ae 1350 lecture #14 introduction to astronautics

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AE 1350 Lecture #14 Introduction to Astronautics

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Page 1: AE 1350 Lecture #14 Introduction to Astronautics

AE 1350

Lecture #14

Introduction to Astronautics

Page 2: 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.

Page 3: AE 1350 Lecture #14 Introduction to Astronautics

Rockets• Thrust depends on two factors:

– rate at which momentum leaves the rocket through the nozzle

– Exit pressure pexit

• 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 anegative quantity.

VJet

Page 4: AE 1350 Lecture #14 Introduction to Astronautics

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 : Vrocket = Vjet loge{mstart/mend}

Page 5: AE 1350 Lecture #14 Introduction to Astronautics

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):

Vrocket = 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.

Page 6: AE 1350 Lecture #14 Introduction to Astronautics

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 V.

• 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.

Page 7: AE 1350 Lecture #14 Introduction to Astronautics

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

Page 8: AE 1350 Lecture #14 Introduction to Astronautics

Specific Impulse

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Page 9: AE 1350 Lecture #14 Introduction to Astronautics

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.

Page 10: AE 1350 Lecture #14 Introduction to Astronautics

Satellites in Circular Orbit

Force due to gravity= GMm/r2

Centrifugal force = mV2/r

GMm/r2 = mV2/r

Here, G = Universal constantM = Mass of the earthV = Velocity of the satelliter = Radius of the orbit

r

Page 11: AE 1350 Lecture #14 Introduction to Astronautics

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 = 2r/v = 5090 sec = 1.41 hours

• A satellite on this low earth orbit will go around the earth once very 1.4 hours.

Page 12: AE 1350 Lecture #14 Introduction to Astronautics

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

Page 13: AE 1350 Lecture #14 Introduction to Astronautics

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

Page 14: AE 1350 Lecture #14 Introduction to Astronautics

Elliptical Orbits

Elliptical Orbits, with earth (or Mars, or Jupiter, or whatever) arealso possible. In such orbits, the object will be at one of the focal points of the ellipse.

Page 15: AE 1350 Lecture #14 Introduction to Astronautics

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 =

• This work is done by the rocket which supplies the payload with energy 1/2mv

• 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.

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Page 16: AE 1350 Lecture #14 Introduction to Astronautics

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

Page 17: AE 1350 Lecture #14 Introduction to Astronautics

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.

Page 18: AE 1350 Lecture #14 Introduction to Astronautics

Ellipse

Major axis: Length 2a

Focus FocusMinor axis

2b

Page 19: AE 1350 Lecture #14 Introduction to Astronautics

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.

Page 20: AE 1350 Lecture #14 Introduction to Astronautics

Kepler’s Third Law

Center-body that provides attraction

Satellite 1Time period 1

Semi-major axis= a1

Satellite 2Time period 2

Semi-major axis= a2

32

31

22

21

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a