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
Before we introduce the two-body problem, we provide the following
short review of necessary concepts and definitions from kinematics and
dynamics of particles. We strongly recommend the reader to read these
sections first before tackling the two-body problem.
2.1.1 Particle Kinematics
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
V
a
r(t)  x(t)i  y(t) j  z(t)k
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)
dr
 r  xi  y j  zk
dt
 vx i  v y j  vz k
v(t) 
and its acceleration will be
r
(2-3)
s
Y
X
Fig. 2-1.
Particle kinematics.
Then, the particle velocity is
P
r
C
(2-1)
o
C
2
2
H A P T E R
T
H E
T
W O
B
O D Y
P
R O B L E M
dv
 v  r
dt
 v x i  v y j  v z k
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. Since a velocity vector describes the
direction of motion (or the future position of the particle), it 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 the unit vectors ut and u n which are tangent and normal to
the particle trajectory at its local position respectively. Since the velocity is
always tangent to the trajectory, then we can write it as
v  vu t
r
s
Y
o
Fig. 2-2.
Particle trajectory and osculating
plane.
v  v  vv
(2-5)
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
d
r   r
dt

(2-7)
You can check this by taking, for example, r  3t i  2t j  5t k and
comparing 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 ut  anun
(2-8)
C
at  v  s, an 
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
v2
r
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
V
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)
dt
 r  (mv)  r  (ma)
(2-10)
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,
Fnet  ma
(2-11)
Therefore, equation Error! Reference source not found. 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
M net O .
Z
(2-9)
Then, for constant m, we can find the rate of change of angular
momentum
H O  d r  (mv)
Fnet
m
r
O
X
Fig. 2-3.
Particle kinetics.
P
Y
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
Then, we can write
r
Fig. 2-4.
Earth and a rotating satellite is a good
approximation of two-body system.
Fg
Z
M
m
O
(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.
Fg
M
H O  Mnet O
m
Fg
Y
X
Z’
Y’
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,
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
X’
Fig. 2-5.
Formulation of the two-body
problem.
Z
In order to mathematically formulate the problem, let us consider the system of two body masses M and m (as shown in Fig. 2-5). Assume X’Y’Z’ is
an inertial frame of reference (frame of reference which is neither
accelerating nor rotating as illustrated in Fig. 2-6). Let XYZ be a nonrotating frame of reference parallel to X’Y’Z’ with its origin O coincident
with the center of mass M.
Fg  G
r
Mm r
r2 r
(2-14)
Moving frame
O
Y
X
Inertial frame
Fig. 2-6.
Inertial frame and moving frame.
2.2.3 Equation of Motion
The position vectors of M and m with respect to X’Y’Z’ are rM and rm
respectively. Then, the position of m relative to M will be
r  rm  rM
(2-15)
C
H A P T E R
2
T
H E
T
W O
B
O D Y
Apply Newton’s second law of motion to m and M to get
mrm  G
Mm r
r2 r
(2-16)
MrM  G
Mm r
r2 r
Or
rm  G
rM  G
Mr
r2 r
(2-17)
mr
r2 r
(2-18)
If we subtract (2-17) from (2-18), we get
rm  rM  r  G
(M  m)
r
r3
(2-19)
Equation (2-18) is the vector differential equation of the relative motion.
Now, if we assume m << M, then G(M  m)  GM and equation (2-19)
will become
r  G
M
r
r3
(2-20)
If we compare equation (2-18) and equation (2-20), we notice that r and
r will measure the same magnitude and direction whether in XYZ or X’Y’Z’
(Recall that the frame XYZ is non-rotating and parallel to the inertial frame
X’Y’Z’.) also, when m << M (e.g. a satellite with respect to Earth), we can
write
G(M  m)  GM  
(2-21)
 is called the gravitational parameter which can be found from the
astronomical data of most known celestial objects. Hence, the equation
of motion will become
P
R O B L E M
5
6
C
H A P T E R
2
T
H E
T
W O
r 
B
O D Y
P
R O B L E M

r0
r3
(2-22)
We notice that the equation of motion (2-22) is a 2nd order differential
equation of position in time which needs to be integrated twice to obtain
the trajectory of a spacecraft for example. Then, two constant vectors are
needed for the solution of the equation of motion, which may be taken as the
initial position and velocity vectors.
2.2.4 Constants of Motion
If we take the dot product of the equation of motion (2-22) with r we get
r  r 

r  r  0
r3
(2-23)
The first term can be mathematically manipulated as
r  r 
1 d
1 d
d  v2 
(r  r ) 
( v  v)   
2 dt
2 dt
dt  2 
(2-24)
We know that
d
(r  r)  2r  r
dt
(2-25)
but
d
d
(r  r)  (r 2 )  2rr
dt
dt
(2-26)
Then, we get the important relation
r  r  rr
(2-27)
 
 , we get
 r
Now, if we take the time derivative of  
d   


    2 r  3 (rr)  3 r  r
dt  r  r
r
r
(2-28)
Now, if we substitute from (2-24) and (2-28) into (2-23) to get
d  v 2  

0
dt  2 r 
(2-29)
C
H A P T E R
2
T
H E
T
W O
B
O D Y
By integration, the quantity in brackets will be a constant. This quantity is
also the total mechanical energy per unit mass of spacecraft, . The total
mechanical energy of spacecraft is the summation of the kinetic energy,
1 mv 2 and
2
potential energy,  m

. We know from dynamics that the
r
potential energy of an object depends upon the selection of the datum at
which the potential energy vanishes. Here, we opt to have the datum for
gravitational potential at infinity. For other selections of datum, at the
center of M, for example, a constant should be added to  m

.
r
Therefore, for the two-body motion, we can write

v2 
  const
2 r
(2-30)
Equation (2-30) is known as the energy equation which is also referred to as
the vis viva equation (vis viva, in Latin, means the living-force). This is the first
constant of motion which complies with the conservation of energy principle.
According to which, a body moving in a conservative field, such as the
gravitational field, has constant mechanical energy.
Now, let us take the cross product of the equation of motion (2-22) with r
from left to get
r  r 

rr  0
r3
(2-31)
The second term on left hand side vanishes by vector identity, since r  r  0 .
So, we have
r  r  0
(2-32)
Now, if we take the time derivative of r  r , we get
d
(r  r)  r  r  r  r  r  r  0
(2-33)
dt
Note that r  r  0 . By integration, the quantity in brackets will be
a constant. This quantity is nothing but the angular momentum of
spacecraft per unit mass, h . So, we can write
P
R O B L E M
7
C
8
H A P T E R
2
T
H E
T
W O
B
O D Y
P
R O B L E M
h  r  r  r  v  const
h
g
O
f
m
Z
M
V
Y
X
Fig. 2-7.
Angular momentum.
(2-34)
This is the second constant of motion which complies with the
conservation of angular momentum principle. According to equation
(2-13), since the gravitational force acting on spacecraft always passes
through the center of the central body (O), it will have zero moment about
O. Hence, the angular momentum of the particle about point O will be
invariant. We also notice that since h  r  v  const , the plane of
motion which contains r and v (the osculating plane) will be fixed in
space.
For convenience, we will usually refer to angular momentum per unit
mass as angular momentum and mechanical energy per unit mass as
mechanical energy.
2.2.5 Trajectory Equation
To find the trajectory, we need to integrate the equation of motion. The
nonlinear equation of motion cannot be integrated directly. Instead, we
will use the following procedure to integrate the equation of motion
indirectly. First, let us take the cross product of equation of motion (2-22)
with h from right, we get
r  h 

rh  0
r3
(2-35)
We can show that
d
r  h   r h  r  h  r h
dt
The term r  h will vanish since h is constant. Also since
(2-36)
r  h  r  (r  v)  r(r  v)  v(r  r)
 r r r  r 2 v
(2-37)
If we substitute into equation , we get
d
v r
r  h      2 r  0
dt
r r 
(2-38)
C
H A P T E R
2
T
d  r
 r r 
v r 
      2 r     2 r 
dt  r 
r r 
r r 
H E
T
W O
B
O D Y
P
9
R O B L E M
(2-39)
Now, if we substitute into equation, we get
d
r

r

h


0
dt 
r 
(2-40)
By integration we get
r
r  h    B
(2-41)
r
where B is a constant vector. Take the dot product of equation with r , to
V
m
get
M

r  (r  h )  r  r  r  B
r
(2-42)
O
From vector identity, the left hand side can written as
r  (r  h )  h  (r  r )  h  h  h 2
(2-43)
Then,
h 2  r  rB cos 
(2-44)
 is the angle between B and r which is called true anomaly. The true
anomaly of spacecraft may be considered as an angular coordinate of the
spacecraft position measured from a fixed direction in the plane of
motion. Now, we can solve equation r , to get
h2 / 
r
1  (B / ) cos 
(2-45)
Defining
e
B

(2-46)
Then, we can write the trajectory equation as
r
h2 / 
1  e cos 
(2-47)
Fig. 2-8.
True anomaly.

B
10
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
2.2.6 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:
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.
The Shape of Orbit
For an ellipse,
a : semi-major axis
b : semi-minor axis
e : eccentricity
a 2  b 2  (ae)2
(2-48)
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.
1. 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.
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
The Position of Spacecraft on the Orbit
Finally, the spacecraft position in orbit is specified by the true anomaly
angle, 𝜈, measured from perigee direction
Note that ℎ, 𝜇 and 𝑒 are all constants.
𝑒 : eccentricity
𝜃 : true anomaly
The above function is solution for 𝑟 as a function of true anomaly, 𝜃.
2.2.7 Orbit Equations
h2 / 
r
1  e cos 
(2-49)
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  r
(2-50)
and
𝑉⊥ =
ℎ
ℎ
= 2
ℎ
1
𝑟
( 𝜇 ) 1 + 𝑒 cos 𝜈
∴ 𝑉⊥ =
𝑉𝑟 = 𝑟̇ =
𝜇
(1 + 𝑒 cos 𝜈)
ℎ
𝑑 ℎ2
1
ℎ2 𝑒 sin 𝜈 ℎ
[
]=
𝑑𝑡 𝜇 1 + 𝑒 cos 𝜈
𝜇 (1 + 𝑒ν)2 𝑟 2
∴ 𝑉𝑟 =
𝜇
𝑒 sin 𝜈
ℎ
( 2-99 )
(2-100)
(2-101)
(2-102)
From the orbital equation, 𝑟 is only minimum when 𝜈 = 0. This is called a
periapsis.
∴ 𝑟𝑝 =
ℎ2 1
𝜇 1+𝑒
(2-103)
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
At periapsis,
𝑉𝑟 = 0 ;
𝑉⊥ =
𝜇
(1 + 𝑒)
ℎ
Flight Path Angle
tan 𝛾 =
𝑉𝑟
𝑒 sin 𝜈
=
𝑉⊥ 1 + 𝑒 cos ν
(2-104)
Since cosine is an even function, therefore the trajectory described by
the orbit equation is symmetric about the apse line.
𝑝 ≡ semi latus rectum ≡ orbit parameter
𝑝=
ℎ2
𝜇
𝑣2 𝜇
− = 𝜀 ≡ 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
2 𝑟
(2-105)
(2-112)
Energy at Periapsis,
𝜀𝑝 =
𝑣𝑝2 𝜇
−
2 𝑟𝑝
(2-113)
But at perigee, 𝑉𝑟 = 0. Therefore 𝑉𝑝 = 𝑉⊥ = ℎ⁄𝑟𝑝
∴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)
2.2.8 Summary of the Two-Body Problem
1. The only possible path for an orbiting satellite in a two-body system is
a conic section (circle, ellipse, parabola, or hyperbola).
C
H A P T E R
2
T
H E
T
W O
B
O D Y
P
13
R O B L E M
2. The focus of the conic orbit is located at the center of the central
body.
3. The mechanical energy () of a satellite (sum of kinetic and potential
energies) does not change as the satellite moves along its conic orbit.
However, the kinetic and potential energies may exchange.
4. The orbital motion takes place in a plane which is fixed in inertial
space.
5. The specific angular momentum (h) of a satellite about the central
body remains constant. Hence, as r and v changes, the flight path
angle (g) must change as well to keep h = r  v constant.
3
V
Keplerian Orbits

We have in the previous section that the solution of the two-body system
results in a planar trajectory which has the shape of a conic section. Since
such orbits agrees with the Kepler’s laws of planetary motion, the twobody orbits are usually referred to as Keplerian orbits. A Keplerian orbit
can be circular, elliptic, parabolic, or hyperbolic.
Fig. 3-1. Circular orbit.
5.1.1 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
∴𝑟=
ℎ2 (𝑟𝑣)2
=
𝜇
𝜇
(3-117)
𝜇
𝑟
(3-118)
∴ 𝑉𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 = √
The time, T required for one orbit is known as the period. Since speed is
constant,
C
14
H A P T E R
2
T
H E
𝑇=
T
W O
B
O D Y
P
R O B L E M
𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒
2𝜋𝑟
=
𝑠𝑝𝑒𝑒𝑑
√𝜇⁄𝑟
∴ 𝑇𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 =
2𝜋 3 2𝜋
𝑟2 =
;
𝑛
√𝜇
𝑛=√
(3-119)
𝜇
𝑟3
(3-120)
Specific energy,
𝜀=−
V
1 𝜇2
2 ℎ2
(3-121)
2
g
Since 𝑟 = ℎ ⁄𝜇,

𝜀𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 = −
apoapsis
periapsis
𝜇
2𝑟
(3-122)
5.1.2 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.
Fig. 3-2. Elliptic orbit.
The minimum value of 𝑟 is
a
a
b
rB
F’
C
𝑟𝑝 =
ℎ2 1
𝜇 1+𝑒
(3-123)
𝑟𝑎 =
ℎ2 1
𝜇 1−𝑒
(3-124)
2𝑎 = 𝑟𝑎 + 𝑟𝑝
(3-125)
and its maximum value is
F
ae
ra
Fig. 3-3. Geometry of an ellipse.
rp
∴ 2𝑎 =
ℎ2
1
1
(
+
)
𝜇 1−𝑒 1+𝑒
(3-126)
ℎ2 1
ℎ2
∴𝑎=
→
= 𝑎(1 − 𝑒 2 )
𝜇 1 − 𝑒2
𝜇
(3-127)
𝑎(1 − 𝑒 2 )
∴𝑟=
1 + 𝑒 cos 𝜈
(3-128)
C
𝑟𝐵 =
H A P T E R
2
T
H E
𝑎(1 − 𝑒 2 )
1 + 𝑒 𝑐𝑜𝑠 𝛽
T
W O
B
O D Y
P
R O B L E M
(3-129)
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
(3-130)
Solving for 𝑒,
∴ 𝑒 = − cos 𝛽
∴ 𝑟𝐵 = 𝑎
∴ 𝑏 2 = 𝑟𝐵2 − (𝑎𝑒)2 = 𝑎2 (1 − 𝑒 2 )
∴ 𝑏 = 𝑎√1 − 𝑒 2
𝜀=−
(3-131)
1 𝜇2
(1 − 𝑒 2 )
2 ℎ2
But for an elliptic orbit, ℎ2 = 𝜇𝑎(1 − 𝑒 2 )
∴𝜀=−
∴
𝜇
2𝑎
𝑣2 𝜇
𝜇
− =−
2 𝑟
2𝑎
Kepler’s law,
𝑑𝐴 ℎ
=
𝑑𝑡 2
Missing figure
∴ ∆𝐴 =
Area of ellipse = 𝜋𝑎𝑏
ℎ
∆𝑡
2
(3-132)
(3-133)
15
16
C
H A P T E R
2
T
H E
T
∴ 𝜋𝑎𝑏 =
∴𝑇=
W O
B
ℎ
𝑇 →
2
O D Y
P
𝑇=
R O B L E M
2𝜋𝑎𝑏
ℎ
2𝜋 2
2𝜋 ℎ2 1
𝑎 √1 − 𝑒 2 =
(
) √1 − 𝑒 2
ℎ
ℎ 𝜇 1 − 𝑒2
(3-134)
(3-135)
3
2𝜋
ℎ
∴𝑇= 2(
)
𝜇 √1 − 𝑒 2
(3-136)
2𝜋
𝜇
; 𝑛=√ 3
𝑛
𝑎
(3-137)
Since ℎ2 = 𝜇𝑎(1 − 𝑒 2 ),
∴𝑇=
𝑟𝑝 1 − 𝑒
=
𝑟𝑎 1 + 𝑒
Solving for 𝑒
∴𝑒=
𝑟𝑎 − 𝑟𝑝
𝑟𝑎 + 𝑟𝑝
(3-138)
Since 𝑟𝑎 + 𝑟𝑝 = 2𝑎, therefore 𝑎 is the average of 𝑟𝑎 and 𝑟𝑝 . The average
value for 𝑟 is:
2𝜋
2𝜋
1
1
𝑑𝜈
𝑟̅𝜈 =
∫ 𝑟(𝜈)𝑑𝜈 =
𝑎(1 − 𝑒 2 ) ∫
2𝜋
2𝜋
1 + 𝑒 𝑐𝑜𝑠 𝜈
(3-139)
∴ 𝑟̅𝜃 = 𝑎√1 − 𝑒 2 = 𝑏 = √𝑟𝑝 𝑟𝑎
(3-140)
0
0
5.1.3 Parabolic Orbits
If 𝑒 = 1,
∴𝑟=
∴𝜀=−
As 𝜈 → 180°, 𝑟 → ∞.
ℎ2
1
𝜇 1 + 𝑐𝑜𝑠 𝜈
1 𝜇2
(1 − 𝑒 2 ) = 0
2 ℎ2
(3-141)
(3-142)
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
𝑣2 𝜇
− =0
2 𝑟
∴
2𝜇
∴𝑣=√
𝑟
(3-143)
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,
2𝜇
𝑟
the escape velocity is: 𝑣𝑒𝑠𝑐 = √
If spacecraft is in a circular orbit,
∴ 𝑣𝑒𝑠𝑐 = √2𝑣𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟
(3-144)
This indicates a required velocity boost of 41.4%
Missing figure
𝑡𝑎𝑛 𝛾 =
𝑠𝑖𝑛 𝜈
1 + 𝑐𝑜𝑠 𝜈
(3-145)
From trigonometric identities,
𝜈
𝜈
sin 𝜈 = 2 sin cos
2
2
𝜈
𝜈
𝜈
cos 𝜈 = cos 2 − sin2 = 2 cos 2 − 1
2
2
2
∴ tan 𝛾 = tan
∴𝛾=
𝜈
2
𝜈
2
(3-146)
5.1.4 Hyperbolic Orbits
If 𝑒 > 1, then the orbit equation
𝑟=
ℎ2
1
𝜇 1 + 𝑒 𝑐𝑜𝑠 𝜈
(3-147)
17
18
C
H A P T E R
2
T
H E
T
B
W O
O D Y
P
R O B L E M
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,
1
𝜈∞ = cos −1 (− ) ;
𝑒
90° < 𝜈∞ < 180°
Where 𝜈∞ corresponds to 𝑟 → ∞.
From trigonometry,
sin 𝜈∞ =
√𝑒 2 − 1
𝑒
For −𝜈∞ < 𝜈 < 𝜈∞ , spacecraft is in hyperbola I while for 𝜈∞ < 𝜈 <
(360° − 𝜈∞ ), vacant orbit in hyperbola II is traced.
For 𝜈 = 0,
ℎ2 1
𝜇 1+𝑒
(3-148)
ℎ2 1
; 𝑟 <0
𝜇 1−𝑒 𝑎
(3-149)
𝑟𝑝 =
For 𝜃 = 𝜋,
𝑟𝑎 =
2𝑎 = |𝑟𝑎 | − 𝑟𝑝 = −𝑟𝑎 − 𝑟𝑝
ℎ2 1
∴𝑎=
𝜇 𝑒2 − 1
(3-150)
𝑎(𝑒 2 − 1)
∴𝑟=
1 + 𝑒 𝑐𝑜𝑠 𝜈
(3-151)
𝑟𝑝 = 𝑎(𝑒 − 1); 𝑟𝑎 = −𝑎(𝑒 + 1)
𝜀=−
1 𝜇2
(1 − 𝑒 2 )
2 ℎ2
C
H A P T E R
∴𝜀=
2
T
H E
T
𝜇
2𝑎
W O
B
O D Y
P
R O B L E M
(3-152)
Hyperbolic excess speed, 𝑣∞ occurs when spacecraft is at ∞, given that:
𝑣2 𝜇
𝜇
2𝜇 𝜇
− =
→ 𝑣2 =
+
2 𝑟 2𝑎
𝑟
𝑎
𝜇
𝑎
∴ 𝑣∞ = √
(3-153)
2𝜇
Recall that 𝑣𝑒𝑠𝑐 = √ 𝑟 ,
∴ 𝑣 2 = 𝑣 2 𝑒𝑠𝑐 + 𝑣 2 ∞
(3-154)
(3-153)
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.
𝑣𝑒𝑠𝑐 = √
2𝜇
= 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⊥
19
20
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
𝑣⊥ = 𝑣 cos 𝛾 = 8.6 cos 50° = 5.528 𝑘𝑚/𝑠
∴ ℎ = 80710 𝑘𝑚2 /𝑠
c) Compute 𝜈
ℎ2
1
𝑟=
𝜇 1 + 𝑒 cos 𝜈
∴ 𝑒 cos 𝜈 = 0.1193
𝑣𝑟 =
𝜇
𝑒 sin 𝜈 → 𝑒 sin 𝜈 = 6.588 𝑘𝑚/𝑠
ℎ
∴
𝑒 cos 𝜈
= tan 𝜈 = 11.18
𝑒 sin 𝛾
∴ 𝜈 = 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
C
H A P T E R
2
T
H E
T
W O
B
O D Y
5.2 References
Valado (2007). Fundamentals of Astrodynamics with Applications.
Curtis (2005). Orbital Mechanics for Engineers.
Ulrich Walter (2008). Astronautics.
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
P
R O B L E M
21