Lecture 01 in the twobody problem

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16.346 Astrodynamics
Fall 2008

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Lecture 1

The Two Body Problem


Newton’s Two-Body Equations of Motion
Force = Mass × Acceleration
Gm1 m2 (r2 − r1 )
d r
= m1 21
2
r
r
dt
Gm2 m1 (r1 − r2 )
d2 r2
=
m
2
r2
r
dt2

2

=⇒

1687

d2

(m r + m2 r2 ) = 0

dt2
1 1
G(m1 + m2 ) (r2 − r1 )
d2

−
=
(r − r1 )

r2
r
dt2 2

Conservation of Total Linear Momentum
d2 rcm
=0
dt2

=⇒

rcm = c1 t + c2

Page 96

or

def

where

Two-Body Equation of Relative Motion
µ
d2 r
+ 3r = 0
2
dt
r

#3.1, #3.3

rcm =

m1 r1 + m2 r2
m1 + m2

Page 108

dv
µ
= − 3r

dt
r

r = r2 − r1
r = |r| = |r2 − r1 |
µ = G(m1 + m2 )

where

Vector Notation
• Position Vectors
r1 = x1 ix + y1 iy + z1 iz
r2 = x2 ix + y2 iy + z2 iz
r = r2 − r1 = x ix + y iy + z iz




x1
r1 =  y1 
z1




x2
r2 =  y2 
z2

 

x

r = r 2 − r1 = y 
z

• Two-Body Equations of Motion in Rectangular Coordinates
µ
d2 x
+ 3x = 0
2
dt
r
• Velocity Vectors

d2 y
µ
+ 3y = 0
2
dt
r

d2 z
µ
+ 3z = 0
2
dt
r




dx/dt
dr
dx
dy
dz
v=
=
ix +
iy +
i =  dy/dt 
dt
dt
dt
dt z
dz/dt

• Polar Coordinates
r = r ir
v=

ir = cos θ ix + sin θ iy

iθ = − sin θ ix + cos θ iy =

dr
dr
di dθ
dθ
dr
=

ir + r r
=
ir + r iθ = vr ir + vθ iθ
dt
dt
dθ dt
dt
dt

16.346 Astrodynamics

Lecture 1

dir
dθ


Kepler’s Second Law

1609

Equal Areas Swept Out in Equal Times

Assume z = 0 so that the motion is confined to the x-y plane
d2 x
d �
dy
dx
�

dy
dx

d2 y
0 = x
2 − y 2 =

x
− y

=

⇒

x
− y
= Constant

dt
dt
dt dt
dt

dt
dt

Using polar coordinates
x = r cos θ
y = r sin θ

=⇒

x

dx
d
dθ
dy
−y
= r2
= Constant ≡ h = 2 × Area
dt
dt
dt
dt

Josiah Willard Gibbs (1839–1908)

Vector Analysis for the Engineer
Appendix B–1

Image removed due to
copyright restrictions.

r1 · r2 = x1 x2 + y1 y2 + z1 z2 = r1 r2 cos
�
�
�i
�
� x iy iz �

r1 × r2 = �� x1 y1 z1 �� = r1 r2 sin in
�
x y z �

2
2
2
�
�
�x y z �
1
1�
� 1
r1 × r2 · r3 = �� x2 y2 z2 ��
�
x y z �

3

3

3

(r1 × r2 ) × r3 = (r1 · r3 )r2 − (r2 · r3 )r1
r1 × (r2 × r3 ) = (r1 · r3 )r2 − (r1 · r2 )r3

Kepler’s Second Law

1609

Conservation of Angular Momentum

d
dv
= (r × v) = 0 =⇒
h = r × v = Constant
dt
dt
Motion takes place in a plane and angular momentum is conserved
r×

In polar coordinates
dr
dθ
dr
=v=
ir + r
i = vr i r + vθ i θ
dt
dt
dt θ
so that the angular momentum of m2 with respect to m1 is
r = r ir

m2 r vθ = m2 r2
• Rectilinear Motion:

16.346 Astrodynamics

For r

dθ def
= m2 h = Constant
dt

v, then h = 0 .

Lecture 1



The quantity h is called the angular momentum but is actually the massless angular
momentum. In vector form h = h i z so that h = r×v and is a constant in both magnitude
and direction. This is called Kepler’s second law even though it was really his first major
result. As Kepler expressed it, the radius vector sweeps out equal areas in equal time since
1 dθ
h
dA
= Constant
= r2
=
dt
2 dt
2
Kepler’s Law is a direct consequence of radial acceleration!
Eccentricity Vector
d
dv
µ
µh

µh
dθ
di
(v × h) =
× h = − 3 r × h = − 2 i r × ih = 2 i θ = µ i θ = µ r
dt
dt
r
r
r
dt
dt
Hence

µe = v × h −

µ
r = Constant
r

The vector quantity µe is often referred to as the Laplace Vector.
We will call the vector e the eccentricity vector because its magnitude e is the eccentricity
of the orbit.
Kepler’s First Law

1609

The Equation of Orbit

If we take the scalar product of the Laplace vector and the position vector, we have

µe · r = v × h · r −
Also

µe · r = µre cos f
r=

µ
r · r = r × v · h − µr = h · h − µr = h2 − µr
r

where f is the angle between r and e

p
1 + e cos f

or

r = p − ex

where

so that

def

p =

h2
µ

is the Equation of Orbit in polar coordinates. (Note that r cos f = x .)
The angle f is the true anomaly and p , called the parameter, is the value of the radius
r for f = ± 90 ◦ .
The pericenter ( f = 0) and apocenter ( f = π ) radii are
rp =

p
1+e

and ra =

p
1−e

If 2a is the length of the major axis, then rp + ra = 2a

16.346 Astrodynamics

Lecture 1

=⇒

p = a(1 − e2 )


Kepler’s Third Law

1619

The Harmony of the World

Archimedes was the first to discover that the area of an ellipse is πab where a and b are
the semimajor and semiminor axes of the ellipse.
Since the radius vector sweeps out equal areas in equal times, then the entire area will be
swept out in the time interval called the period P . Therefore, from Kepler’s Second Law
πab
h
= =
P
2

�
√
µp
µa(1 − e2 )
=
2
2

√
Also, from the elementary properties of an ellipse, we have b = a 1 − e2 so that the
Period of the ellipse is
�
P = 2π

a3
µ

Other expressions and terminology are used

Mean Motion

2π
=
n=
P

�

µ
a3

or

µ = n2 a3

or


a3
= Constant
P2

The last of these is known as Kepler’s third law.
• Kepler made the false assumption that µ is the same for all planets.
Units for Numerical Calculations
A convenient choice of units is
Length
Time
Mass

The astronomical unit (Mean distance from Earth to the Sun)
The year (the Earth’s period)
The Sun’s mass (Ignore other masses compared to Sun’s mass)

Then
µ = G(m1 + m2 ) = G(msun + mplanet ) = G(msun ) = G
so that, from Kepler’s Third Law, we have
µ = G = 4π 2

or

k=

where G is the Universal Gravitation Constant.

16.346 Astrodynamics

Lecture 1

√
G = 2π


Josiah Willard Gibbs (1839–1908) was a professor of mathematical physics
at Yale College where he inaugurated the new subject — three dimensional
vector analysis. He had printed for private distribution to his students a small
pamphlet on the “Elements of Vector Analysis” in 1881 and 1884.
Gibbs’ pamphlet became widely known and was finally incorporated in the
book “Vector Analysis” by J. W. Gibbs and E. B. Wilson and published in

1901.
Gibb’s Method of Orbit Determination

Pages 131–133

• Given r1 , r2 , r3 with r1 × r2 · r3 = 0
• To determine the eccentricity vector e and the parameter p
r ×r ·n
r ×r ·n
r2 = αr1 + βr3 with n = r1 × r3 =⇒ α = 2 23
and β = 1 22
n
n
αr − r2 + βr3
p= 1
0 = e · (r2 − αr1 − βr3 ) = p − r2 − α(p − r1 ) − β(p − r3 ) =⇒
α−1+β
• To determine the eccentricity vector, we observe that:
n × e = (r1 × r3 ) × e = (e · r1 )r3 − (e ·
r3 )r1 = (p − r1 )r3 − (p − r3 )r1
Then, since

(n × e) × n = n2 e
e=

it follows that


1
[(p − r1 )r3 × n − (p − r3 )r1 × n]

n2