𝒊
𝑑𝒋
𝑑𝒌
= 𝑄̇𝑥 𝒊 + 𝑄̇𝑦 𝒋 + 𝑄̇𝑧 𝒌 + 𝑄𝑥 + 𝑄𝑦 + 𝑄𝑧
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑡
( 2-25 )
Assume the angular velocity of the moving frame is Ω, then:
𝑑𝒊
𝑑𝒋
𝑑𝒌
= 𝜴𝑥𝒊;
= 𝜴𝑥𝒋;
= 𝜴𝑥𝒌
𝑑𝑡
𝑑𝑡
𝑑𝑡
∴
𝑑𝑸
= 𝑄̇𝑥 𝒊 + 𝑄̇𝑦 𝒋 + 𝑄̇𝑧 𝒌 + 𝑄𝑥 (𝜴 𝑥 𝒊) + 𝑄𝑦 (𝜴 𝑥 𝒋)
𝑑𝑡
+ 𝑄𝑧 (𝜴 𝑥 𝒌)
( 2-26 )
5
6
C
2
H A P T E R
T
H E
∴
T
W O
B
O D Y
P
R O B L E M
𝑑𝑸 𝑑𝑸
=
) +𝜴𝑥𝑸
𝑑𝑡
𝑑𝑡 𝑟𝑒𝑙
( 2-27 )
Where,
𝑑𝑄𝑦
𝑑𝑸
𝑑𝑄𝑥
𝑑𝑄𝑧
) =
𝒊+
𝒋+
𝒌
𝑑𝑡 𝑟𝑒𝑙
𝑑𝑡
𝑑𝑡
𝑑𝑡
( 2-28 )
is the time derivative of Q in the moving frame.
Higher Order Time Derivatives
𝑑2 𝑸 𝑑 𝑑𝑸
𝑑𝜴
𝑑𝑸
=
)
+
𝑥
𝑸
+
𝜴
𝑥
𝑑𝑡 2 𝑑𝑡 𝑑𝑡 𝑟𝑒𝑙 𝑑𝑡
𝑑𝑡
( 2-29 )
𝑑2 𝑸 𝑑 𝑑𝑸
𝑑𝜴
𝑑𝑸
=
)
+
𝑥
𝑸
+
𝜴
𝑥
[
) + 𝜴 𝑥 𝑸]
𝑑𝑡 2 𝑑𝑡 𝑑𝑡 𝑟𝑒𝑙 𝑑𝑡
𝑑𝑡 𝑟𝑒𝑙
( 2-30 )
𝑑 𝑑𝑸
𝑑2 𝑸
𝑑𝑸
) = 2) +𝜴𝑥
)
𝑑𝑡 𝑑𝑡 𝑟𝑒𝑙
𝑑𝑡 𝑟𝑒𝑙
𝑑𝑡 𝑟𝑒𝑙
( 2-31 )
𝑑2𝑸
) = 𝑄̈𝑥 𝒊 + 𝑄̈𝑦 𝒋 + 𝑄̈𝑧 𝒌
𝑑𝑡 2 𝑟𝑒𝑙
( 2-32 )
Also,
Where
𝑑2𝑸
𝑑2𝑸
𝑑𝑸
𝑑𝜴
∴ 2 = [ 2) +𝜴𝑥
) ]+
𝑥𝑸
𝑑𝑡
𝑑𝑡 𝑟𝑒𝑙
𝑑𝑡 𝑟𝑒𝑙
𝑑𝑡
𝑑𝑸
+ 𝜴 𝑥 [ ) + 𝜴 𝑥 𝑸]
𝑑𝑡 𝑟𝑒𝑙
∴
𝑑2𝑸 𝑑2𝑸
𝑑𝑸
= 2 ) + 𝜴̇ 𝑥 𝑸 + 𝜴 𝑥 (𝜴 𝑥 𝑸) + 2𝜴 𝑥
2
𝑑𝑡
𝑑𝑡 𝑟𝑒𝑙
𝑑𝑡
( 2-33 )
( 2-34 )
Relative Motion
Missing figure
𝒓 = 𝒓𝒐 + 𝒓𝒓𝒆𝒍
( 2-35 )
C
H A P T E R
2
T
H E
T
W O
𝒓𝒓𝒆𝒍 = 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌
𝒗 = 𝒗𝑜 +
𝑑
𝒓
𝑑𝑡 𝒓𝒆𝒍
B
O D Y
( 2-37 )
( 2-38 )
𝑑
𝒓 ) = 𝒗𝑟𝑒𝑙 = 𝑥̇ 𝒊 + 𝑦̇ 𝒋 + 𝑧̇ 𝒌
𝑑𝑡 𝒓𝒆𝒍 𝑟𝑒𝑙
( 2-39 )
∴ 𝒗 = 𝒗𝒐 + 𝒗𝑟𝑒𝑙 + 𝛀 x 𝒓𝒓𝒆𝒍
( 2-40 )
𝒂𝒐 =
𝑑2 𝒓𝒓𝒆𝒍
𝑑𝑡 2
𝑑
𝒗
𝑑𝑡 𝒐
R O B L E M
( 2-36 )
𝑑
𝑑
𝒓𝒓𝒆𝒍 = 𝒓𝒓𝒆𝒍 ) + 𝛀 x 𝒓𝒓𝒆𝒍
𝑑𝑡
𝑑𝑡
𝑟𝑒𝑙
𝒂 = 𝒂𝒐 +
P
( 2-41 )
( 2-42 )
Where 𝒂𝒐 is the absolute acceleration of O.
We showed before that:
𝑑 2 𝒓𝒓𝒆𝒍 𝑑2 𝒓𝒓𝒆𝒍
=
) + 𝛀̇ 𝑥 𝒓𝒓𝒆𝒍 + 𝛀 x (𝛀 x 𝒓𝒓𝒆𝒍 )
𝑑𝑡 2
𝑑𝑡 2 𝑟𝑒𝑙
𝑑𝒓𝒓𝒆𝒍
+ 𝟐𝛀 x
)
𝑑𝑡 𝑟𝑒𝑙
( 2-43 )
But
𝒗𝒓𝒆𝒍 =
∴
𝑑𝒓𝒓𝒆𝒍
)
;
𝑑𝑡 𝑟𝑒𝑙
𝒂𝒓𝒆𝒍
𝑑2 𝒓𝒓𝒆𝒍
)
𝑑𝑡 2 𝑟𝑒𝑙
𝑑2 𝒓𝒓𝒆𝒍
= 𝒂𝒓𝒆𝒍 + 𝛀̇ 𝑥 𝒓𝒓𝒆𝒍 + 𝛀 x (𝛀 x 𝒓𝒓𝒆𝒍 ) + 𝟐𝛀 x 𝒗𝒓𝒆𝒍
𝑑𝑡 2
( 2-44 )
Missing figure

𝑋𝑌𝑍 is a non-rotating inertial frame, with its origin at the center of the
Earth, C
7
8
C
H A P T E R









2
T
H E
T
W O
B
O D Y
P
R O B L E M
𝑥’𝑦’𝑧’ is rotating with the Earth, also centered at C. 𝑥’ intersects the
equator at the prime meridian.
𝜃𝑔̇ = Ω ≡ angular velocity of Earth
P is particle (i.e. airplane, spacecraft, etc.)
𝒓𝒓𝒆𝒍 ≡ position vector of P relative to C in the rotating 𝑥’𝑦’𝑧’ system.
P is always over point “O”, which lies on the Earth’s surface at longitude Λ
and latitude 𝜙.
Point “O” is the origin of the topocentric-horizon coordinate system 𝑥𝑦𝑧.
𝑥 – along local latitude eastward
𝑦 – along local meridian northward
Tangent plane to the Earth surface at “O” is the local horizontal
The 𝑧-axis is the local vertical
As P moves, “O” does as well.
Absolute velocity and acceleration of P:
𝒓𝒓𝒆𝒍 = (𝑅𝐸 + 𝑧)𝒌
𝒗𝒓𝒆𝒍 =
𝑑𝒓𝒓𝒆𝒍
𝑑𝒌
= 𝑧̇ 𝒌 + (𝑅𝐸 + 𝑧)
𝑑𝑡
𝑑𝑡
𝑑𝒌
=𝝎𝑥𝒌
𝑑𝑡
( 2-45 )
( 2-46 )
( 2-47 )
Where, 𝝎 is the angular velocity of the 𝑥𝑦𝑧 frame.
𝝎 = −𝜙̇𝒊 + Λ̇ cos 𝜙 𝐣 + Λ̇ sin 𝜙 𝐤
∴
𝑑𝒌
= Λ̇ cos 𝜙 𝐢 + 𝜙̇𝒋
𝑑𝑡
∴ 𝒗𝒓𝒆𝒍 = (𝑅𝐸 + 𝑧)Λ̇ cos 𝜙 𝐢𝛎 + (𝑅𝐸 + 𝑧)𝜙̇𝒋 + 𝑧̇ 𝒌 𝝂
( 2-48 )
( 2-49 )
( 2-50 )
Or
𝒗𝒓𝒆𝒍 =
𝑑𝒓𝒓𝒆𝒍 𝑑𝒓𝒓𝒆𝒍
=
) + 𝝎 𝑥 𝒓𝒓𝒆𝒍
𝑑𝑡
𝑑𝑡 𝑟𝑒𝑙
= 𝑧̇ 𝒌 + 𝝎(𝑅𝐸 + 𝑧)𝒌
( 2-51 )
C
2
H A P T E R
T
H E
T
W O
B
O D Y
P
R O B L E M
= 𝑧̇ 𝒌 + (𝑅𝐸 + 𝑧)𝜙̇𝒋 + (𝑅𝐸 + 𝑧)Λ̇ cos 𝜙 𝐢
𝒂𝒓𝒆𝒍 =
𝑑𝒗𝒓𝒆𝒍
) + 𝝎 𝑥 𝒗𝒓𝒆𝒍
𝑑𝑡 𝑟𝑒𝑙
∴ 𝒂𝒓𝒆𝒍 = [𝑧̇ Λ̇ cos 𝜙 + (𝑅𝐸 + 𝑧)Λ̈ cos 𝜙 − (𝑅𝐸 + 𝑧)𝜙̇Λ̇ sin 𝜙]𝒊
+ [𝑧̇ 𝜙̇ + (𝑅𝐸 + 𝑧)𝜙̈]𝒋 + 𝑧̈ 𝒌
+ (𝑅𝐸 + 𝑧)Λ̇ cos 𝜙 (𝝎 𝑥 𝒊) + (𝑅𝐸 + 𝑧)𝜙̇(𝝎 𝑥 𝒋)
+ 𝑧̇ (𝝎 𝑥 𝒌)
( 2-52 )
( 2-53 )
Note that
( 2-54 )
𝒗𝑟𝑒𝑙 = 𝑥̇ 𝒊 + 𝑦̇ 𝒋 + 𝑧̇ 𝒌
Comparing equations (50) and (54),
∴ 𝜙̇ =
𝑦̇
;
𝑅𝐸 + 𝑧
Λ̇ =
𝑥̇
(𝑅𝐸 + 𝑧) cos 𝜙
Taking the time derivatives,
(𝑅𝐸 + 𝑧)𝑦̈ − 𝑦̇ 𝑧̇
(𝑅𝐸 + 𝑧)2
( 2-55 )
(𝑅𝐸 + 𝑧)𝑥̈ cos 𝜙 − (ż cos 𝜙 − ẏ sin 𝜙)ẋ
(𝑅𝐸 + 𝑧)2 cos 2 𝜙
( 2-56 )
𝜙̈ =
Λ̈ =
Substitute into equation (53),
∴ 𝒂𝒓𝒆𝒍 = [𝑥̈ +
𝑥̇ (𝑧̇ − 𝑦̇ tan 𝜙)
𝑦̇ 𝑧̇ + 𝑥̇ 2 tan 𝜙
] 𝒊 + [𝑦̈ +
]𝒋
𝑅𝐸 + 𝑧
𝑅𝐸 + 𝑧
𝑥̇ 2 + 𝑦̇ 2
+ [𝑧̈ −
]𝒌
𝑅𝐸 + 𝑧
( 2-57 )
2.2 The Two-Body Problem
2.2.1 Problem Description
The two-body problem is the dynamic problem to find the trajectory of
motion of a system composed of two body masses M and m , for instance,
9
C
10
m
Fg
Fg
M
r
H A P T E R
2
T
H E
T
W O
B
O D Y
P
R O B L E M
in the absence of any effect other than the mutual gravitational force –
given some initial condition on the positions and velocities of these body
masses. From this description, we notice that an actual two-body system
does not exist in reality, but as we will see later, the trajectory of motion of
many body pairs in space can be approximated, to a sufficiently highdegree of accuracy, by a two-body motion.
2.2.2 Problem Formulation
Fig. 2-7. Earth and a rotating
satellite is a good approximation of two-body system.
Fg
Z
M
In order to mathematically formulate the problem, let us consider the system of two body masses M and m (as shown in Fig. 2-8). Assume X’Y’Z’ is an
inertial frame of reference (frame of reference which is neither accelerating
nor rotating). Let XYZ be a non-rotating frame of reference parallel to X’Y’Z’
with its origin O coincident with the center of mass M.
m
O
Y
X
Z’
Fg  G
Mm r
r2 r
(2-19)
2.2.3 Equations of Motion
Y’
X’
Fig. 2-8. Formulation of the
two-body problem.
Two point masses acted upon only by the mutual force of gravity between
them. The position vector of the center of mass G is:
𝑹𝐺 =
𝑚1 𝑹1 + 𝑚2 𝑹2
𝑚1 + 𝑚2
(2-20)
𝑭𝟏𝟐 is the force acting on 𝑚1 by 𝑚2 .
𝒗𝐺 = 𝑹̇𝐺 =
𝑚1 𝑹̇1 + 𝑚2 𝑹̇2
𝑚1 + 𝑚2
(2-21)
𝒂𝐺 = 𝑹̈𝐺 =
𝑚1 𝑹̈1 + 𝑚2 𝑹̈2
𝑚1 + 𝑚2
(2-22)
𝒓 = 𝑹2 − 𝑹1
̂𝑟 =
𝑼
𝒓
𝑟
C
𝑭21 =
H A P T E R
2
T
H E
T
W O
B
O D Y
𝐺𝑚1 𝑚2
𝐺𝑚1 𝑚2
̂ 𝑟) = −
̂ 𝑟 = 𝑚2 𝑹̈2
(−𝑼
𝑼
2
𝑟
𝑟2
( 2-61 )
𝐺𝑚1 𝑚2
̂ 𝑟 ) = 𝑚1 𝑹̈1
(𝑼
𝑟2
( 2-62 )
P
R O B L E M
Similarly,
𝑭12 =
From equations (61) and (62),
𝑚1 𝑹̈1 + 𝑚2 𝑹̈2 = 0
∴ 𝒂𝐺 = 0
Therefore G moves with constant velocity 𝑉𝐺 in straight line
∴ 𝑹𝐺 = 𝑹𝐺 𝑜 + 𝒗𝐺 𝑡
( 2-63 )
Where 𝑹𝐺 𝑜 is position of G at 𝑡 = 0.
Equations of Relative Motion
−
𝐺𝑚1 𝑚2
̂ 𝑟 = 𝑚2 𝑹̈2
𝑼
𝑟2
𝐺𝑚1 𝑚2
̂ 𝑟 = 𝑚1 𝑹̈1
𝑼
𝑟2
∴ 𝑚1 𝑚2 (𝑹̈2 − 𝑹̈1 ) = −
𝒓̈ = −
𝐺𝑚1 𝑚2
̂𝑟
(𝑚1 + 𝑚2 )𝑼
𝑟2
𝐺(𝑚1 + 𝑚2 )
̂𝑟
𝑼
𝑟2
( 2-64 )
( 2-65 )
( 2-66 )
( 2-67 )
𝜇 = 𝐺(𝑚1 + 𝑚2 ) ≡ Gravitational constant
∴ 𝒓̈ = −
𝜇
𝒓
𝑟3
( 2-68 )
This is the second order differential equation that governs the motion of
𝑚2 relative to 𝑚1 . In order to integrate equation (68) twice to compute r at
any time, we will have 2 vector constants of integration, each has 3 scalars.
11
12
C
H A P T E R
2
T
H E
T
W O
B
O D Y
P
R O B L E M
Orbit propagator:
Let
𝒓
𝒙=[ ]
𝒓̇
( 2-69 )
𝒙 (4: 6)
𝒓̇
∴ 𝒙̇ = [ ] = [ 𝜇
]
− 3 𝒙 (1: 3)
𝒓̈
𝑟
( 2-70 )
So, we can integrate 𝒙 given 𝒙 at initial time (i.e. given 𝒓, 𝒓̇ at initial time)
2.2.4 Orbital Elements
The position of a spacecraft can be specified using six parameters (recall 𝒓,
𝒓̇ at 𝑡 = 0). Another set of six parameters are commonly used in
specifying spacecraft positions in space, the orbital elements. In order to
specify a spacecraft position in space completely, we need to specify:
1. The orbit plane
In general a plane can be defined in space using 2 parameters.
These 2 parameters could be two components of a unit vector
normal to the plane, or two angles measured from a reference
frame. In the standard orbital elements, the orbit plane is defined
using two angles:
i)
The inclination, i, of the orbit plane to the equatorial plane
ii)
The right ascension of ascending node, Ω.
Missing figure
2. The shape of orbit
Missing figure
In an ellipse:
C
H A P T E R
2
T
H E
T
W O
B
O D Y
P
R O B L E M
a : semi-major axis
b : semi-minor axis
e : eccentricity
𝑎2 = 𝑏 2 + (𝑎𝑒)2
( 2-71 )
The shape of orbit is specified by any 2 of the above 3 parameters.
The equation above can be used to compute the 3rd parameter. In
standard orbital elements, 𝑎 and 𝑒 are used to describe the shape
of orbit.
3. The orientation of orbit in plane
Once we have specified the ellipse shape and the orbit plane, then
we need to describe how to place the ellipse in the plane. This is
done by specifying the angle 𝜔, argument of perigee, between the
line of nodes and the perigee point vector.
Missing figure
4. The position of spacecraft on the orbit
Finally, the spacecraft position in orbit is specified by the true
anomaly angle, 𝜃, measured from perigee direction
Missing figure
Angular momentum
Missing figure
Angular momentum of 𝑚2 with respect to 𝑚1 is:
𝑯21 = 𝒓 𝑥 𝑚2 𝒓̇
Where 𝒓̇ = 𝒗 is the velocity of 𝑚2 relative to 𝑚1 .
Let
( 2-72 )
13
14
C
H A P T E R
2
T
H E
T
W O
𝒉=
but
B
O D Y
P
R O B L E M
𝑯21
𝑚2
( 2-73 )
∴ 𝒉 = 𝒓 𝑥 𝒓̇
( 2-74 )
𝒉̇ = 𝒓̇ 𝑥 𝒓̇ + 𝒓 𝑥 𝒓̈
( 2-75 )
𝒓̇ 𝑥 𝒓̇ = 0
and
𝒓 𝑥 𝒓̈ = 𝒓 𝑥 −
𝜇
𝒓=0
𝑟3
( 2-76 )
∴ 𝒉̇ = 0 → 𝒉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Since 𝒉 = 𝒓 𝑥 𝒓̇ is constant, the plane containing 𝒓 and 𝒓̇ is constant. This
plane is the “orbit plane”. So 𝑚2 will be moving around 𝑚1 in a single plane,
the orbit plane.
Consider the orbit plane:
Missing figure
𝒉
̂ ⊥ + 𝑉𝒓 𝑼
̂𝒓
𝒓̇ = 𝒗 = 𝑽⊥ + 𝑽𝒓 = 𝑉⊥ 𝑼
( 2-77 )
̂
̂ 𝒓 𝑥 (𝑉⊥ 𝑼
̂ ⊥ + 𝑉𝒓 𝑼
̂ 𝒓 ) = 𝑟𝑉⊥ 𝒉
∴ 𝒉 = 𝑟𝑼
( 2-78 )
𝒉
̂=
Where 𝒉
=ℎ
‖𝒉‖
∴ ℎ = 𝑟𝑉⊥
( 2-79 )
2.2.5 Integration of Equations of Motion
𝒓̈ = −
𝜇
𝒓
𝑟3
We will now solve this equation for 𝒓.
( 2-80 )
C
H A P T E R
2
T
H E
T
W O
B
O D Y
Take the cross product of 𝒉 for both sides.
∴ 𝒓̈ 𝑥 𝒉 = (−
𝜇
𝒓) 𝑥 𝒉
𝑟3
( 2-81 )
Note that,
𝑑
(𝒓̇ 𝑥 𝒉) = 𝒓̈ 𝑥 𝒉 + 𝒓̇ 𝑥 𝒉̇ = 𝒓̈ 𝑥 𝒉
𝑑𝑡
∴
(𝒉̇ = 0)
𝑑
𝜇
(𝒓̇ 𝑥 𝒉) = (− 3 𝒓) 𝑥 𝒉
𝑑𝑡
𝑟
( 2-82 )
But,
1
1
𝒓 𝑥 𝒉 = 3 [𝒓 𝑥 (𝒓 𝑥 𝒓̇ )]
3
𝑟
𝑟
=
=
1
[𝒓(𝒓. 𝒓̇ ) − 𝒓̇ (𝒓. 𝒓)]
𝑟3
1
[𝒓(𝒓. 𝒓̇ ) − 𝑟 2 𝒓̇ ]
𝑟3
( 2-83 )
𝒓 . 𝒓 = 𝑟2
∴
𝑑
(𝒓. 𝒓) = 2𝑟𝒓̇
𝑑𝑡
∴ 𝒓. 𝒓̇ + 𝒓̇ . 𝒓 = 2𝑟𝒓̇
∴ 𝒓̇ . 𝒓 = 𝑟𝒓̇
( 2-84 )
Substituting the previous equation into equation (83),
∴
1
1
𝒓̇ 𝒓 − 𝑟𝒓̇
𝒓 𝑥 𝒉 = 3 [𝑟𝒓𝒓̇ − 𝑟 2 𝒓̇ ] =
3
𝑟
𝑟
𝑟2
Note also that,
𝑑 𝒓 𝑟𝒓̇ − 𝒓̇ 𝒓
=
𝑑𝑡 𝑟
𝑟2
( 2-85 )
P
R O B L E M
15
16
C
H A P T E R
2
T
H E
∴
T
W O
B
O D Y
P
R O B L E M
1
𝑑𝒓
𝒓𝑥𝒉=−
3
𝑟
𝑑𝑡 𝑟
( 2-86 )
Substituting in equation (82),
𝑑
𝑑
𝒓
(𝒓̇ 𝑥 𝒉) = (𝜇 )
𝑑𝑡
𝑑𝑡 𝑟
( 2-87 )
𝑑
𝒓
[(𝒓̇ 𝑥 𝒉) − (𝜇 )] = 0
𝑑𝑡
𝑟
( 2-88 )
∴
∴
∴ 𝒓̇ 𝑥 𝒉 − 𝜇
𝒓
=𝒄
𝑟
( 2-89 )
Note that 𝒄 has the units of 𝜇. Taking the dot product with 𝒉,
𝒓
(𝒓̇ 𝑥 𝒉). 𝒉 − 𝜇 . 𝒉 = 𝒄. 𝒉
𝑟
Since 𝒉 ⊥ 𝒓̇ , 𝒓,
∴ (𝒓̇ 𝑥 𝒉). 𝒉 = 0 𝑎𝑛𝑑 𝒓. 𝒉 = 0
∴ 𝒄. 𝒉 = 0
( 2-90 )
∴ 𝒄 ⊥ 𝒉 i.e. 𝒄 is in orbit plane
Defining
𝒆=
𝒄
𝜇
( 2-91 )
Therefore equation (89) can be written as,
𝒓
𝒓̇ 𝑥 𝒉
+𝒆=
𝑟
𝜇
( 2-92
( 2-91
) )
Taking the dot product with 𝒓,
∴
𝒓. 𝒓
𝒓. (𝒓̇ 𝑥 𝒉)
+ 𝒓. 𝒆 =
𝑟
𝜇
( 2-93 )
C
H A P T E R
2
T
H E
T
W O
B
O D Y
P
R O B L E M
Recall that 𝑨. (𝑩 𝑥 𝑪) = (𝑨 𝑥 𝑩). 𝑪
∴ 𝒓. (𝒓̇ 𝑥 𝒉) = (𝒓 𝑥 𝒓̇ ). 𝒉 = 𝒉. 𝒉 = ℎ2
𝒓. 𝒓
ℎ2
+ 𝒓. 𝒆 =
𝑟
𝜇
∴ 𝑟 + 𝒓. 𝒆 =
ℎ2
𝜇
( 2-94 )
𝒓. 𝒆 = 𝑟𝑒 cos 𝜃; where 𝜃 is the angle between 𝒓 and 𝒆.
Missing figure
ℎ2
𝜇
( 2-95 )
ℎ2
1
𝜇 1 + 𝑒 cos 𝜃
( 2-96 )
∴ 𝑟 + 𝑟𝑒 cos 𝜃 =
𝑟=
Note that ℎ, 𝜇 and 𝑒 are constants.
𝑒 : eccentricity
𝜃 : true anomaly
The above function is solution for 𝑟 as a function of true anomaly, 𝜃.
2.2.6 Orbit Equations
𝑟=
ℎ2
1
𝜇 1 + 𝑒 cos 𝜃
( 2-97 )
This is called the “orbit equation”. It defines the path of a spacecraft with
respect to the central body. The velocity that is perpendicular to 𝒓 is:
𝑉⊥ = 𝑟𝜃̇
and
( 2-98 )
17
18
C
H A P T E R
2
T
H E
𝑉⊥ =
T
W O
O D Y
P
R O B L E M
ℎ
ℎ
= 2
ℎ
1
𝑟
( 𝜇 ) 1 + 𝑒 cos 𝜃
∴ 𝑉⊥ =
𝑉𝑟 = 𝑟̇ =
B
𝜇
(1 + 𝑒 cos 𝜃)
ℎ
𝑑 ℎ2
1
ℎ2
𝑒 sin 𝜃
ℎ
[
]=
𝑑𝑡 𝜇 1 + 𝑒 cos 𝜃
𝜇 (1 + 𝑒 cos 𝜃)2 𝑟 2
∴ 𝑉𝑟 =
𝜇
𝑒 sin 𝜃
ℎ
( 2-99 )
(2-100)
(2-101)
(2-102)
From the orbital equation, 𝑟 is only minimum when 𝜃 = 0. This is called a
periapsis.
ℎ2 1
∴ 𝑟𝑝 =
𝜇 1+𝑒
(2-103)
At periapsis,
𝑉𝑟 = 0 ;
𝑉⊥ =
𝜇
(1 + 𝑒)
ℎ
Flight Path Angle
Missing figure
tan 𝛾 =
𝑉𝑟
𝑒 sin 𝜃
=
𝑉⊥ 1 + 𝑒 cos 𝜃
(2-104)
Since cosine is an even function, therefore the trajectory described by the orbit
equation is symmetric about the apse line.
Missing figure
𝑃 ≡ semi latus rectum ≡ orbit parameter
𝑃=
Energy Law
ℎ2
𝜇
(2-105)
C
H A P T E R
𝒓̈ . 𝒓̇ = −
2
T
H E
T
𝜇
𝒓. 𝒓̇
𝑟3
W O
B
O D Y
P
(2-106)
but
1𝑑
1𝑑
𝑑 𝑣2
(𝒓̇
)
(𝒗.
𝒓̈ . 𝒓̇ =
. 𝒓̇ =
𝒗) = ( )
2 𝑑𝑡
2 𝑑𝑡
𝑑𝑡 2
(2-107)
𝑑 1
1 𝑑𝑟
𝑟̇
( )=− 2
=− 2
𝑑𝑡 𝑟
𝑟 𝑑𝑡
𝑟
(2-108)
and
∴
𝜇
𝜇
𝜇
𝑑
𝑑 𝜇
𝒓. 𝒓̇ = 3 𝑟𝑟̇ = 2 𝑟̇ = −𝜇 𝑟 = − ( )
3
𝑟
𝑟
𝑟
𝑑𝑡
𝑑𝑡 𝑟
(2-109)
Substituting in equation (106)
𝑑 𝑣2
𝑑 𝜇
∴ ( )= ( )
𝑑𝑡 2
𝑑𝑡 𝑟
(2-110)
𝑑 𝑣2 𝜇
∴ ( − )=0
𝑑𝑡 2 𝑟
(2-111)
𝑣2 𝜇
− = 𝜀 ≡ 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
2 𝑟
(2-112)
Or
This is known as the vis-viva (or living force) equation. Where:
𝑣2
2
𝜇
𝑟
: Kinetic energy/unit mass
: Potential energy/unit mass
Energy at perigee,
𝜀𝑝 =
𝑣𝑝2 𝜇
−
2 𝑟𝑝
But at perigee, 𝑉𝑟 = 0. Therefore 𝑉𝑝 = 𝑉⊥ = ℎ⁄𝑟𝑝
(2-113)
R O B L E M
19
20
C
H A P T E R
2
T
H E
T
W O
B
P
O D Y
R O B L E M
∴𝜀=
1 ℎ2 𝜇
−
2 𝑟𝑝2 𝑟𝑝
(2-114)
𝑟𝑝 =
ℎ2 1
𝜇 1+𝑒
(2-115)
and
𝜀=−
1 𝜇2
(1 − 𝑒 2 )
2 ℎ2
(2-116)
Circular Orbits
If we let 𝑒 = 0 in the orbit equation,
ℎ2
∴𝑟=
≡ 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝜇
This proves that the orbit is circular. Since 𝑟̇ = 0, therefore 𝑉𝑟 = 0 and 𝑉 = 𝑉⊥.
ℎ = 𝑟𝑉⊥ = 𝑟𝑉
ℎ2 (𝑟𝑣)2
∴𝑟=
=
𝜇
𝜇
(2-117)
𝜇
𝑟
(2-118)
∴ 𝑉𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 = √
The time, T required for one orbit is known as the period. Since speed is
constant,
𝑇=
𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒
2𝜋𝑟
=
𝑠𝑝𝑒𝑒𝑑
√𝜇⁄𝑟
∴ 𝑇𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 =
2𝜋 3 2𝜋
𝑟2 =
;
𝑛
√𝜇
(2-119)
𝜇
𝑟3
𝑛=√
(2-120)
Specific energy,
1 𝜇2
𝜀=− 2
2ℎ
2
Since 𝑟 = ℎ ⁄𝜇 ,
(2-121)
C
H A P T E R
𝜀𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 = −
2
T
H E
𝜇
2𝑟
T
W O
B
O D Y
P
R O B L E M
(2-122)
Elliptical Orbits
If 0 < 𝑒 < 1, then the orbit equation is bounded. The initial orbit equation
describes an ellipse if it is bounded from 0 < 𝑒 < 1.
The minimum value of 𝑟 is
ℎ2 1
𝑟𝑝 =
𝜇 1+𝑒
(2-123)
ℎ2 1
𝜇 1−𝑒
(2-124)
2𝑎 = 𝑟𝑎 + 𝑟𝑝
(2-125)
and its maximum value is
𝑟𝑎 =
Missing figure
ℎ2
1
1
(
+
)
𝜇 1−𝑒 1+𝑒
(2-126)
ℎ2 1
ℎ2
→
= 𝑎(1 − 𝑒 2 )
𝜇 1 − 𝑒2
𝜇
(2-127)
∴𝑟=
𝑎(1 − 𝑒 2 )
1 + 𝑒 cos 𝜃
(2-128)
𝑟𝐵 =
𝑎(1 − 𝑒 2 )
1 + 𝑒 𝑐𝑜𝑠 𝛽
(2-129)
∴ 2𝑎 =
∴𝑎=
Missing figure
Projection of 𝑟𝐵 on apse line is 𝑎𝑒.
∴ 𝑎𝑒 = 𝑟𝐵 cos(180 − 𝛽) = −𝑟𝐵 cos 𝛽
= −
𝑎(1 − 𝑒 2 )
𝑐𝑜𝑠 𝛽
1 + 𝑒 𝑐𝑜𝑠 𝛽
(2-130)
21
22
C
H A P T E R
2
T
T
H E
W O
B
O D Y
P
R O B L E M
Solving for 𝑒,
∴ 𝑒 = − cos 𝛽
∴ 𝑟𝐵 = 𝑎
∴ 𝑏 2 = 𝑟𝐵2 − (𝑎𝑒)2 = 𝑎2 (1 − 𝑒 2 )
(2-131)
∴ 𝑏 = 𝑎√1 − 𝑒 2
1 𝜇2
(1 − 𝑒 2 )
2 ℎ2
𝜀=−
But for an elliptic orbit, ℎ2 = 𝜇𝑎(1 − 𝑒 2 )
∴𝜀=−
∴
𝜇
2𝑎
(2-132)
𝑣2 𝜇
𝜇
− =−
2 𝑟
2𝑎
(2-133)
Kepler’s law,
𝑑𝐴 ℎ
=
𝑑𝑡 2
Missing figure
∴ ∆𝐴 =
ℎ
∆𝑡
2
Area of ellipse = 𝜋𝑎𝑏
∴ 𝜋𝑎𝑏 =
∴𝑇=
ℎ
𝑇 →
2
𝑇=
2𝜋𝑎𝑏
ℎ
(2-134)
2𝜋 2
2𝜋 ℎ2 1
𝑎 √1 − 𝑒 2 =
(
) √1 − 𝑒 2
ℎ
ℎ 𝜇 1 − 𝑒2
(2-135)
3
2𝜋
ℎ
(
)
𝜇 2 √1 − 𝑒 2
(2-136)
∴𝑇=
Since ℎ2 = 𝜇𝑎(1 − 𝑒 2 ),
C
∴𝑇=
2
H A P T E R
T
H E
2𝜋
𝜇
; 𝑛=√ 3
𝑛
𝑎
T
W O
B
O D Y
P
R O B L E M
(2-137)
𝑟𝑝 1 − 𝑒
=
𝑟𝑎 1 + 𝑒
Solving for 𝑒
∴𝑒=
𝑟𝑎 − 𝑟𝑝
𝑟𝑎 + 𝑟𝑝
(2-138)
Since 𝑟𝑎 + 𝑟𝑝 = 2𝑎, therefore 𝑎 is the average of 𝑟𝑎 and 𝑟𝑝 . The average value
for 𝑟 is:
2𝜋
2𝜋
1
1
𝑑𝜃
𝑟̅𝜃 =
∫ 𝑟(𝜃)𝑑𝜃 =
𝑎(1 − 𝑒 2 ) ∫
2𝜋
2𝜋
1 + 𝑒 𝑐𝑜𝑠 𝜃
(2-139)
∴ 𝑟̅𝜃 = 𝑎 √1 − 𝑒 2 = 𝑏 = √𝑟𝑝 𝑟𝑎
(2-140)
0
0
Parabolic Orbits
If 𝑒 = 1,
∴𝑟=
∴𝜀=−
ℎ2
1
𝜇 1 + 𝑐𝑜𝑠 𝜃
1 𝜇2
(1 − 𝑒 2 ) = 0
2 ℎ2
(2-141)
(2-142)
As 𝜃 → 180°, 𝑟 → ∞.
∴
𝑣2 𝜇
− =0
2 𝑟
∴𝑣=√
2𝜇
𝑟
(2-143)
If a spacecraft is on a parabolic orbit, it will reach to ∞ with zero velocity.
Parabolic orbits are called escape trajectories. At a distance 𝑟 from Earth, the
escape velocity is:
23
24
C
H A P T E R
2
T
H E
T
W O
B
O D Y
P
𝑣𝑒𝑠𝑐 = √
R O B L E M
2𝜇
𝑟
If spacecraft is in a circular orbit,
(2-144)
∴ 𝑣𝑒𝑠𝑐 = √2𝑣𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟
This indicates a required velocity boost of 41.4%
Missing figure
𝑡𝑎𝑛 𝛾 =
𝑠𝑖𝑛 𝜃
1 + 𝑐𝑜𝑠 𝜃
(2-145)
From trigonometric identities,
𝜃
𝜃
sin 𝜃 = 2 sin cos
2
2
cos 𝜃 = cos2
𝜃
𝜃
𝜃
− sin2 = 2 cos2 − 1
2
2
2
∴ tan 𝛾 = tan
∴𝛾=
𝜃
2
𝜃
2
(2-146)
Hyperbolic Orbits
If 𝑒 > 1, then the orbit equation
𝑟=
ℎ2
1
𝜇 1 + 𝑒 𝑐𝑜𝑠 𝜃
(2-147)
describes a hyperbola. Two symmetric curves, one is occupied by the
spacecraft and the other one is empty.
Missing figure
The true anomaly of asymptotes,
C
H A P T E R
1
𝜃∞ = cos −1 (− ) ;
𝑒
2
T
H E
T
W O
B
O D Y
P
R O B L E M
90° < 𝜃∞ < 180°
Where 𝜃∞ corresponds to 𝑟 → ∞.
From trigonometry,
sin 𝜃∞ =
√𝑒 2 − 1
𝑒
For −𝜃∞ < 𝜃 < 𝜃∞ , spacecraft is in hyperbola I while for 𝜃∞ < 𝜃 < (360° −
𝜃∞ ), vacant orbit in hyperbola II is traced.
For 𝜃 = 0,
ℎ2 1
𝜇 1+𝑒
(2-148)
ℎ2 1
; 𝑟 <0
𝜇 1−𝑒 𝑎
(2-149)
𝑟𝑝 =
For 𝜃 = 𝜋,
𝑟𝑎 =
2𝑎 = |𝑟𝑎 | − 𝑟𝑝 = −𝑟𝑎 − 𝑟𝑝
ℎ2 1
𝜇 𝑒2 − 1
(2-150)
𝑎(𝑒 2 − 1)
1 + 𝑒 𝑐𝑜𝑠 𝜃
(2-151)
∴𝑎=
∴𝑟=
𝑟𝑝 = 𝑎(𝑒 − 1); 𝑟𝑎 = −𝑎(𝑒 + 1)
1 𝜇2
𝜀 = − 2 (1 − 𝑒 2 )
2ℎ
∴𝜀=
𝜇
2𝑎
(2-152)
Hyperbolic excess speed, 𝑣∞ occurs when spacecraft is at ∞, given that:
𝑣2 𝜇
𝜇
2𝜇 𝜇
− =
→ 𝑣2 =
+
2 𝑟 2𝑎
𝑟
𝑎
25
26
C
H A P T E R
2
T
H E
T
W O
B
O D Y
P
R O B L E M
𝜇
𝑎
∴ 𝑣∞ = √
(2-153)
2𝜇
Recall that 𝑣𝑒𝑠𝑐 = √ 𝑟 ,
∴ 𝑣 2 = 𝑣 2 𝑒𝑠𝑐 + 𝑣 2 ∞
(2-154)
(2-153)
Characteristic energy ≡ 𝐶3 = 𝑣 2 ∞
𝑣∞ represents excess kinetic energy over that which is required to simply
escape from the center of attraction.
Example 2.2
At a given point of a spacecraft’s geocentric trajectory, 𝑟 = 14600 𝑘𝑚, 𝑣 =
8.6 𝑘𝑚/𝑠 and 𝛾 = 50°, show that the path is hyperbolic.
2𝜇
𝑣𝑒𝑠𝑐 = √ = 7.389 𝑘𝑚/𝑠
𝑟
Since 𝑣 > 𝑣𝑒𝑠𝑐 , therefore it is hyperbolic.
a) Compute 𝐶3
𝐶3 = 𝑣 2 ∞ = 𝑣 2 − 𝑣 2 𝑒𝑠𝑐 = 19.36 𝑘𝑚2 /𝑠 2
b) Compute ℎ
ℎ = 𝑟𝑣⊥
𝑣⊥ = 𝑣 cos 𝛾 = 8.6 cos 50° = 5.528 𝑘𝑚/𝑠
∴ ℎ = 80710 𝑘𝑚2 /𝑠
c) Compute 𝜃
C
𝑟=
H A P T E R
2
T
H E
T
W O
B
O D Y
P
R O B L E M
ℎ2
1
𝜇 1 + 𝑒 cos 𝜃
∴ 𝑒 cos 𝜃 = 0.1193
𝑣𝑟 =
𝜇
𝑒 sin 𝜃 → 𝑒 sin 𝜃 = 6.588 𝑘𝑚/𝑠
ℎ
∴
𝑒 cos 𝜃
= tan 𝜃 = 11.18
𝑒 sin 𝛾
∴ 𝜃 = 84.89°
d) Compute 𝑒
𝑒 = 1.339
e) Compute 𝑟𝑝
𝑟𝑝 =
ℎ2 1
= 6986 𝑘𝑚
𝜇 1+𝑒
f) Compute 𝑎
ℎ2 1
𝑎=
= 20590 𝑘𝑚
𝜇 𝑒2 − 1
Perifocal Frame
Missing figure
̂ is shown in figure. 𝑥̅ 𝑦̅ plane is the orbit plane
The perifocal frame frame 𝑃̂𝑄̂ 𝑊
frame centered at focus of orbit.
𝑥̅ -axis is directed from focus to periapsis
𝑦̅-axis is at 𝜃 = 90°
𝑧̅-axis is normal to orbit plane in direction of 𝒉
𝑤
̂=
𝒓 = 𝑥̅ 𝒑 + 𝑦̅𝒒
𝒉
ℎ
27
28
C
H A P T E R
2
T
H E
T
W O
B
O D Y
P
R O B L E M
= 𝑟 cos 𝜃 𝒑 + 𝑟 sin 𝜃 𝒒
ℎ2
1
∴𝒓=
(𝑐𝑜𝑠 𝜃 𝒑 + 𝑠𝑖𝑛 𝜃 𝒒)
𝜇 1 + 𝑒 𝑐𝑜𝑠 𝜃
(2-155)
̅̇𝒑 + 𝒚
̅̇𝒒
𝒗 = 𝒓̇ = 𝒙
(2-156)
𝜇
𝑒 𝑠𝑖𝑛 𝜃
ℎ
(2-156)
𝑥̅̇ = 𝑟̇ cos 𝜃 − 𝑟𝜃̇ sin 𝜃
𝑦̅̇ = 𝑟̇ sin 𝜃 + 𝑟𝜃̇ cos 𝜃
But
𝑟̇ = 𝑣𝑟 =
and
𝑟𝜃̇ = 𝑣⊥ =
𝜇
(1 + 𝑒 𝑐𝑜𝑠 𝜃)
ℎ
(2-157)
𝜇
𝜇
∴ 𝑥̅̇ = − 𝑠𝑖𝑛 𝜃 ; 𝑦̅̇ = (𝑒 + 𝑐𝑜𝑠 𝜃)
ℎ
ℎ
∴𝒗=
𝜇
[− 𝑠𝑖𝑛 𝜃 𝒑 + (𝑒 + 𝑐𝑜𝑠 𝜃)𝒒]
ℎ
2.3 References
Kit, S. T. (2007). Astrodynamics Premier.
(2-158)
C
H A P T E R
2
T
H E
T
W O
B
O D Y
P
R O B L E M
Unsöld, A., Baschek, B., & Brewer, W. (2001). The New Cosmos: An Introduction to
Astronomy and Astrophysics. Berlin: Springer.
Satellite Tool Kit Tutorial. STK version 8.0
29