ASEN 5050
SPACEFLIGHT DYNAMICS
Two-Body Motion
Prof. Jeffrey S. Parker
University of Colorado – Boulder
Lecture 3: The Two Body Problem
1
Announcements
• Homework #1 is due Friday 9/5 at 9:00 am
– Either handed in or uploaded to D2L
– Late policy is 10% per school day, where a “day” starts at
9:00 am.
• Homework #2 is due Friday 9/12 at 9:00 am
• Concept Quiz #3 will be available starting soon after
this lecture.
• Reading: Chapters 1 and 2
Lecture 3: The Two Body Problem
2
Space News
• Dawn is en route to Ceres, following a beautiful visit at
Vesta. It will arrive at Ceres in January.
• Left: Dawn’s 1-month descent from RC3 to Survey.
Center: Dawn’s expected 6-week descent from the survey
orbit to HAMO.
• Right: Dawn’s 8-week descent from HAMO to LAMO.
Lecture 3: The Two Body Problem
3
Concept Quiz #2
• Great job! 1/3 of the class missed one problem; 2/3
got ‘em all right.
Lecture 3: The Two Body Problem
4
Concept Quiz #2
Energy is a scalar
Lecture 3: The Two Body Problem
5
Concept Quiz #2
3 position, 3 velocity per body
6 per body X 3 bodies = 18
Lecture 3: The Two Body Problem
6
Concept Quiz #2
TOTAL angular momentum is
always conserved in a
conservative system.
Lecture 3: The Two Body Problem
7
Concept Quiz #2
Lecture 3: The Two Body Problem
8
Challenge #1
• The best submission for the “New Earth”
observations of the alternate Solar System earned
Leah 1 bonus extra credit point.
– What’s that worth? Something more than zero and less
than a full homework assignment 
– I’ll only mention names when given permission.
– I’ll try to have more of these challenges in the future!
Lecture 3: The Two Body Problem
9
Homework #2
• This homework will use information presented today and
Friday, hence its due date will be a week from Friday.
• HW2 has a lot of math, but this is the good stuff in
astrodynamics. An example problem:
– A satellite has been launched into a 798 x 816 km orbit (perigee
height x apogee height), which is very close to the planned orbit
of 795 x 814 km. What is the error in the semi-major axis,
eccentricity, and the orbital period?
• I suggest starting to build your own library of code to
perform astrodynamical computations. We will be doing
this a lot in this course.
Lecture 3: The Two Body Problem
10
Today’s Lecture Topics
• Kepler’s Laws
• Properties of conic orbits
• The Vis-Viva Equation! You will fall in love with
this equation.
• Next time: Converting between the anomalies
• Then: More two-body orbital element computations
Lecture 3: The Two Body Problem
11
Kepler’s 1st Law
“Conic Section” is the intersection of a plane with a cone. m is
at the primary focus of the ellipse.
p
a(1- e2 )
r=
=
1+ ecos n 1+ ecos n
Lecture 3: The Two Body Problem
12
Conic Sections
Lecture 3: The Two Body Problem
13
Challenge #2
• For those of you who are very familiar with the properties of
conic sections:
• Consider planar orbits (elliptical, parabolic, hyperbolic)
• What do you get if you plot vx(t) vs. vy(t)?
vy
vx
Lecture 3: The Two Body Problem
Send me an email with the subject:
Challenge #2
No computers (no cheating!)
What do you get if you plot this for:
• Circles
• Ellipses
• Parabolas
14
• Hyperbolas
Geometry of Conic Sections
Lecture 3: The Two Body Problem
(Vallado, 2013)
15
Geometry of Conic Sections
a = semimajor axis
b = semiminor axis
Elliptical Orbits 0 < e < 1
c
e= ,
a
b = a 1- e2
2
also, e =
a -b
a
2
Sometimes flattening is also used
a -b
f =
a
e2 = 2 f - f 2
Lecture 3: The Two Body Problem
16
Elliptic Orbits
p = semiparameter or semilatus rectum
2
b2
h
p=
= a(1- e2 ) =
a
m
Earth
Sun
Moon
ra apoapsis  apogee  aphelion  aposelenium  etc.
rp periapsis perigee  perihelion  periselenium  etc.
Lecture 3: The Two Body Problem
p
ra =
= a(1+ e)
1- e
p
rp =
= a(1- e)
1+ e
17
Geometry of Conic Sections
Elliptical Orbits 0 < e < 1
Check: what’s
What is
Hmmmm, so what is
Lecture 3: The Two Body Problem
18
Elliptic Orbits
• What is the velocity of a satellite at each point along
an elliptic orbit?
r =-
m
r
3
r
v2 m
e= 2 r
Þ
2m
v = 2e +
r
v m
e= 2 r
2
Lecture 3: The Two Body Problem
19
Geometry of Conic Sections
Parabolic Orbit
Lecture 3: The Two Body Problem
(Vallado, 2013)
20
Parabolic Orbit
e = 1,
e = 1+
e = 0,
2e h 2
m
2
r=
p
1+ cosn
=1
v2 m
e= - =0
2 r
Þ
v=
2m
r
Note: As n  180
r ∞
v 0
A parabolic orbit is a borderline case between an open hyperbolic
orbit and a closed elliptic orbit
Lecture 3: The Two Body Problem
21
Geometry of Conic Sections
Hyperbolic Orbit
Lecture 3: The Two Body Problem
(Vallado, 2013)
22
Hyperbolic Orbit
e > 1,
e>0
v2 m
e= - ,
2 r
v 2 = 2e +
2m
r
as r ® ¥, v 2 ® 2e = v 20 or v¥2 = 2e
2m
for any other point on the hyperbola
r0
Excess Velocity at r = ¥
m
m
2
and we will show that e = Þ v¥ = - = hyperbolic excess velocity
2a
a
Lecture 3: The Two Body Problem
23
Hyperbolic Orbit
• Interplanetary transfers use hyperbolic orbits everywhere
–
–
–
–
Launch
Gravity assists
Arrivals
Probes
24
Lecture 3: The Two Body Problem
(Vallado, 2013)
Properties of Conic Sections
Since
Lecture 3: The Two Body Problem
h2
m
= a(1- e2 ) is positive
25
Flight Path Angle
This is also a good time to define the flight path angle, ffpa, as the
angle from the local horizontal to the velocity vector.
+ from periapsis to apoapsis
- from apoapsis to periapsis
0 at periapsis and apoapsis
Always 0 for circular orbits
Lecture 3: The Two Body Problem
Only elliptic
orbits
(h=rava=rpvp)
26
Flight Path Angle
Another useful relationship is:
h = rv cos f fpa
Lecture 3: The Two Body Problem
27
Specific Energy
Recall the energy equation:
Note at periapse
p=
So, at periapse :
Thus
h=rpvp
h2
m
v2 m
e= 2 r
rp=a(1-e)
= a(1- e 2 ) Þ h = m a(1- e2 )
v 2p
h 2 m a(1- e 2 ) m (1+ e)
= 2 = 2
=
2
a(1- e)
rp a (1- e)
1
m (1- e)
1 é m (1+ e) ù
m
m
2
e= ê
=
=
ú
2 ë a(1- e) û a(1- e)
a(1- e)
2a
e =Lecture 3: The Two Body Problem
m
2a
28
Vis-Viva Equation
The energy equation:
v2 m
m
e = - =2 r
2a
Solving for v yields the Vis-Viva Equation!
2 1ö
æ
v = mç - ÷
èr aø
2
or
v=
Lecture 3: The Two Body Problem
Vis - Viva Equation
2m m
r a
29
Vis-Viva Equation
The energy equation:
v2 m
m
e = - =2 r
2a
Solving for v yields the Vis-Viva Equation!
2 1ö
æ
v = mç - ÷
èr aø
2
or
v=
2m m
r a
Vis - Viva Equation
m
r =- 3r
r
v2 m
e= 2 r
2m m
v=
r a
Lecture 3: The Two Body Problem
30
Additional derivables
vc =
m
r
2m
v=
= v escape
r
vp =
va =
m ( 1 + e)
a ( 1 - e)
m ( 1 - e)
a ( 1 + e)
And any number of other things. I’m sure I’ll find an interesting
way to stretch your imagination on a quiz / HW / test.
Lecture 3: The Two Body Problem
31
Proving Kepler’s 2nd and 3rd Laws
r
dA = ò rdvdr
= 1r
2
2
0
dv
dA 1 2 dv h
= r
=
dt 2 dt 2
(h = r 2q)
t2
2 ò dA = h ò dt
Proves 2nd Law
t1
area of ellipse = pab
P = t 2 - t1 = period
thus P =
2pab
h
b 2 = a 2( 1-e 2 ) = ap Þ b =
h=
3
P=
2pa p
m
1
2
2
1
2
a( 1 - e 2 )
Lecture 3: The Two Body Problem
=
ap
ma( 1 - e 2 )
2p
a
3
m
=P
Expression of
Kepler’s 3rd Law
éP2 a 3 ù
ê 12 = 1 3 ú
a2 û
ë P2
32
Proving Kepler’s 2nd and 3rd Laws
r
dA = ò rdvdr
= 1r
2
2
0
dv
dA 1 2 dv h
= r
=
dt 2 dt 2
(h = r 2q)
t2
2 ò dA = h ò dt
Proves 2nd Law
t1
area of ellipse = pab
P = t 2 - t1 = period
thus P =
2pab
h
b 2 = a 2( 1-e 2 ) = ap Þ b =
h=
3
P=
2pa p
m
1
2
2
1
2
a( 1 - e 2 )
Lecture 3: The Two Body Problem
=
ap
ma( 1 - e 2 )
2p
a
3
m
=P
Expression of
Kepler’s 3rd Law
éP2 a 3 ù
ê 12 = 1 3 ú
a2 û
ë P2
33
Proving Kepler’s 2nd and 3rd Laws
r
dA = ò rdvdr
= 1r
2
2
0
dv
dA 1 2 dv h
= r
=
dt 2 dt 2
(h = r 2q)
t2
2 ò dA = h ò dt
Proves 2nd Law
t1
area of ellipse = pab
P = t 2 - t1 = period
thus P =
2pab
h
b 2 = a 2( 1-e 2 ) = ap Þ b =
h=
3
P=
2pa p
m
1
2
2
1
2
a( 1 - e 2 )
Lecture 3: The Two Body Problem
=
ap
ma( 1 - e 2 )
2p
a
3
m
=P
Expression of
Kepler’s 3rd Law
éP2 a 3 ù
ê 12 = 1 3 ú
a2 û
ë P2
34
Proving Kepler’s 2nd and 3rd Laws
r
dA = ò rdvdr
= 1r
2
2
0
dv
dA 1 2 dv h
= r
=
dt 2 dt 2
(h = r 2q)
t2
2 ò dA = h ò dt
Proves 2nd Law
t1
area of ellipse = pab
P = t 2 - t1 = period
thus P =
2pab
h
b 2 = a 2( 1-e 2 ) = ap Þ b =
h=
3
P=
2pa 2 p
m
1
2
1
2
a( 1 - e 2 )
Lecture 3: The Two Body Problem
=
ap
ma( 1 - e 2 )
2p
a3
m
=P
Expression of
Kepler’s 3rd Law
éP2 a 3 ù
ê 12 = 1 3 ú
a2 û
ë P2
35
Proving Kepler’s 2nd and 3rd Laws
If we define the mean motion, n, to be :
2p
m
n=
then n =
P
a3
Mean angular
n 2 a 3 = m rate of change of
or,
Shuttle (300km)
Earth Obs (800 km)
GPS (20,000 km)
GEO (36,000 km)
90 min
101 min
~12 hrs
~24 hrs
the object in orbit
Lecture 3: The Two Body Problem
36
Final Statements
• Homework #1 is due Friday 9/5 at 9:00 am
– Either handed in or uploaded to D2L
– Late policy is 10% per school day, where a “day” starts at
9:00 am.
• Homework #2 is due Friday 9/12 at 9:00 am
• Concept Quiz #3 will be available starting soon after
this lecture.
• Reading: Chapters 1 and 2
Lecture 3: The Two Body Problem
37