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8/10/2019 MAE 2201 FA13 Lecture 24 Orbital Eq
http://slidepdf.com/reader/full/mae-2201-fa13-lecture-24-orbital-eq 1/15
Rockets-Micro thrusters
State of the artmicrothrusters…
DARPA + Tanner
research
Micro-thrusters are
miniature actuators…
8/10/2019 MAE 2201 FA13 Lecture 24 Orbital Eq
http://slidepdf.com/reader/full/mae-2201-fa13-lecture-24-orbital-eq 2/15
Rockets-Micro thrusters
Micro-thrusters are miniature actuators created using smallcavities, rocket propellant, and low energy igniters.
Work like Space Shuttle solid boosters, but are 1000’s of times smaller
Individual micro-electromechanical system (MEMS) thrusters
size: poppy seed-sized cell fueled with lead styphnate propellant, fired
more than 20 times at 1-second intervals each thruster delivered 0.0001 Newton seconds of impulse.
MEMS design, based on silicon chip fabrication technology, offers
several advantages over conventional thrusters
no moving parts
utilizes a variety of propellants
is scalable
eliminates the need for tanks, fuel lines and valves, and fully
integrates the structure of the satellite with the propulsion to power it.
8/10/2019 MAE 2201 FA13 Lecture 24 Orbital Eq
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Rockets-Micro thrusters
MEMS micro-thruster arrays are fabricated as a three-layersilicon and glass sandwich,
middle layer consisting of multiple small propellant cells sealed with
a rupturable diaphragm on one side and an ignitor on the other.
Each cell is a separate thruster, and when ignited, delivers one
impulse bit.
Delivering propulsion in discrete increments by igniting thrusters in
controlled sequences has lent the technology the name "digital
propulsion."
8/10/2019 MAE 2201 FA13 Lecture 24 Orbital Eq
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Rockets-Micro thrusters
State of art Cavity now built of aluminum to withstand higher pressures
Cavity is 4mm per side
Latest micro-thruster design may generate up to 40 mN-seconds of
impulse by combusting 35 milligrams of propellant primarily composed
of black powder.
Isp=116 s
Will be used for small satellites, guided munitions etc.
8/10/2019 MAE 2201 FA13 Lecture 24 Orbital Eq
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Orbital Mechanics
Last time we discussed burnout velocity…why isthat important?
Velocities on order of 7.9 km/s are needed to place
satellites in Earth orbit
Interplanetary vehicles need 11 km/s to get out of Earth’s
gravity
8/10/2019 MAE 2201 FA13 Lecture 24 Orbital Eq
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8/10/2019 MAE 2201 FA13 Lecture 24 Orbital Eq
http://slidepdf.com/reader/full/mae-2201-fa13-lecture-24-orbital-eq 7/15
8/10/2019 MAE 2201 FA13 Lecture 24 Orbital Eq
http://slidepdf.com/reader/full/mae-2201-fa13-lecture-24-orbital-eq 8/15
8/10/2019 MAE 2201 FA13 Lecture 24 Orbital Eq
http://slidepdf.com/reader/full/mae-2201-fa13-lecture-24-orbital-eq 9/15
8/10/2019 MAE 2201 FA13 Lecture 24 Orbital Eq
http://slidepdf.com/reader/full/mae-2201-fa13-lecture-24-orbital-eq 10/15
Orbital Mechanics
Starts with Kepler who developed the three laws of planetarymotion
First: The orbit of each planet is an ellipse, with the sun at
a focus
Second: The line joining the planet to the sun sweeps out
equal areas in equal times
8/10/2019 MAE 2201 FA13 Lecture 24 Orbital Eq
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Parameters of Elliptic Orbit
8/10/2019 MAE 2201 FA13 Lecture 24 Orbital Eq
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Orbital Mechanics
Third: The square of the period, (time for 1 orbit) of aplanet is proportional to the cube of its mean distance
from the sun (semi-major axis, “a”, of the ellipse)
Review Example Problem 8.2 on pg. 675 for period
calculations
Newton:
Law of gravity:
r mma F
r
r
r
GMm F
2
8/10/2019 MAE 2201 FA13 Lecture 24 Orbital Eq
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Orbital Mechanics
Law of gravity: G is gravitational constant
Vector r is distance
M, m are masses of the two bodies
The potential energy of a body of mass m foundsome distance away from a body of mass M (>m) is:
The kinetic energy of the small body is (in a polarcoordinate system):
r
r
r
GMm
F
2
r
GMm
222
2
1
2
1 r r mmV T
8/10/2019 MAE 2201 FA13 Lecture 24 Orbital Eq
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Orbital Mechanics
The motion of the space vehicle in the direction is the conservation ofangular momentum
The orbit equation : gives the orbital or trajectory motion of satellite or planet
e eccentricity e=0; path is circle
e<1; path is an ellipse
e>1; path is hyperbola
e=1; path is parabola
h is angular momentum per unit mass is a constant =
p=h2 /k2 e=A(h2 /k2) all these are determined from burnout conditions
G*M=k2
Review Example Problem 8.1 on page 668 for trajectory calculations
I momentumangular
const r m
_
.2
)cos(1 C e pr
2r
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Orbital Mechanics
e eccentricity e=0; path is circle
e<1; path is an ellipse
e>1; path is hyperbola
e=1; path is parabola
e