Space and Rocket Propulsion

1. Orbits and space flight, types of orbits, basic orbital
equations, elliptical transfer orbits, launch trajectories,
the velocity increment needed for launch, the thermal
rocket engine, concepts of vertical takeoff and landing,
SSTO and TSTO, launch assists
1

2. Orbits and space flight
• Conic Sections
Conic Section Eccentricity, e Semi-major axis Energy
Circle 0 = radius < 0
Ellipse 0 < e < 1 > 0 < 0
Parabola 1 infinity 0
Hyperbola > 1 < 0 > 0
2

3. 3

4. Orbital Elements
• To mathematically describe an orbit one must define six quantities,
called orbital elements. They are
• The semi-major axis is one-half of the major axis and represents a
satellite's mean distance from its primary.
4

5. 5

6. • Inclination is the angular distance between a satellite's orbital plane
and the equator of its primary (or the ecliptic plane in the case of
heliocentric, or sun centered, orbits).
• An inclination of zero degrees indicates an orbit about the primary's
equator in the same direction as the primary's rotation, a direction
called prograde (or direct). An inclination of 90 degrees indicates a
polar orbit. An inclination of 180 degrees indicates a retrograde
equatorial orbit. A retrograde orbit is one in which a satellite moves
in a direction opposite to the rotation of its primary.
6

7. • Periapsis is the point in an orbit closest to the primary. The opposite
of periapsis, the farthest point in an orbit, is called apoapsis.
• Periapsis and apoapsis are usually modified to apply to the body
being orbited, such as perihelion and aphelion for the Sun, perigee
and apogee for Earth, perijove and apojove for Jupiter, perilune and
apolune for the Moon, etc.
• The argument of periapsis is the angular distance between the
ascending node and the point of periapsis. The time of periapsis
passage is the time in which a satellite moves through its point of
periapsis.
7

8. • Nodes are the points where an orbit crosses a plane, such as a
satellite crossing the Earth's equatorial plane.
• If the satellite crosses the plane going from south to north, the node
is the ascending node; if moving from north to south, it is
the descending node.
• The longitude of the ascending node is the node's celestial longitude.
Celestial longitude is analogous to longitude on Earth and is
measured in degrees counter-clockwise from zero with zero longitude
being in the direction of the vernal equinox.
8

9. • In general, three observations of an object in orbit are required
to calculate the six orbital elements. Two other quantities often
used to describe orbits are period and true anomaly.
• Period, P, is the length of time required for a satellite to
complete one orbit.
• True anomaly, ʋ, is the angular distance of a point in an orbit
past the point of periapsis, measured in degrees.
9

10. 10

11. Types of orbits
• For a spacecraft to achieve Earth orbit, it must be launched to an elevation above
the Earth's atmosphere and accelerated to orbital velocity. The most energy
efficient orbit, that is one that requires the least amount of propellant, is a direct
low inclination orbit. To achieve such an orbit, a spacecraft is launched in an
eastward direction from a site near the Earth's equator. The advantage being that
the rotational speed of the Earth contributes to the spacecraft's final orbital
speed.
• At the United States' launch site in Cape Canaveral (28.5 degrees north latitude) a
due east launch results in a "free ride" of 1,471 km/h (914 mph). Launching a
spacecraft in a direction other than east, or from a site far from the equator,
results in an orbit of higher inclination. High inclination orbits are less able to take
advantage of the initial speed provided by the Earth's rotation, thus the launch
vehicle must provide a greater part, or all, of the energy required to attain orbital
velocity. Although high inclination orbits are less energy efficient, they do have
advantages over equatorial orbits for certain applications.
11

12. • Geosynchronous orbits (GEO) are circular orbits around the Earth having a period
of 24 hours. A geosynchronous orbit with an inclination of zero degrees is called
a geostationary orbit. A spacecraft in a geostationary orbit appears to hang
motionless above one position on the Earth's equator. For this reason, they are
ideal for some types of communication and meteorological satellites. A spacecraft
in an inclined geosynchronous orbit will appear to follow a regular figure, pattern
in the sky once every orbit. To attain geosynchronous orbit, a spacecraft is first
launched into an elliptical orbit with an apogee of 35,786 km (22,236 miles)
called a geosynchronous transfer orbit (GTO). The orbit is then circularized by
firing the spacecraft's engine at apogee.
• Polar orbits (PO) are orbits with an inclination of 90 degrees. Polar orbits are
useful for satellites that carry out mapping and/or surveillance operations
because as the planet rotates the spacecraft has access to virtually every point on
the planet's surface.
12

13. • Walking orbits: An orbiting satellite is subjected to a great many gravitational
influences. First, planets are not perfectly spherical and they have slightly uneven
mass distribution. These fluctuations have an effect on a spacecraft's trajectory.
Also, the sun, moon, and planets contribute a gravitational influence on an
orbiting satellite. With proper planning it is possible to design an orbit which
takes advantage of these influences to induce a precession in the satellite's
orbital plane. The resulting orbit is called a walking orbit, or precessing orbit.
• Sun synchronous orbits (SSO) are walking orbits whose orbital plane precesses
with the same period as the planet's solar orbit period. In such an orbit, a satellite
crosses periapsis at about the same local time every orbit. This is useful if a
satellite is carrying instruments which depend on a certain angle of solar
illumination on the planet's surface. In order to maintain an exact synchronous
timing, it may be necessary to conduct occasional propulsive maneuvers to adjust
the orbit.
13

14. • Molniya orbits are highly eccentric Earth orbits with periods of approximately 12
hours (2 revolutions per day). The orbital inclination is chosen so the rate of
change of perigee is zero, thus both apogee and perigee can be maintained over
fixed latitudes. This condition occurs at inclinations of 63.4 degrees and 116.6
degrees. For these orbits the argument of perigee is typically placed in the
southern hemisphere, so the satellite remains above the northern hemisphere
near apogee for approximately 11 hours per orbit. This orientation can provide
good ground coverage at high northern latitudes.
• Hohmann transfer orbits are interplanetary trajectories whose advantage is that
they consume the least possible amount of propellant. A Hohmann transfer orbit
to an outer planet, such as Mars, is achieved by launching a spacecraft and
accelerating it in the direction of Earth's revolution around the sun until it breaks
free of the Earth's gravity and reaches a velocity which places it in a sun orbit
with an aphelion equal to the orbit of the outer planet. Upon reaching its
destination, the spacecraft must decelerate so that the planet's gravity can
capture it into a planetary orbit.
14

15. 15

16. Orbit types
16

17. Kepler's laws of planetary motion
1.The orbit of a smaller body about a larger body is always an ellipse,
with the centre of mass of the larger body coinciding with on of the
two foci of the ellipse.
17

18. 2. The orbit of the smaller body sweeps out equal areas in equal time.
This is graphically depicted in Figure, whereby the the areas A12 and
A34 are equal if the time differences t2 - t1 and t4- t3 are the same.
18

19. 3. The square of the period of revolution T of the smaller body is equal
to a constant multiplied by the 3rd power of the semi-major axis length
a, i.e.
19

20. Planet Mach 6
Period
(days)
Mercury 87.77
Venus 224.70
Earth 365.25
Mars 686.95
Jupiter 4332.62
Saturn 10759.2
20

21. Basic orbital equations
• The "centripetal" force required to keep a particle with a mass, m, travelling on a circular
path of radius, R is:
where v is the velocity of the particle.
• This equation can be partially derived intuitively by noting that if you swing a ball
around your head on a string...
• If you increase the mass of the ball you will have to pull harder on the string.
• If you swing the ball around your head at a faster speed you will have to pull (much) harder
on the string.
• If you lengthen the string while keeping the speed of the ball constant your tug on the
string decreases.
21

22. • Equating this force and gravitational force we get:
• In the equation above M is the mass of the object being orbited (e.g. the Sun)
and m is the mass of the object in orbit. R is the separation between the two
objects (i.e. the length of the string).
• Rearranging this equation you can get:
22

23. v = the orbital velocity of an object (m/s)
G = the universal gravitational constant, G = 6.673x10(-11) N∙m2/kg2
mE = the mass of the Earth (5.98 x 1024 kg)
r = the distance from the object to the center of the Earth
23

24. These relations show that you can:
• find the circular orbital velocity needed to launch a satellite into Earth
orbit.
• determine the mass of a planet (or star or galaxy...) which has a
satellite with a known velocity and separation from the planet.
• Applying Newton's Laws permits us to determine the mass of just
about any structure in the Universe.
24

25. 25

26. Numerical
1. The International Space Station orbits at an altitude of 400 km
above the surface of the Earth. What is the space station's orbital
velocity?
2. A satellite is orbiting the Earth with an orbital velocity of 3200 m/s.
What is the orbital radius?
v=7672 m/s
r = 3.897 x 107m
26

27. 3. Calculate the velocity of an artificial satellite orbiting the Earth in a circular orbit
at an altitude of 200 km above the Earth's surface.
v = 7,784 m/s
4. Calculate the period of revolution for the satellite in problem 3.
P2 = 4 × π 2 × r3 / GM
5. Calculate the radius of orbit for a Earth satellite in a geosynchronous orbit,
where the Earth's rotational period is 86,164.1 seconds.
27
P = 5,310 s
r = 42,164,170 m

28.  2
-
r
GMm
F
r
Newton's Law of Gravitation
To find the velocity increment required for various missions, we
must calculate trajectories and orbits. This is done using Newton's
Law of Gravitation:
Here the lhs is the "radial force" of attraction due to gravitation,
between two bodies; the big one of mass M,
and the little one of mass m.
The universal gravitational constant G is 6.670 E-11 Nm2/kg2.
28

29. Keplerian Orbits
Johannes Kepler (1609: first law, 1619: third law)
Kepler’s First Law: The orbit of each planet is an ellipse, with the sun at
one focus.
Second Law: The line joining the planet to the sun sweeps out equal areas
in equal times.
Third Law: The square of the period of a planet is proportional to the cube of
its mean distance from the sun
29

30. Kepler's Laws applied to satellite of mass m, orbiting a
much bigger object of mass M. i.e, m << M):
1. The satellite travels in an elliptical path around its center of
attraction, which is located at one of focii of the ellipse. The orbit
must lie in a plane containing the center of attraction.
30

31. Implication: The radius vector from the center of attraction sweeps equal areas
of the orbit per unit time. As the satellite moves away, its speed decreases. As it
nears the center of attraction, its speed increases.
Second Law: The line joining the planet to the sun sweeps out
equal areas in equal times.
Found on the web at a
secondary site. Primary
developer of this application is
unknown
31

32. Third Law: The square of the period of a planet is proportional to the cube of its
mean distance from the sun
Implication: The ratio of the squares of the orbital periods of any two satellites
about the same body equals the ratio of the cubes of the semi-major axes of
the respective orbits.



2 3
1 1
2 3
2 2
a
a
32

33. Satellite Equations of Motion
Newton’s Second Law
rmF 
Law of Gravitation
F=magnitude of force due to gravity (N)
G=Universal constant of gravity =
M=mass of Earth (kg)
M =satellite mass
R =distance from center of Earth to the satellite (km)
= GM for Earth = 398,600.5
Equation for satellite acceleration (from the above two Newton laws)
(2-body equation of motion)
22
r
m
r
GMm
F




2
17
1067.6
Kg
Nkm
x 
 23
sec
km
03
 r
r
r

The solution to this is the polar equation of a conic section.
33

34. Source: http://www.bartleby.com/61/imagepages/A4conics.html
Courtesy Houghton-Miflin Co.


Conic Energy Semi-major
axis a
e
Circle <0 = radius 0
Ellipse <0 >0 0<e<1
Parabola 0 1
Hyperbola >0 <0 >1
Total specific mechanical Energy of an orbit:
(mechanical energy per unit mass)
Why Conic Sections are called conic
sections:
ar
V
22
2




Conic Section Orbits / Trajectories
where a is the semi major axis and V is magnitude of velocity.
34

35. A Few Commonly-Used Results on Orbits
Circular orbit around Earth:
rr
R
r
V E
c
3481.631
905366.7 






 r in km is radius of orbit
rr
R
r
V E
escape
8611.892
17988.11
2

 V in km/s; r in km
Escape Velocity from Earth:
Speed at any point in an elliptical orbit:  
  
 
2
0 0
2 1
v g r
r a
 
 

 

2
0 0 min
1
1
apogee
e
v g r v
e
 
 

 

2
0 0 max
1
1
perigee
e
v g r v
e
Time for one orbit is:

 
3
0 0
2 a
r g
Specific angular momentum: is constant for the 2-body problem, per Kepler
(no plane change)
Vxrh 
35

36. Orbit Maneuvers
currentneed VVV 
where V is the velocity at that point in the corresponding orbit.
The assumption made usually in simplifying orbit maneuvers is that we are
dealing with a 2-body problem. For transfers between earth orbit and say,
Mars orbit, we assume initially an Earth-vehicle problem. When the vehicle
leaves the vicinity of Earth, we assume a Sun-vehicle problem, and finally a
Mars-vehicle problem
Hohmann transfer orbit:
Transfer orbit’s ellipse is tangent to initial and final
orbits, at the transfer orbit’s pengee and apogee
respectively
 Most efficient transfer between two circular
orbits.
Two-burn transfer between circular orbits at altitudes
: BA rr &
rA
rB
VB
VA
36

37. Hohmann Transfer
BATotal VVV 













































2
1
2
1
2
1
2
1
112112
BtxBAtxA rarrar

Example:
skmV
skmVskmV
kmr
kmr
Total
BA
B
A
/95.3
/49.1;/46.2
42160
6567




V for a two-burn transfer between circular orbits at altitudes . BA rr &
For Earth, with delta-v in km/s,   631.3481
 

2
A B
tx
r r
a
When the radii of the two orbits are close, the delta-v’s are approximately equal.
37

38. One-Tangent Burn
This is a quicker orbit transfer than the Hohman transfer ellipse, but is more costly in delta-v.
VA
VB
38

39. Launch from Earth
Note that the objective is to achieve the required tangential velocity for orbit – the need to gain height
above Earth is mainly dictated by the need to get out of the atmosphere and minimize drag losses.
As the vehicle trajectory begins to slant over, gravity pulls the trajectory down. At low speeds, this
requires a substantial pitch angle between the instantaneous velocity vector and the thrust direction
to counter gravity.
0
0
Gravity Turn
1. Begin ascent with short vertical rise, during which it starts usually and completes a roll from
launch-pad azimuth to desired flight azimuth.
2. Pitch-over (pitch because angular change of axis occurs faster than change of velocity
vector from vertical)
3. Adjust altitude to keep in trajectory through aerodynamic load region.
39

40. Launch Trajectories
40

41. 41

42. Acceleration During Launch
42

43. Values for Sample Missions: Garrison & Stuhler
V m/s
Earth- LEO (270km above Earth’s surface) 7600
LEO – GEO (42,227 km radius) 4200
LEO – Earth escape 3200
LEO – Lunar Orbit (7 days) 3900
LEO – Mars orbit (0.7 yrs) 5700
LEO – Mars orbit (40 days) 85000
LEO – Neptune Orbit (50 yr) 13,400
LEO – Neptune Orbit (29.9 yr) 8,700
LEO – Solar escape 70,000
LEO – 1000 AU (50yr)
LEO – Alpha Centauri (50yr)
142,000
30 Million
1 Astronautical Unit = mean distance from Earth to Sun = 149,558,000km
Impulsive burns assumed in all the above.
43

44. Gravity Assist Maneuvers
http://science.howstuffworks.com/framed.htm?
parent=question102.htm&url=http://www.jpl.nasa.gov
/basics/bsf4-1.html
Vehicle approaches another planet with velocity
greater than escape velocity for that planet.
Planet’s gravity adds momentum to the vehicle
in desired direction.
With respect to the planet, vehicle gains speed on
approach and decelerates as it moves away.
But when the planet’s velocity is considered, there
is net gain in angular momentum with respect
to the Sun.
44

45. 45

46. Thermal Rocket Engine
46