Present:
Sea Launch/Zenit
Thrust:
8,180,000 N
Fueled Weight:
450,000 kg
Payload to LEO:
13,740 kg
Cost per launch:
$100,000,000
Cost per kg:
$7,300
Launches:
31/28
Gateway To Space
ASEN 1400 / ASTR 2500
Class #20
Colorado Space Grant Consortium
T-30
Today:
- Announcements
- One minute Report Questions
- Mid Semester Team Evaluations
- Orbits and Mission Design – Part II
- Launch is in 30 days
Announcements…
pCDR peer reviews…
- 3rd place is Team #5
- 2nd place is Team #7
- 1st place is Team #4
DD Rev A/B Grades
HW #8 Due 4:00 PM November 9th
Office Hours and Questions in Class
4
Mid Semester Team Evaluations…
Please pass them forward now
New grades posted next Tuesday
Community Service project will be included
5
Next Tuesday…
Guest Lecture on ADCS
Colorado Space Grant Consortium
One Minute Reports:
Geostationary
VS.
Geosynchronous
One Minute Reports:
- What types of orbits do they do around other
planets?
- Is there a polar orbit that is also geosynchronous?
- Could spacecraft ever be launched from Colorado?
- How do you get on elliptical orbit?
- What is the advantage of elliptical orbit vs. a circular
orbit around the Earth?
- Does the angle at which you launch a satellite affect
its eccentricity?
- How many different orbits are there?
- Do you have launch a satellite at an angle to get it
into orbit or can you shoot it straight up?
Types of Rockets:
One Minute Reports:
- Who owns the geosync orbit space?
UN through the International Telecommunications
Union
- When is our Movie Night?
- What is the amount of time between turning on the
Sat at launch?
- Do you have to write a journal about every chapter?
- How will the in-class simulation work?
- Do we need to have all the satellite building and
testing done before the in-class simulation?
- What chances do students have to go to those big
conferences?
One Minute Reports:
- Where does Tom Kelly work now?
One Minute Reports:
-- What is an acoustic test?
- Are vibration tests done with mass models or the
actual products?
- Did they use Velcro on floor to keep them in place?
- Has an emergency ever occurred on an EVA?
- Is Grumman still making space vehicles?
- Arduino is beginning to look like a next of wires?
- Why is water blue?
- Why is this class so awesome?
- What is the craziest thing I ever did…
Orbits and Mission
Design – Part 2
ASEN 1400 / ASTR 2500
Class #19
Colorado Space Grant Consortium
Orbits:
A Brief Historical Look
Earth, the Moon, Mars, and the
Stars Beyond
A Brief Discussion on Mission Design
Universal Gravitation, Applied:
•What is an orbit?
Newton’s Laws:
Newton Continued...
• 1687, Principia Published
• Law of Universal Gravitation (Attraction)
m2V
F ma 
r
2
m1 m 2 G
F
2
r
Orbit History:
Kepler’s 3 Laws of Planetary Motion:
1. All planets move in elliptical orbits, sun at
one focus
Orbit History:
Kepler’s 3 Laws of Planetary Motion:
2. A line joining any planet to the sun, sweeps
out equal areas in equal times
Orbit History:
Kepler’s 3 Laws of Planetary Motion:
3. The square of the period of any planet about the sun is
proportional to the cube of the of the planet’s mean distance
from the sun.
Planet P (yr) a (AU) T2 R3
2
3
T =R
Mercury 0.24
0.39 0.06 0.06
2
3
T aR
a3
Venus
Earth
Mars
Jupiter
0.62
1.00
1.88
11.9
0.72
1.00
1.52
5.20
0.39
1.00
3.53
142
m
Saturn
29.5
9.54
870 868
2
4
p
T2=
a3
G(M1 + M 2 )
T =2p
0.37
1.00
3.51
141
If you can observe the period of rotation, you can determine
the distance
Types of Orbits:
Orbits are conic sections:
• Circle
• Ellipse
• Parabola
• Hyperbola
From Kepler’s Law, the
central body is at a focus of
the conic section
2MG MG
V

r
a
Kepler:
Kepler’s Laws...Orbits described by conic sections
Velocity of an orbit described by following equation
2    
v
  
r
For a circle (a=r):
v
For a ellipse (a>0):
GM
a
 
r
2    
v
  
r
For a parabola (a=):
v
2  
r
a
Earth, the Moon, Mars, and the
Stars Beyond
A Brief Discussion on Mission Design
Orbit Introduction:
What is an orbit?
- The path of a satellite around the Earth
(or any central body)
What shape is it?
- Orbits are conic sections
- Circles, Ellipses, Parabolas, Hyperbolas
How are orbits described?
- Position and Velocity at any one time
- Keplerian Elements (from Kepler’s Laws)
Orbit Definition:
Velocity & Position
- Given position and velocity of a satellite at
time t, you can calculate the position and
velocity at any other time
Orbit Definition:
Keplerian Elements
- Semi major axis (a)
- Size
- Eccentricity (e)
- Shape
Orbit Definition:
Keplerian Elements
- Inclination (i)
- Angle to the Equator
Orbit Definition:
Orbit Definition:
Keplerian Elements
- Right Ascension of Ascending Node (RAAN, Ω)
- Rotation about the Earth’s Spin Axis
Orbit Definition:
Keplerian Elements
- Argument of Perigee (ω)
- Rotation of the conic section in the plane
Orbit Definition:
Keplerian Elements
- True Anomaly (θ)
- Defines the position of a body in orbit
- Angle between the Position Vector and
the vector to Perigee
- Elliptical only
Types of Orbits (cont.)
•
Geosynchronous/Geostationary (equator)
Types of Orbits (cont.)
•
Critical Inclination
Types of Orbits (cont.)
•
Repeating Ground Trace
Types of Orbits (cont.)
•
Polar/ Sun Synchronous
Types of Orbits (cont.)
•
Molniya
Circular Orbit:
For a 250 km circular
Earth Orbit
Orbital Velocity
v
v
 
r
398600.4
(250  6378.14)
km
v7.75
 17,347 mph
sec
Circular Orbit:
Orbital Period
P =a
circumference
P=
velocity
2
3
P = 2p
r
3
m
(250 + 6378.14)
P = 2p
398600.4
P = 5, 370 sec = 89.5 min
3
Circular Orbit:
For a 500 km circular
Earth Orbit
Orbital Velocity
v
v
 
r
398600.4
(500  6378.14)
km
v7.61
 17,028 mph
sec
Circular Orbit:
For a 500 km circular
Earth Orbit
Orbital Period
P  2
r3

(500  6378.14)3
P  2
398600.4
P  5,676 sec  94.6 min
Conclusions???
Changing Orbits:
How about 250 km to 500 km
How would you do it?
Changing Orbits:
Changing orbits usually involves an elliptical
orbit or Transfer Orbit
Perigee = close
Apogee = far
v1  v per  vi
v2  v f  vapo
Changing Orbits:
1) Velocity of initial orbit
km
vi = 7.75
sec
v1  v per  vi
v2  v f  vapo
2) Velocity of final orbit
km
v f  7.61
sec
3) Velocity at perigee
4) Velocity at apogee
Changing Orbits:
Since orbit is elliptical at Vper and Vapo a > 0,
so
2    
v
  
r
a
where
a 
r1  r2 
2
(250  6378.14)  (500  6378.14)
a 
2
a  6753 km
Changing Orbits:
So back to our V’s
3) Velocity at perigee
v per 
v per
v per
2     
r
a 
2* 398600.4
398600.4 


(250  6378.14)  6753 
km
 7.83
sec
Changing Orbits:
So back to our V’s
4) Velocity at apogee
v per =
( 2m ) - æ m ö
r
ç ÷
èaø
( 2 * 398600.4)
æ 398600.4 ö
v per =
-ç
÷
(500 + 6378.14) è 6753 ø
km
v per = 7.54
sec
Changing Orbits:
1) Velocity of initial orbit
km
vi = 7.75
sec
v1  v per  vi
v2  v f  vapo
2) Velocity of final orbit
km
v f  7.61
sec
3) Velocity at perigee
km
v per = 7.83
sec
4) Velocity at apogee
v apo
km
 7.54
sec
Changing Orbits:
Therefore:
V1 is to start transfer
Dv1 = v per - vi
Dv1 = 7.83- 7.75
km
Dv1 = .08
sec
Dv1 = 178.9 mph
v1
v2
Changing Orbits:
V2 is to circularize
orbit
Dv2 = v f - vapo
Dv2 = 7.61- 7.54
km
Dv2 = .07
sec
Dv2 =156.6 mph
v1
v2
Changing Orbits:
What if we did the
whole thing in reverse?
Go from 500 to 250
km?
What happens to the
answer?
Dv1 = vapo - vi
Dv2 = v f - v per
Dv2
Dv1
Changing Orbits:
1) Velocity of initial orbit
km
vi = 7.61
sec
Dv1 = vapo - vi
Dv2 = v f - v per
2) Velocity of final orbit
km
v f = 7.75
sec
3) Velocity at perigee
km
v per = 7.83
sec
4) Velocity at apogee
v apo
km
 7.54
sec
Changing Orbits:
Therefore:
V1 is to start transfer
Dv1 = vapo - vi
Dv1 = 7.54 - 7.61
km
Dv1 = -.07
sec
Dv1 = -156.5mph
Dv2
Dv1
Changing Orbits:
V2 is to circularize
orbit
Dv2 = v f - v per
Dv2 = 7.75- 7.83
km
Dv2 = -.08
sec
Dv2 =-178.9 mph
Dv2
Dv1
Changing Orbits:
Time to do transfer is
the same
a
P = 2p
*.5
m
3
v1
3
(6753)
P = 2p
*.5
398600.4
P = 2, 761 sec
P = 46 min
v2
How well do you understand Hohmann Transfers?
• 1 to 2?
• 2 to 3?
• 3 to 1?
3
• 1 to 3?
2
1
Circular Orbit:
Changing Orbits:
Also something called
“Fast Transfer”
• It is more direct and
quicker
• However it takes more
fuel
• V1 and V2 are
much bigger