# Lecture 14

Information about Lecture 14

Published on January 13, 2008

Author: Tibald

Source: authorstream.com

AE 1350:  AE 1350 Lecture #14 Introduction to Astronautics Astronautics is a broad field:  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:  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 a negative quantity. VJet Velocity of the Rocket:  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?:  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:  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:  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:  Specific Impulse Question..:  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:  Satellites in Circular Orbit Force due to gravity= GMm/r2 Centrifugal force = mV2/r GMm/r2 = mV2/r Here, G = Universal constant M = Mass of the earth V = Velocity of the satellite r = Radius of the orbit r Low Earth Orbits:  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:  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:  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 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:  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 1/2mv2 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:  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:  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:  Ellipse Major axis: Length 2a Focus Focus Minor axis 2b Graphical Explanation of Kepler’s Second Law:  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:  Kepler’s Third Law Center-body that provides attraction Satellite 1 Time period t1 Semi-major axis= a1 Satellite 2 Time period t2 Semi-major axis= a2

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