Transcript AE 1350
AE 1350
Lecture #14
Introduction to Astronautics
Astronautics is a broad field
• It includes a study of
– Payload Design and Integration (e.g. satellites, space
lab, Mars rover, etc.)
– Mission Design and Analysis (Selection of Trajectories)
– Launch Vehicle Design and Analysis (Rockets, space
shuttle, X- vehicles, etc.)
– Reentry Systems
• Here, we will briefly look at rockets as launch vehicles,
and some simple missions.
Rockets
• Thrust depends on two factors:
– rate at which momentum leaves the rocket
through the nozzle
– Exit pressure pexit
•
•
•
•
VJet
T = -dm/dt Vjet + (pexit-patmosphere) Aexit
In well designed rockets, pexit = patmosphere
T = -dm/dt Vjet
Notice the negative sign. The mass m of
the rocket decreases. dm/dt is thus a
negative quantity.
Velocity of the Rocket
•
From the previous slide, Thrust T = - dm/dt Vjet
•
This thrust is used to accelerate the rocket, and the
payload it carries.
•
From Newton’s second law,
m dV/dt = T = - dm/dt Vjet
•
We can write the above equation as:
dV = - Vjet dm/m
•
Integrate :
DVrocket = Vjet loge{mstart/mend}
How can the velocity of the
rocket be maximized?
• From the previous slide:
Change in the speed of the rocket (and
payload it carries is given by):
DVrocket = Vjet loge{mstart/mend}
• We must increase mass of the rocket at the start by loading
it up with fuel.
• We must decrease the mass of the rocket at the end of each
stage or leg of the the mission by discarding used up stages
of the rocket.
Single Stage vs. Multistage
• A multi-stage rocket discards earlier stages
as soon as they are used up. This decreases
mass of the rocket and the end of each
stage, and increases DV.
• A multi-stage rocket is more complex and is
expensive to build.
• A single stage rocket may is conceptually
cheaper, but may not be able to achieve high
velocities.
Newton’s Law of Gravitation
• On the earth surface, gravity is nearly a constant.
• Gravitational forces decrease as we move away from the
center of earth.
• According to Newton, two objects of mass M (the earth)
and m (the satellite) separated by a distance r will attract
each other with a force: GMm/r2 where G is a universal
constant.
• The gravitational acceleration on the satellite is, thus,
GM/ r2.
• Here r is the distance from the center of the earth.
• G = 6.67 x 10-11 m3/kg/sec2
• GM = 3.956 x 1014 m3/sec2
Specific Impulse
From earlier slides,
dm
Ve
dt
where Ve is the exhaust ve locity of the rocket jet.
T
Since W mg
dm 1 dW
g dt
dt
V
T
e
Thus,
dW
g
dt
The quantity Ve /g is called the Specific Impulse.
Rockets are rated on how high a specific impulse they can generate.
Question..
• Chemical rockets typically have specific impulses around 250-400
seconds and a thrust/weight ratio of 100-200.
• Nuclear electric rockets can have a specific impulse of 20,000 seconds
and a thrust to weight ratio of 0.0001.
• For example,
– the Saturn V main engine produced 265 seconds
– The shuttle 455 seconds
– NERVA (Nuclear Engine for Rocket Vehicle Applications ) has a specific
impulse of 900 seconds
– Orion (detonation of nuclear bombs in a controlled manner) would have
produced 2000 - 6000 seconds, with 10,000 seconds expected from a full
production vehicle.
• A rocket with a high specific impulse doesn't need as much fuel as a
rocket with a low specific impulse.
• In your own words, Explain what this information means.
Satellites in Circular Orbit
Centrifugal force = mV2/r
r
Force due to gravity= GMm/r2
GMm/r2 = mV2/r
Here,
G = Universal constant
M = Mass of the earth
V = Velocity of the satellite
r = Radius of the orbit
Low Earth Orbits
• Consider a satelite in a low earth orbit.
• r = 6.4 Million meters (essentially earth’s radius which equals 6.378
Million meters).
• Gravitational Force = Centrifugal Force
• GMm/r2 = mv2/r
• Use GM = 3.956 x 1014 m3/sec2 from the previous slides.
• v = {GM/r}1/2 = 7900 m/s
• We need to accelerate a satellite from zero velocity to 7900 m/s in
order for it to stay on a low earth circular orbit.
• Time period = 2pr/v = 5090 sec = 1.41 hours
• A satellite on this low earth orbit will go around the earth once very
1.4 hours.
Troposphere ground to 6 miles
Stratosphere 6 miles-50 miles
Ionosphere 50-300 miles
Exosphere 300-600 miles
Space Shuttle 150 miles
Other manned spacecraft 90-300 miles
Earth-observing satellites 500 miles
Navigational satellites 6,200-13,000 miles
Geostationary satellites 22,000 miles
Moon 238,857 miles
Geostationary Orbits
• A geostationary orbit is one where the
satellite revolves around the earth once per
day.
• It appears to stay stationary above the earth,
facilitating TV communications.
• Using the procedure identical to the one for
the low earth orbit, we can show that a
geostationary orbit will have r = 22,000
miles
Elliptical Orbits
Elliptical Orbits, with earth (or Mars, or Jupiter, or whatever) are
also possible. In such orbits, the object will be at one of the focal
points of the ellipse.
Escape Velocity
• Consider a payload of mass m.
• Gravitational Force on this mass = GMm/r2
• Work done if we move this satellite from earth surface (r=6372 km) to
infinitely far away =
r
r 6372km
GMm
r
2
dr
GMm
R earth
• This work is done by the rocket which supplies the payload with energy
1/2mv2
• Solve for v to get escape velocity. v = 11.2 km/s
• If we accelerate a payload to 11.2 km/s, it will escape earth’s gravity.
• Lower velocities will lead to circular or elliptical orbits.
• At very low velocities, the centrifugal force is too small, and the payload
will spiral into the earth’s atmosphere.
Example #1
Calculate the velocity of an artificial satellite orbiting the earth
in a circular orbit at an altitude of 150 miles above the earth's
surface.
SOLUTION:
Given: r = (3,960 + 150) x 5,280 = 21,700,800 ft
v = SQRT[ GM / r ]
v = SQRT[ 1.408x1016 / 21,700,800 ]
v = 25,470 ft/s
Kepler’s Laws
• A satellite describes an elliptical path around its
center of attraction, which is located at one of the
two focus points of the ellipse.
• Per unit time, area swept by the radius vector
joining the satellite and the center of attraction is
constant.
• The periods of any two satellites revolving around
the same center body are proportional to their
(3/2) power of their semi-major axes.
Ellipse
Minor axis
2b
Focus
Focus
Major axis: Length 2a
Graphical Explanation of
Kepler’s Second Law
Area swept per unit time is a constant,
no matter where the satellite is on its orbit.
These two areas are equal per unit time.
Kepler’s Third Law
2
3
t 1 a1
3 Satellite 1
Satellite 2
2
Time period t
t 2 a2 Time period t
2
Semi-major axis= a2
1
Semi-major axis= a1
Center-body that provides attraction