calculate escape velocities

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Conceptual Experiment

For this activity, you are going to calculate escape velocities for several exoplanets and compare them with our major planets.

In order to find escape velocities for the given exoplanets, enter the appropriate values of their masses and radii in unit of Earth's mass and Earth's radius in the provided template. Then, the escape velocity and gravitational acceleration will be automatically evaluated. In the case of our major planets, insert data from Table 10.1 on p. 199 in the textbook. After completing the table, answer the questions directly on the worksheet. You will save and upload your work on the provided template and submit it when you are complete.

Escape Velocities on Exoplanets Are we alone? Is the solar system unique in the Universe? No. It is just difficult to find planets because they are so tiny and dark compared to stars. Maybe direct observation is impossible even if the planet is larger than Jupiter because of the brightness of stars. However, there are some indirect methods to find them, and so far about 2,000 planetary systems have been found. In 1992 for the first time, two planets circling around Pulsar PSR 1257+12 were founded by radio astronomers using the pulsar timing method. In 1995, a planet orbiting around a main sequence star like the sun, 51 Pegasi, was discovered using radial velocity method. After that, many extra solar planets (exoplanets) were discovered using numerous indirect methods. For more further information, you may visit the following websites: http://exoplanetarchive.ipac.caltech.edu/ http://planetquest.jpl.nasa.gov/ http://planetquest.jpl.nasa.gov/news/239# For this activity, we are going to calculate escape velocities for several exoplanets and compare them with our major planets. Escape velocity, Ve, is defined to be the minimum velocity an object must have in order to escape the gravitational field of planet, that is, escape the planet without ever falling back. It can be evaluated by ve = 2 GM R = 2 gR where M is the mass of the planet, G is the gravitational constant, g is acceleration of gravity on the planet's surface, the radius of the planet. The selected exoplanets with some physical properties are as follows. Kepler- 452b is located about 1,400 light years away from earth. Its size is 1.6 times of Earth's radius and it has 5 times Earth's mass. 51 Pegasi b is about 41 light years from Earth. Its mass is about half of that of Jupiter. Its size is about twice of that of Jupiter's mass is 318 times Earth's mass and Jupiter's size is about 11 times Earth's size. Kepler-78b is located about 400 light-years from Earth. Its mass is double of Earth's mass and its size is 1.2 times of Earth's radius. The distance from "Super-Earth" exoplanet, OGLE-2005-BLG-390 Lb, is about 22,000 light years. of Earth. It is five times heavier than that of Earth. The distance between WASP-18b and Earth is about 325 light years. The mass of WASP-18b is 10 times of Jupiter's mass, that is, about 3,180 times the mass of Earth. Its size is 11 times bigger than that of Earth's radius. Look at the provided table below. The first column is the name of planet, the second column is the mass, the third column is the radius, the fourth column is the escape velocity, Ve, and the fifth column is gravitational acceleration, g. escape velocities for the given exoplanets, enter the appropriate values of their masses and radii in unit of Earth's mass and Earth's radius in the table below. Then, the escape velocity and gravitational acceleration will be automatically calculated. the case of our major planets, insert data from Table 10.1 on p. 199 in the textbook. Then, you will get them, too. After completing the table, answer the questions. Microsoft excel software is required to use the table below for automated calculation. However, if you do not have this, you can use your own calculator. It should be no problem to answer the questions. 1. Fill out Mass and Radius columns below using above information. Exoplanet name 51 Pegasi b Kepler -78b Kepler-452b Mass [ME] Radius [RE] Ve[km/s] g [m/s/s] WASP-18b OGLE-2005-BLG-390 Lb Our Planets Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune 2.Which planet (including both exoplanets and our major planets) is the most difficult to escape? 3.Which planet (including both exoplanets and our major planets) has the largest gravitational field, and which planet has th gravitational field? 4.Which exoplanet is most like the earth? Justify your answer. Your response should be at least 50 words in length. 5.Which factor affects the escape speed? Mass and/or radius? Your response should be at least 50 words in length. 6. If one of the planets becomes a black hole, what would the escape speed be? Your response should be at least 50 words in length. ult to find planets because they are so tiny and lanet is larger than Jupiter because of the and so far about 2,000 planetary systems have R 1257+12 were founded by radio astronomers quence star like the sun, 51 Pegasi, was (exoplanets) were discovered using numerous lanets and compare them with our major planets. ave in order to escape the gravitational field of the eration of gravity on the planet's surface, and R is .6 times of Earth's radius and it has 5 times of Jupiter. Its size is about twice of that of Jupiter. Earth's mass and its size is 1.2 times of Earth's out 22,000 light years. Its size is about half of that 18b is 10 times of Jupiter's mass, that of Earth's radius. second column is the mass, the third column is is gravitational acceleration, g. In order to find the eir masses and radii in unit of Earth's mass and acceleration will be automatically calculated. In extbook. Then, you will get them, too. After However, if you do not have this, ost difficult to escape? gravitational field, and which planet has the smallest
UNIT IV STUDY GUIDE Gravity and Orbital Motion Course Learning Outcomes for Unit IV Upon completion of this unit, students should be able to: 3. Explain Newton's laws of motion at work in common phenomena. 3.1 Illustrate the relation of the universal law of gravitation to Newton’s second law. 3.2 Distinguish between gravitational acceleration (g) and gravitational constant (G). 3.3 Evaluate gravitational field strength when mass and radius of an object are given. 4. Explain the concepts and applications of momentum, work, mechanical energy, and general relativity. 4.1 Apply total mechanical energy conservation for orbital motion. 4.2 Calculate escape velocity when gravitational potential energy is balanced with kinetic energy. 4.3 Describe the escape velocity in a black hole, a consequence of Einstein’s general relativity. Reading Assignment Chapter 9: Gravity Chapter 10: Projectile and Satellite Motion Unit Lesson Projectile Motion When an object moves with a curved path near the earth’s surface under the influence of gravity, its motion is called projectile motion. For example, look at Figures 10.6 and 10.8 on pages 185 to 186 in the textbook. If we ignore air resistance, the horizontal motion of the projectile does not slow down; its velocity is constant. In other words, the horizontal component of the acceleration is zero. However, the vertical component of the velocity is not constant, but changes. In addition, the vertical component of the acceleration is downward acceleration, gravitational acceleration, (g). Weightlessness and Free Fall Suppose you are in an elevator. If the elevator is not accelerating, your weight (W) is just your mass (m) times the gravitational acceleration (g). In fact, two forces are acting on you; the weight (W) and the normal force (F). According to Newton’s second law, in the vertical direction, ma=F-W=F-mg. That is, normal force F=m(g+a). Here, g is positive, but a may be either positive for upward acceleration or negative for downward acceleration of the elevator. If the elevator is in upward motion, apparent weight (or normal force) is greater than your true weight. On the other hand, if the elevator is in downward motion, the apparent weight is smaller than your true weight. In a special case, when the acceleration is equal to g, that is, a=-g, or free fall, the apparent weight becomes zero: weightless. Please look at Figure 9.9 on p.166 in the textbook for an example of this. The same phenomena occur when an object is circling around the earth. The orbiting satellite, which accelerates toward the center of the earth, is also in free fall. See Figure 9.10 on p.167 in the textbook. Over a long period of time, the weightlessness is harmful for humans, and thus, a rotating space station in a wheel shape is provided to create artificial gravity. It is balanced with the centripetal force, mv2/r, of the system. That is mg=mv2/r. Here, m is the mass of an astronaut, r is the distance from axis to the surface of the station, and v is the rotating speed. For instance, if r is given 1 km, then v=(rg)1/2= 100 m/s. PHS 1110, Principles of Classical Physical Science 1 Newton’s Law of Universal Gravitation Newton speculated about the highest reachable point by the force of gravity on the earth. He realized that there is a limit and concluded that the orbital motion of the moon around the earth is maintained by the gravitational force (Hewitt, 2015). Suppose you throw a stone horizontally from a high place (See Figure 10.16 on p. 190 in the textbook). The stone falls to the ground because of gravity. However, if you throw the stone with great speed, it will move further and further away from where you are standing before falling to the ground. When the speed is great enough, the stone will eventually circle around the earth. This is the projectile motion, where the projectile falls in the gravitational field but never touches the ground. This logical consideration can be applied to explain the orbital motion of the moon. Newton concluded that the moon is falling in its pathway around the Earth because of the gravitational acceleration. Newton extended the above idea to any two objects in the universe in order to explain the interaction between them. Newton’s law of universal gravitation postulates that there is an attractive force between the two objects (Hewitt, 2015). The force between two objects in the universe is proportional to the product of two masses m and M and is inversely proportional to the square of distance r between two objects; F=GmM/r2 , where G= (6.6710-11 N m2/kg2) is the universal gravitational constant. This is the case when the gravitational acceleration (a) is equal to g in the second law of Newton; a=g, and thus, g=GM/r2. The constant, G was measured by Cavendish 100 years after Newton announced his theory. It was not an easy task because of the extremely small value of gravitation attraction. The detailed story is in Section 9.2 on pp. 163–164 in the textbook. Example: What is the magnitude of the gravitational force between the sun and the earth? The distance between the sun and the earth is 1AU= 1.501011 m. The mass of the earth is m = 5.9810 24 kg and the mass of the Sun is M=1.991030 kg. Solution: From F=GmM/r2= 6.6710-11 x 5.981024 x 1.991030 / (1.50 x1011)2 = 3.51025 N Kepler’s Three Empirical Laws for Planetary Motion Johannes Kepler (1571 - 1630) was a German astronomer and had an endless enthusiasm for researching the solar system. It took him more than 20 years to realize through his calculations the exact shape of the planets’ orbitals. He tested many different kinds of models using his teacher Tyco Brache’s enormous data set. Brache had accumulated very exact planetary data without even the use of telescopes. Kepler established three important empirical laws of planetary motion: the law of elliptical orbit, the law of areas, and the law of the relation between period and distance. This is what he used to describe and understand the motion of the Solar System (Zeilik & Smith, 1987). Mechanical motion of our solar system obeys gravitational law, and planets are orbiting around the sun, which is the heaviest mass in the solar system. The orbital shape is not circular, but elliptical. Some comets have parabolic or hyperbolic orbits. These well-known mechanics were not easily discovered. Since the ancient times, the sky was considered a realm of gods, so perfectness was assumed. The notion that the orbits of planets should be a perfect circle was widely accepted, and no scholar would be able to prove otherwise to the people, even Kepler. For these reasons, it took a long time to accurately describe planetary motion in our solar system. The famous three empirical laws of planetary motion describe the motion of the solar system as follows:  First Law—The law of ellipses: The orbit of each planet is an ellipse with the sun at one foci. The shape of a planet's orbit is an ellipse.  Second Law—The law of areas: The radius vector to a planet sweeps out equal areas in equal intervals of time. When a planet is closer to the sun, it revolves faster, and, on the other hand, when a planet is farther away from the sun, it revolves slower. PHS 1110, Principles of Classical Physical Science 2  Third Law—The law of harmony: The squares of the sidereal periods of the planets are proportional to the cubes of the semi-major axes (mean radii) of their orbits. Here, the sidereal period is the time it takes the planet to complete one orbit of the Sun with respect to the stars (Zeilik & Smith, 1987). Thus, Kepler's laws and Newton's laws taken together imply that a force holds a planet in its orbit by continuously changing the planet's velocity, so that it follows an elliptical path. The force is directed toward the sun from the planet and is proportional to the product of the masses of the sun and the planets. Also, the force is inversely proportional to the square of the planet-sun separation. This is precisely the form of the gravitational force postulated by Newton. Newton's laws of motion, with a gravitational force used in the second law, imply Kepler's laws, and the rest of the planets obey the same laws of motion as objects on the surface of the earth. Conic Sections and Gravitational Orbits Hypatia (360 - 415) visualized various shapes of geometric equations using conic sections for the first time in Alexandria, Egypt (Larson & Edwards, 2010). Conic sections are formed when a cone is cut with a plane at various angles. For a more detailed description, visit the website about this in the Suggested Reading section of this unit. There are various orbits in a gravitational system. The circular orbit is a special case of ellipse. The ellipse can be formed when the plane intersects opposite “edges” of the cone. In the case of the parabola orbit, the plane is parallel to one edge of the cone. On the other hand, the hyperbola orbit does not intersect opposite edges of the cone, and the plane is not parallel to the edge. Planets in our solar system have elliptical orbits with various eccentricities. The orbital eccentricity (e) determines the shape of orbits. If e=0 (E<0), the orbit is circular, if 01(E>0), the orbit is hyperbolic. Here, E is the total energy. Some comets have elliptical orbits, too. However, other comets have parabolic orbits, and once they pass the Sun, they will never come back. In the case of two interacting stars, they show a hyperbolic orbit. Like the parabolic orbit, the hyperbolic one is a one-time encounter. Geostationary Orbits Some satellites revolve around the earth at the same speed of earth’s rotation, which is very useful for communication purposes. They are located at about 36,000 km above the earth’s surface. This position does not depend on the mass of the satellite. The centripetal force is equal to gravitational force, mv2/r= GmM/r2, where m is the mass of satellite, M is the mass of the earth, and r is the distance from the center of the earth’s surface to the satellite. The speed of the satellite is v=(GM/r)1/2. Also, v=distance/time=the circumference of the circular orbit/the orbital period=2πr/T. That is, the orbital period is T=2πr3/2/(GM)1/2, or T2=4π2r3/GM. Notice that this is exactly the same with Kepler’s third law. In other words, the harmonic law is a consequence when the centripetal force is in balance with the gravitational force. T is about 24 hours. One hour equals 60 minutes. One minute equals 60 seconds. Therefore, 24 hours are 24x3600=86400 seconds. The earth’s mass is 5.98 x 1024 kg and G = 6.67 x 10-11N m2/kg2. By substituting all the appropriate values to the above equation, you obtain r=42300 km. The radius of the earth is about 6400 km, so subtract 6400 from 42300: 42300-6400=35900 km, which is about 22300 miles. Energy Conservation and Escape Velocity For a circular orbit at distance (r) from the center of the Earth (r = RE + h, if h is the altitude of the orbit), the circular speed (vc) can be found by equating the centripetal force (mv2/r) and gravitational (GmM/r2) force, or v = (GM/r)1/2 . Also, from the conservation of total energy, TE = (KE + PE)ground= (KE+PE)h =constant. By evaluating TE at h = 0 and at maximum height (h), we find that mv2/2=mgh, or h=v2/2g. Notice that when h= RE, v = 11.2 km/s, the projectile escapes to h = . This critical speed is called the escape speed. PHS 1110, Principles of Classical Physical Science 3 Escape velocity is defined as the minimum velocity an object must have in order to escape the gravitational field of the earth, to escape the earth without ever falling back. The object must have greater energy than its gravitational binding energy to escape the earth's gravitational field. Black Holes and Einstein’s Relativity A black hole is the final stage of massive stellar evolution. A direct observation of a black hole is impossible because no light can escape from it. If you get too close to a black hole, the speed you would need to escape from it would exceed the speed of light. Because nothing can travel faster than light, nothing—not even light— can escape from a black hole. The existence of black holes was predicted by Einstein’s general theory of relativity, which is the best theory to describe what gravity is and how it behaves. In 1915, Einstein published the general theory of relativity. Click here for more information on the general theory of relativity. The curvature of space and time is influenced by gravity. The more massive, the more distortions of space and time. References Hewitt, P. G. (2015). Conceptual physics (12th ed.). Upper Saddle River, NJ: Pearson. Larson, R., & Edwards, B. H. (2010). Calculus (9th ed.). Belmont, CA: Brooks/Cole. Zeilik, M., & Smith, E. I. (1987). Introductory astronomy & astrophysics (2nd ed.). Philadelphia, PA: Saunders College Publishing. Suggested Reading This link will take you to the website mentioned in the unit lesson. It provides more information about conic sections. Nave, R. (n.d.). Conic sections. Retrieved from http://hyperphysics.phy-astr.gsu.edu/hbase/math/consec.html The following document further explains black holes and Einstein’s theory of relativity. Click here to access a pdf version of the document. Learning Activities (Nongraded) Nongraded Learning Activities are provided to aid students in their course of study. You do not have to submit them. If you have questions, contact your instructor for further guidance and information. To practice what you have learned in this unit, complete the following problems and questions from the textbook. The answers to each problem can be found in the “Odd-numbered Answers” section in the back of the textbook. The question number from the textbook is indicated in parentheses after each question. 1. Calculate the force of Earth’s gravity on a 1-kg mass at Earth’s surface. The mass of Earth is 6.0x1024kg and its radius is 6.4 x 106m. Does the result surprise you? (Textbook #33 on p. 178) 2. Suppose you stood atop a ladder so tall that you were three times as far from Earth’s center as you presently are. Show that your weight would be one ninth of its present value. (Textbook #39 on p. 178) 3. An astronaut lands on a planet that has the same mass as Earth, but twice the diameter. How does the astronaut’s weight differ from that on Earth? (Textbook #59 on p. 179) PHS 1110, Principles of Classical Physical Science 4 4. The force due to gravity on you is mg. Under what condition is mg also your weight? (Textbook #71 on p. 159) 5. If our Sun shrank in size to become a black hole, discuss and show from the gravitational force equation that Earth’s orbit would not be affected. (Textbook #107 on p. 181) 6. With your friends, whirl a bucket of water in a vertical circle fast enough so that water doesn’t spill out. As it happens, the water in the bucket is falling, but with less speed that you give to the bucket. Tell how your bucket swing relates to satellite motion — that satellites in orbit continuously fall toward Earth, but not with enough vertical speed to get closer to the curved Earth below. Remind your friends that physics is about finding the connections in nature! (Textbook #25 on page 201) 7. At a particular point in its orbit, a satellite in an elliptical orbit has a gravitational potential energy of 5000 MJ with respect to Earth’s surface and a kinetic energy of 4500 MJ. Later in its orbit, the satellite’s potential energy is 6000 MJ. What is its kinetic energy at that point? (Textbook #31 on page 202) 8. Fragments of fireworks beautifully illuminate the night sky. (a) What specific path is ideally traced by each fragment? (b) What paths would be fragments trace in a gravity-free region? (Textbook #45 on page 203) 9. Which planets have a more-than-one-Earth-year period: planets nearer than Earth to the Sun, or planets farther from the Sun than Earth? (Textbook #55 on page 204) 10. Can a satellite maintain an orbit in the plane of the Arctic Circle? Why or why not? (Textbook #95 on page 205) PHS 1110, Principles of Classical Physical Science 5

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Good stuff. Would use again.

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