Introduction Newton s law of gravitation. Copyright Kinetic Books Co. Chapter 13

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1 Introduction The topic of gravity has had a starring role in some of the most famous tales in the history of physics. Galileo Galilei was studying the acceleration due to the Earth s gravity when he dropped two balls from the Leaning Tower of Pisa. A theory to explain the force of gravity came to Isaac Newton shortly after an apple fell from a tree and knocked him in the head. Although historians doubt whether these events actually occurred, the stories have come to symbolize how a simple experiment or a sudden moment of insight can lead to important and lasting scientific progress. Galileo is often said to have dropped two balls of different masses from the tower so that he could see if they would land at the same time. Most scientists of his era would have predicted that the heavier ball would hit the ground first. Instead, the balls landed at the same instant, showing that the acceleration due to gravity is a constant for differing masses. While it is doubtful that Galileo actually dropped balls off that particular tower, his writings show that he performed many experiments studying the acceleration due to gravity. In another famous story, Newton formed his theory of gravity after an apple fell and hit him on the head. At least one person (the daughter of French philosopher Voltaire) said that Newton mentioned that watching a falling apple helped him to comprehend gravity. Falling apple or no, his theory was not the result of a momentary insight; Newton pondered the topic of gravity for decades, relying on the observations of contemporary astronomers to inform his thinking. Still, the image of a scientist deriving a powerful scientific theory from a simple physical event has intrigued people for generations. The interactive simulations on the right will help you begin your study of gravity. Interactive 1 permits you to experiment with the gravitational forces between objects. In its control panel, you will see five point masses. There are three identical red masses of mass m. The green mass is twice as massive as the red, and the blue mass is four times as massive. You can start your experimentation by dragging two masses onto the screen. The purple vectors represent the gravitational forces between them. You can move a mass around the screen and see how the gravitational forces change. What is the relationship between the magnitude of the forces and the distances between the masses? Experiment with different masses. Drag out a red mass and a green mass. Do you expect the force between these two masses to be smaller or larger than it would be between two red masses situated the same distance apart? Make a prediction and test it. You can also drag three or more masses onto the screen to see the gravitational force vectors between multiple bodies. There is yet more to do: You can also press GO and see how the gravitational forces cause the masses to move. The other major topic of this chapter is orbital motion. The force of gravity keeps bodies in orbit. Interactive 2 is a reproduction of the inner part of our solar system. It shows the Sun, fixed at the center of the screen, along with the four planets closest to it: Mercury, Venus, Earth and Mars. Press GO to watch the planets orbit about the Sun. You can experiment with our solar system by changing the position of the planets prior to pressing GO, or by altering the Sun s mass as the planets orbit. In the initial configuration of this system, the period of each planet s orbit (the time it takes to complete one revolution around the Sun) is proportional to its actual orbital period. Throughout this chapter, as in this simulation, we will often not draw diagrams to scale, and will speed up time. If we drew diagrams to scale, many of the bodies would be so small you could not see them, and taking 365 days to show the Earth completing one revolution would be asking a bit much of you Newton s law of gravitation Newton s law of gravitation: The attractive force of gravity between two particles is proportional to the product of their masses and inversely proportional to the square of the distance between them. Newton s law of gravitation states that there is a force between every pair of particles, of any mass, in the universe. This force is called the gravitational force, and it causes objects to attract one another. The force does not require direct contact. The Earth attracts the Sun, and the Sun attracts the Earth, yet meters of distance separate the two. The strength of the gravitational force increases with the masses of the objects, and weakens proportionally to the square of the distance between them. Two masses exert equal but opposite attractive forces on each other. The forces act on a line between the two objects. The 346 Copyright Kinetic Books Co. Chapter 13

2 magnitude of the force is calculated using the equation on the right. The symbol G in the equation is the gravitational constant. It took Newton about 20 years and some false starts before he arrived at the relationship between force, distance and mass. Later scientists established the value for G, which equals N m 2 /kg 2. This small value means that a large amount of mass is required to exert a significant gravitational force. The example problems on the right provide some sense of the magnitude of the gravitational force. First, we calculate the gravitational force the Earth exerts on the Moon (and the Moon exerts on the Earth). Although separated by a vast distance (on average, their midpoints are separated by about 384,000,000 meters), the Earth and the Moon are massive enough that the force between them is enormous: N. In the second example problem, we calculate the gravitational force between an kg car and a 2200-kg truck parked 15 meters apart. The force is N. When you press a button on a telephone, you press with a force of about one newton, 1,400,000 times greater than this force. Newton s law of gravitation Gravitational force Proportional to masses of bodies Inversely proportional to square of distance Newton s law of gravitation F = force of gravity M, m = masses of objects r = distance between objects G = gravitational constant G = N m 2 /kg 2 What is the gravitational force between the Earth and the Moon? F = N Each body attracts the other Copyright Kinetic Books Co. Chapter

3 How much gravitational force do the car and the truck exert on each other? F = N G and g Newton s law of gravity includes the gravitational constant G. In this section, we discuss the relationship between G and g, the rate of freefall acceleration in a vacuum near the Earth s surface. The value of g used in this textbook is , an average value that varies slightly by location on the Earth. Both a 10-kg object and a 100-kg object will accelerate toward the ground at The rate of freefall acceleration does not vary with mass. Newton s law of gravitation, however, states that the Earth exerts a stronger gravitational force on the more massive object. If the force on the more massive object is greater, why does gravity cause both objects to accelerate at the same rate? The answer becomes clear when Newton s second law of motion, F = ma, is applied. The acceleration of an object is proportional to the force acting on it, divided by the object s mass. The Earth exerts ten times the force on the 100-kg object that it does on the 10-kg object. But that tenfold greater force is acting on an object with a mass ten times greater, meaning the object has ten times more resistance to acceleration. The result is that the mass term cancels out and both objects accelerate toward the center of the Earth at the same rate, g. G and g G = gravitational constant everywhere in universe g = freefall acceleration at Earth s surface If the gravitational constant G and the Earth s mass and radius are known, then the acceleration g of an object at the Earth s surface can be calculated. We show this calculation in the following steps. The distance r used below is the average distance from the surface to the center of the Earth, that is, the Earth s average radius. We treat the Earth as a particle, acting as though all of its mass is at its center. Variables gravitational constant G = N m 2 /kg 2 mass of the Earth mass of object Earth-object distance acceleration of object Strategy M = kg m r = m g G and g 1. Set the expressions for force from Newton s second law and his law of gravitation equal to each other. 2. Solve for the acceleration of the object, and evaluate it using known values for the other quantities in the equation. g = free fall acceleration at Earth s surface G = gravitational constant M = mass of Earth r = distance to center of Earth 348 Copyright Kinetic Books Co. Chapter 13

4 Physics principles and equations We will use Newton s second law of motion and his law of gravitation. In this case, the acceleration is g. Step-by-step derivation Here we use two of Newton s laws, his second law (F = ma) and his law of gravitation (F = GMm/r 2 ). We use g for the acceleration instead of a, because they are equal. We set the right sides of the two equations equal and solve for g. Step Reason 1. Newton s laws 2. simplify 3. substitute known values 4. g = evaluate Our calculations show that g equals The value for g varies by location on the Earth for reasons you will learn about later. In the steps above, the value for G, the gravitational constant, is used to calculate g, with the mass of the Earth given in the problem. However, if g and G are both known, then the mass of the Earth can be calculated, a calculation performed by the English physicist Henry Cavendish in the late 18 th century. Physicists believe G is the same everywhere in the universe, and that it has not changed since the Big Bang some 13 billion years ago. There is a caveat to this statement: some research indicates the value of G may change when objects are extremely close to each other Shell theorem Shell theorem: The force of gravity outside a sphere can be calculated by treating the sphere s mass as if it were concentrated at the center. Newton s law of gravitation requires that the distance between two particles be known in order to calculate the force of gravity between them. But applying this to large bodies such as planets may seem quite daunting. How can we calculate the force between the Earth and the Moon? Do we have to determine the forces between all the particles that compose the Earth and the Moon in order to find the overall gravitational force between them? Fortunately, there is an easier way. Newton showed that we can assume the mass of each body is concentrated at its center. Newton proved mathematically that a uniform sphere attracts an object outside the sphere as though all of its mass were concentrated at a point at the sphere s center. Scientists call this the shell theorem. (The word shell refers to thin shells that together make up the sphere and which are used to mathematically prove the theorem.) The shell theorem Consider sphere s mass to be concentrated at center r is distance between centers of spheres Consider the groundhog on the Earth s surface shown to the right. Because the Earth is approximately spherical and the matter that makes up the planet is distributed in a spherically symmetrical fashion, the shell theorem can be applied to it. To use Newton s law of gravitation, three values are required: the masses of two objects and the distance between them. The mass of the groundhog is 5.00 kg, and the Earth s mass is kg. The distance between the groundhog and the center of the Earth is the Earth s radius, which averages meters. In the example problem to the right, we use Newton s law of gravitation to calculate the gravitational force exerted on the groundhog by the Earth. The force equals the groundhog s weight (mg), as it should. How much gravitational force does the Earth exert on the Copyright Kinetic Books Co. Chapter

5 groundhog? F = 48.9 N Shell theorem: inside the sphere Another section discussed how to calculate the force of gravity exerted on an object on the surface of a sphere (a groundhog on the Earth). Now imagine a groundhog burrowing halfway to the center of the Earth, as shown to the right. For the purposes of calculating the force of gravity, what is the distance between the groundhog and the Earth? And what mass should be used for the Earth in the equation? The first question is easier: The distance used to calculate gravity s force remains the distance between the groundhog and the Earth s center. Determining what mass to use is trickier. We use the sphere defined by the groundhog s position, as shown to the right. The mass inside this new sphere, and the mass of the groundhog, are used to calculate the gravitational force. (Again, we assume that the Earth s mass is symmetrically distributed.) If the groundhog is 10 meters from the Earth s center, the mass enclosed in a sphere with a radius of 10 meters is used in Newton s equation. If the animal moves to the center of the Earth, then the radius of the sphere is zero. At the center, no mass is enclosed, meaning there is no net force of gravity. The groundhog is perhaps feeling a little claustrophobic and warm, but is effectively weightless at the Earth s center. The volume of a sphere is proportional to the cube of the radius, as the equation to the right shows. If the groundhog burrows halfway to the center of the Earth, then the sphere encloses one-eighth the volume of the Earth and one-eighth the Earth s mass. Inside a sphere To calculate gravitational force Use mass inside the new shell r is distance between object, sphere s center Let s place the groundhog at the Earth s center and have him burrow back to the planet s surface. The gravitational force on him increases linearly as he moves back out to the Earth s surface. Why? The force increases proportionally to the mass enclosed by the sphere, which means it increases as the cube of his distance from the center. But the force also decreases as the square of the distance. When the cube of a quantity is divided by its square, the result is a linear relationship. If the groundhog moves back to the Earth s surface and then somehow moves above the surface (perhaps he boards a plane and flies to an altitude of 10,000 meters), the force again is inversely proportional to the square of the groundhog s distance from the Earth s center. Since the mass of the sphere defined by his position no longer varies, the force is computed using the full mass of the Earth, the mass of the groundhog, and the distance between their centers. Volume of a sphere V = volume r = radius Sample problem: gravitational force inside the Earth What is the gravitational force on the groundhog after it has burrowed halfway to the center of the Earth? Assume the Earth s density is uniform. The Earth s mass and radius are given in the table of variables below. 350 Copyright Kinetic Books Co. Chapter 13

6 Variables gravitational force mass of Earth radius of Earth mass of inner sphere distance between groundhog and Earth s center mass of groundhog F M E = kg r E = m M s r s = m m = 5.00 kg gravitational constant G = N m 2 /kg 2 What is the strategy? 1. Use the shell theorem. Compute the ratio of the volume of the Earth to the volume of the sphere defined by the groundhog s current position. Use this comparison to calculate the mass of the inside sphere. 2. Use the mass of the inside sphere, the mass of the groundhog and the distance between the groundhog and the center of the Earth to calculate the gravitational force. Physics principles and equations Newton s law of gravitation Mathematics principles The equation for the volume of a sphere is Step-by-step solution First, we calculate the mass enclosed by the sphere. Step Reason 1. ratio of masses proportional to ratio of volumes 2. rearrange 3. volume of a sphere 4. cancel common factors 5. enter values 6. M s = kg evaluate Copyright Kinetic Books Co. Chapter

7 Now that we know the mass of the inner sphere, the shell theorem states that we can use it to calculate the gravitational force on the groundhog using Newton s law of gravitation. Step Reason 7. law of gravitation 8. enter values 9. F = 24.4 N solve for the force This calculation also confirms a point made previously: Inside the Earth, the groundhog s weight increases linearly with distance from the planet s center. At half the distance from the center, his weight is half his weight at the surface Earth s composition and g The force of gravity differs slightly at different locations on the Earth. These variations mean that g, the acceleration due to the force of gravity, also differs by location. Why do the force of gravity and g vary? First, the surface of the Earth can be slightly below sea level (in Death Valley, for example), and it can be more than 8000 meters above it (on peaks such as Everest and K2). Compared to an object at sea level, an object at the summit of Everest is 0.14% farther from the center of the Earth. This greater distance to the Earth s center means less force (Newton s law of gravitation), which in turn reduces acceleration (Newton s second law). If you summit Mt. Everest and jump with joy, the force of gravity will accelerate you toward the ground about slower than if you were jumping at sea level. Second, the Earth has a paunch of sorts: It bulges at the equator and slims down at the poles, making its radius at the equator about 21 km greater than at the poles. This is shown in an exaggerated form in the illustration for Concept 3. The bulge is caused by the Earth s rotation and the fact that it is not entirely rigid. This bulge means that at the equator, an object is farther from the Earth s center than it would be at the poles and, again, greater distance means less force and less acceleration. Value of g about at sea level Finally, the density of the planet also varies. The Earth consists of a jumble of rocks, minerals, metals and water. It is denser in some regions than in others. The presence of materials such as iron that are denser than the average increases the local gravitational force by a slight amount. Geologists use gravity gradiometers to measure the Earth s density. Variations in the density can be used to prospect for oil or to analyze seismic faults. Value for g varies: Due to altitude Value for g varies: Because the Earth is not a perfect sphere 352 Copyright Kinetic Books Co. Chapter 13

8 Value for g varies: Because the Earth is not uniformly dense Interactive checkpoint: gravitation A planet and its moon are attracted to each other by a gravitational force of N at a distance of m. The planet is 155 times as massive as the moon. What is the mass of the planet? Answer: M = kg Newton s cannon In addition to noting that the Earth exerts a force on an apple, Newton also pondered why the Moon circles the Earth. He posed a fundamental question: Is the force the Earth exerts on the Moon the same type of force that it exerts on an apple? He answered yes, and his correct answer would forever change humanity s understanding of the universe. Comparing the orbit of the distant Moon to the fall of a nearby apple required great intellectual courage. Although the motion of the Moon overhead and the fall of the apple may not seem to resemble one another, Newton concluded that the same force dictates the motion of both, leading him to propose new ways to think about the Moon s orbit. To explain orbital motion, Newton conducted a thought experiment: What would happen if you used a very powerful cannon to fire a stone from the top of a very tall mountain? He knew the stone would obey the basic precepts of projectile motion, as shown in the diagrams to the right. But, if the stone were fired fast enough, could it just keep going, never touching the ground? (Factors such as air resistance, the Earth s rotation, and other mountains that might block the stone are ignored in Newton s thought experiment.) Newton concluded that the stone would not return to the Earth if fired fast enough. As he wrote in his work, Principia, published in 1686: Newton s cannon Newton imagined a powerful cannon The faster the projectile, the farther it travels At ~8,000, projectile never touches ground... the greater the velocity with which [a stone] is projected, the farther it goes before it falls to the earth. We may therefore suppose the velocity to be so increased, that it would describe an arc of 1, 2, 5, 10, 100, 1000 miles before it arrived at earth, till at last, exceeding the limits of the earth, it should pass into space without touching. Newton correctly theorized that objects in orbit moons, planets and, today, artificial satellites are in effect projectiles that are falling around a central body but moving fast enough that they never strike the ground. He could use his theory of gravity and his knowledge of circular motion to explain orbits. (In this section, we focus exclusively on circular orbits, although orbits can be elliptical, as well.) Why is it that the stone does not return to the Earth when it is fired fast enough? Why can it remain in orbit, forever circling the Earth, as shown to the right? First, consider what happens when a cannon fires a cannonball horizontally from a mountain at a relatively slow speed, say 100. In the Copyright Kinetic Books Co. Chapter

9 vertical direction, the cannonball accelerates at g toward the ground. In the horizontal direction, the ball continues to move at 100 until it hits the ground. The force of gravity pulls the ball down, but there is neither a force nor a change in speed in the horizontal direction (assuming no air resistance). Now imagine that the cannonball is fired much faster. If the Earth were flat, at some point the ball would collide with the ground. But the Earth is a sphere. Its approximate curvature is such that it loses five meters for every 8000 horizontal meters, as shown in Concept 2. At the proper horizontal (or more properly, tangential) velocity, the cannonball moves in an endless circle around the planet. For every 8000 meters it moves forward, it falls 5 meters due to gravity, resulting in a circle that wraps around the globe. In this way, satellites in orbit actually are falling around the Earth. The reason astronauts in a space shuttle orbiting close to the Earth can float about the cabin is not because gravity is no longer acting on them (the Earth exerts a force of gravity on them), but rather because they are projectiles in freefall. Close-up of Newton s cannon At high speeds, Earth s curvature affects whether projectile lands At ~8,000, ground curves away at same rate that object falls Objects in orbit Move fast enough to never hit the ground Continually fall toward the ground, pulled by force of gravity Interactive problem: Newton s cannon Imagine that you are Isaac Newton. It is the year 1680, and you are staring up at the heavens. You see the Moon passing overhead. You think: Perhaps the motion of the Moon is related to the motion of Earth-bound objects, such as projectiles. Suppose you threw a stone very, very fast. Is there a speed at which the stone, instead of falling back to the Earth, would instead circle the planet, passing around it in orbit like the Moon? You devise an experiment in your head, a type of experiment called a thought experiment. A thought experiment is a way physicists can test or explain valuable concepts even though they cannot actually perform the experiment. You ask: What if I had an extremely powerful cannon mounted atop a mountain. Could I fire a stone so fast it would never hit the ground? Try Newton s cannon in the simulation to the right. You control the initial speed of the cannonball by clicking the up and down buttons in the control panel. The cannon fires horizontally, tangent to the surface of the Earth. See if you can put the stone into orbit around the Earth. You can create a nearly perfect circular orbit, as well as orbits that are elliptical. (In this simulation we ignore the rotation of the Earth, as Newton did in his thought experiment. When an actual satellite is launched, it is fired in the same direction as the Earth s rotation to take advantage of the tangential velocity provided by the spinning Earth.) Circular orbits The Moon orbits the Earth, the Earth orbits the Sun, and today artificial satellites are propelled into space and orbit above the Earth s surface. (We will use the term satellite for any body that orbits another body.) These satellites move at great speeds. The Earth s orbital speed around the Sun averages about 30,000 (that is about 67,000 miles per hour!) A communications satellite in circular orbit 250 km above the surface of the Earth moves at In this section, we focus on circular orbits. Most planets orbit the Sun in roughly circular paths, and artificial satellites typically travel in circular orbits around the Earth as well. 354 Copyright Kinetic Books Co. Chapter 13

10 The force of gravity is the centripetal force that along with a tangential velocity keeps the body moving in a circle. We use an equation for centripetal force on the right to derive the relationship between the mass of the body being orbited, orbital radius, and satellite speed. As shown in Equation 1, we first set the centripetal force equal to the gravitational force and then we solve for speed. This equation has an interesting implication: The mass of the satellite has no effect on its orbital speed. The speed of an object in a circular orbit around a body with mass M is determined solely by the orbital radius, since M and G are constant. Satellite speed and radius are linked in circular orbits. A satellite cannot increase or decrease its speed and stay in the same circular orbit. A change in speed must result in a change in orbital radius, and vice versa. At the same orbital radius, the speed of a satellite increases with the square root of the mass of the body being orbited. A satellite in a circular orbit around Jupiter would have to move much faster than it would if it were in an orbit of the same radius around the Earth. Circular orbits Satellites in circular orbit have constant speed Orbital speed and radius Satellite speed and radius are linked The smaller the orbit, the greater the speed Speed in circular orbit m = mass of satellite v = satellite speed G = gravitational constant M = mass of body being orbited r = orbital radius Copyright Kinetic Books Co. Chapter

11 Sample problem: speed of an orbiting satellite What is the speed of a satellite in circular orbit 500 kilometers above the Earth's surface? The illustration above shows a satellite in circular orbit 500 km above the Earth s surface. The Earth s radius is stated in the variables table. Variables satellite speed satellite orbital radius satellite height Earth s radius Earth s mass v r h = 500 km r E = m M E = kg gravitational constant G = N m 2 /kg 2 What is the strategy? 1. Determine the satellite s orbital radius, which is its distance from the center of the body being orbited. 2. Use the orbital speed equation to determine the satellite s speed. Physics principles and equations The equation for orbital speed is Step-by-step solution We start by determining the satellite s orbital radius. Careful: this is not the satellite s height above the surface of the Earth, but its distance from the center of the planet. Step Reason 1. r = r E + h equation for satellite s orbital radius 2. r = m m enter values 3. r = m add Now we apply the equation for orbital speed. Step Reason 4. orbital speed equation 5. enter values 6. v = 7610 evaluate Interactive problem: intercept the orbiting satellite In this simulation, your mission is to send up a rocket to intercept a rogue satellite broadcasting endless Barney reruns. You can accomplish this by putting the rocket into an orbit with the same radius and speed as that of the satellite, but traveling in the opposite direction. The resulting collision will destroy the satellite (and your rocket, but that is a small price to pay to save the world s sanity). 356 Copyright Kinetic Books Co. Chapter 13

12 The rogue satellite is moving in a counterclockwise circular orbit 40,000 kilometers ( m) above the center of the Earth. Your rocket will automatically move to the same radius and will move in the correct direction. You must do a calculation to determine the proper speed to achieve a circular orbit at that radius. You will need to know the mass of the Earth, which is kg. Enter the speed (to the nearest 10 ) in the control panel and press GO. Your rocket will rise from the surface of the Earth to the same orbital radius as the satellite, and then go into orbit with the speed specified. You do not have to worry about how the rocket gets to the orbit; you just need to set the speed once the rocket is at the radius of the satellite. If you fail to destroy the satellite, check your calculations, press RESET and try a new value Interactive problem: dock with an orbiting space station In this simulation, you are the pilot of an orbiting spacecraft, and your mission is to dock with a space station. As shown in the diagram to the right, your ship is initially orbiting in the same circular orbit as the space station. However, it is on the far side of the Earth from the space station. In order to dock, your ship must be in the same orbit as the space station, and it must touch the space station. To dock, your speed and radius must be very close to that of the space station. A high speed collision does not equate to docking! You have two buttons to control your ship. The Forward thrust button fires rockets out the back of the ship, accelerating your spacecraft in the direction of its current motion. The other button, labeled Reverse thrust, fires retrorockets in the opposite direction. To use more professional terms, the forward thrust is called prograde and the reverse thrust is called retrograde. Using these two controls, can you figure out how to dock with the space station? To assist in your efforts, the current orbital paths for both ships are drawn on the screen. Your rocket s path is drawn in yellow, and the space station s is drawn in white. This simulation requires no direct mathematical calculations, but some thought and experimentation are necessary. If you get too far off track, you may want to press RESET and try again from the beginning. There are a few things you may find worth observing. You will learn more about them when you study orbital energy. First, what is the change in the speed of the rocket after firing its rear (Forward thrust) engine? What happens to its speed after a few moments? If you qualitatively consider the total energy of the ship, can you explain you what is going on? You may want to consider it akin to what happens if you throw a ball straight up into the sky. If you cannot dock but are able to leave and return to the initial circular orbit, you can consider your mission achieved Kepler s first law The law of orbits: Planets move in elliptical orbits around the Sun. The reason Newton s comparison of the Moon s motion to the motion of an apple was so surprising was that many in his era believed the orbits of the planets and stars were divine circles: arcs across the cosmic sky that defied scientific explanation. Newton used the fact that the force of gravity decreases with the square of the distance to explain the geometry of orbits. Scientists had been proposing theories about the nature of orbits for centuries before Newton stated his law of gravitation. Numerous theories held that all bodies circle the Earth, but subsequent observations began to point to the truth: the Earth and other planets orbit the Sun. The conclusion was controversial; in 1633 the Catholic Church forced Galileo to repudiate his writings that implied this conclusion. Kepler s first law Planets move in elliptical orbits with the Sun at one focus Even earlier, in 1609, the astronomer and astrologer Johannes Kepler began to propose what are now three basic laws of astronomy. He developed these laws through careful mathematical analysis, relying on the detailed observations of his mentor, Tycho Brahe, a talented and committed observational astronomer. Kepler and Brahe formed one of the most productive teams in the history of astronomy. Brahe had constructed a state of the art observatory on an island off the coast of Denmark. State of the art is always a relative term the telescope had not yet been invented, and Brahe might well Copyright Kinetic Books Co. Chapter

13 have traded his large observatory for a good pair of current day binoculars. However, Brahe s records of years of observations allowed Kepler, with his keen mathematical insight, to derive the fundamental laws of planetary motion. He accomplished this decades before Newton published his law of gravitation. Kepler, using Brahe s observations of Mars, demonstrated what is now known as Kepler s first law. This law states that all the planets move in elliptical orbits, with the Sun at one focus of the ellipse. The planets as well as other solar system bodies move in elliptical motion. The distinctly elliptical orbit of Pluto, formerly a planet, is shown to the right, with the Sun located at one focus of the ellipse. Had Kepler been able to observe Pluto, the elliptical nature of orbits would have been more obvious. The other planets in the solar system, some of which he could see, have orbits that are very close to circular. (If any of them moved in a perfectly circular orbit, they would still be moving in an ellipse, since a circle is an ellipse with both foci at its center.) Some of the orbits of these other planets are shown in Concept 2. Solar system Most planetary orbits are nearly circular More on ellipses and orbits The ellipse shape is fundamental to orbits and can be described by two quantities: the semimajor axis a and the eccentricity e. Understanding these properties of an ellipse proves useful in the study of elliptical orbits. The semimajor axis, represented by a, is one-half the width of the ellipse at its widest, as shown in Concept 1. You can calculate the semimajor axis by averaging the maximum and minimum orbital radii, as shown in Equation 1. The eccentricity is a measure of the elongation of an ellipse, or how much it deviates from being circular. (The word eccentric comes from ex-centric, or off-center.) Mathematically, it is the ratio of the distance d between the ellipse s center and one focus to the length a of its semimajor axis. You can see both these lengths in Equation 2. Since a circle s foci are at its center, d for a circle equals zero, which means its eccentricity equals zero. Pluto has the most eccentric orbit in our solar system, with an eccentricity of 0.25, as calculated on the right. By comparison, the eccentricity of the Earth s orbit is Most of the planets in our solar system have nearly circular orbits. Elliptical orbits Semimajor axis: one half width of orbit at widest Eccentricity: elongation of orbit Comets have extremely eccentric orbits. This means their distance from the Sun at the aphelion, the point when they are farthest away, is much larger than their distance at the perihelion, the point when they are closest to the Sun. (Both terms come from the Greek, far from the Sun and near the Sun respectively.) Halley s comet has an eccentricity of We show the comet s orbit in Example 1, but for visual clarity it is not drawn to scale. The comet s orbit is so eccentric that its maximum distance from the Sun is 70 times greater than its minimum distance. In Example 1, you calculate the perihelion of this object in AU. The AU (astronomical unit) is a unit of measurement used in planetary astronomy. It is equal to the average radius of the Earth s orbit around the Sun: about meters. Semimajor axis a = semimajor axis r min = minimum orbital distance r max = maximum orbital distance 358 Copyright Kinetic Books Co. Chapter 13

14 Eccentricity e = eccentricity d = distance from center to focus a = semimajor axis What is the perihelion distance of Halley s Comet? The comet s semimajor axis is 17.8 AU. r min = 2a r max r min = 2(17.8 AU) 35.1 AU r min = 0.5 AU What is the eccentricity of Pluto s orbit? e = 0.25 Copyright Kinetic Books Co. Chapter

15 Interactive checkpoint: elliptical orbit An asteroid orbits a star in an elliptical orbit with a periapsis (closest approach) of 13.5 AU and an apoapsis (farthest distance) of 98.7 AU. What are the semimajor axis and the eccentricity of the asteroid s orbit? Answer: a = AU e = Kepler s second law The law of areas: An orbiting body sweeps out equal areas in equal amounts of time. In his second law, Kepler used a geometrical technique to show that the speed of an orbiting planet is related to its distance from the Sun. (We use the example of a planet and the Sun; this law applies equally well for a satellite orbiting the Earth, or for Halley s comet orbiting the Sun.) Kepler used the concept of a line connecting the planet to the Sun, moving like a second hand on a watch. As shown to the right, the line sweeps out slices of area over time. His second law states that the planet sweeps out an equal area in an equal amount of time in any part of an orbit. In an elliptical orbit, planets move slowest when they are farthest from the Sun and move fastest when they are closest to the Sun. Kepler s second law Planets in orbit sweep out equal areas in equal times Kepler established his second law nearly a century before Newton proposed his theory of gravitation. Although Kepler did not know that gravity varied with the inverse square of the distance, using Brahe s data and his own keen quantitative insights he determined a key aspect of elliptical planetary motion Kepler s third law The law of periods: The square of the period of an orbit is proportional to the cube of the semimajor axis of the orbit. Kepler s third law, proposed in 1619, states that the period of an orbit around a central body is a function of the semimajor axis of the orbit and the mass of the central body. The semimajor axis a is one half the width of the orbit at its widest. In a circular orbit, the semimajor axis is the same as the radius r of the orbit. We illustrate this in Concept 1 using the Earth and the Sun. Given the scale of illustrations in this section, the Earth s nearly circular orbit appears as a circle. The length of the Earth s period a year, the time required to complete a revolution about the Sun is solely a function of the mass of the Sun and the distance a shown in Concept 2. Orbital period Time of one revolution Kepler s third law states that the square of the period is proportional to the cube of the semimajor axis, and inversely proportional to the mass of the central body. The law is shown in Equation 1. For the equation to hold true, the mass of the central body must be much greater than that of the satellite. This law has an interesting implication: The square of the period divided by the cube of the semimajor axis has the same value for all the bodies orbiting the Sun. In our solar system, that ratio equals about years 2 /meters 3 (where years are Earth years) or s 2 /m 3. This is demonstrated in the graph in Concept 3. The horizontal and vertical scales of the coordinate system are logarithmic, with semimajor axis measured in AU and period measured in Earth years. 360 Copyright Kinetic Books Co. Chapter 13

16 Kepler s third law Square of orbital period proportional to cube of semimajor axis Graph of Kepler s third law Orbital size versus period for planets orbiting Sun Kepler s third law T = satellite period (in seconds) G = gravitational constant M = mass of central body a = semimajor axis Copyright Kinetic Books Co. Chapter

17 Sample problem: the period of the Moon What is the period of the Moon's orbit, in days? The illustration shows the Moon in its slightly elliptical orbit around the Earth, with a semimajor axis of m. In this problem, we calculate the sidereal period, the time it takes for the Moon to return to the same position in the sky, using a background of fixed stars as a reference. The Moon s synodic period, the time it takes to return to the same position relative to the Sun, is longer. Variables Moon period Moon s orbital semimajor axis Earth s mass T a = m M E = kg gravitational constant G = N m 2 /kg 2 What is the strategy? Use Kepler s third law to find the Moon s orbital period in seconds and convert the result to days. Physics principles and equations We will use Kepler s third law Step-by-step solution Enter the given values into the equation for Kepler s third law. Be careful with units: The equation for Kepler s third law gives the period in seconds. Step Reason 1. Kepler s third law 2. enter values 3. T = s evaluate 4. T in days = convert seconds to days 5. T in days = 27.4 days evaluate Interactive problem: geosynchronous satellite In this simulation, your mission is to put a satellite into circular geosynchronous orbit around the Earth. A geosynchronous orbit is one in which the satellite s orbital period is 24 hours, the same as the Earth s rotational period. If situated over the equator, a geosynchronous satellite stays positioned over the same location, a requirement for some communications satellites. What orbital radius and orbital speed must you set for the satellite so that it achieves a circular, geosynchronous orbit? You need to determine a speed that will cause the satellite to remain over a fixed point above the Earth. (Hint: There are 24 hours in a day.) Kepler s third law will help you determine the orbital radius. Once you have determined the required radius, use that value to determine the orbital speed for a satellite in a circular orbit. This simulation contains an added complication: You need to factor in the Earth s motion. The Earth supplies a tangential velocity of 460 to 362 Copyright Kinetic Books Co. Chapter 13

18 satellites launched from the equator, and the rocket launches in the direction that allows it to take advantage of that boost. In short, you must subtract 460 from whatever you have determined is the orbital speed needed for geosynchronous orbit. The ship will keep that speed and maneuver appropriately as it moves to the radius orbit you specified. Enter an orbital radius (in kilometers) and a launch speed (in ) for your satellite in the entry boxes in the control panel. When you press GO, the satellite will launch according to your specifications. If the satellite lines up with the rotating red dot on Earth, you have succeeded. (A yellow circle indicates the size of the necessary orbit.) If you do not succeed at first, press RESET, redo your calculations, and blast off again. Note that we are showing a view of the Earth from above the South Pole, and from that view the Earth rotates in a clockwise direction Orbits and energy Satellites have both kinetic energy and potential energy. The KE and PE of a satellite in elliptical orbit both change as it moves around its orbit. This is shown in Concept 1. Energy gauges track the satellite s changing PE and KE. When the satellite is closer to the body it orbits, Kepler s second law states that it moves faster, and greater speed means greater KE. The PE of the system is less when the satellite is closer to the body it orbits. When discussing gravitational potential energy, we must choose a reference point that has zero PE. For orbiting bodies, that reference point is usually defined as infinite separation. As two bodies approach each other from infinity, potential energy decreases and becomes increasingly negative as the value declines from zero. Concept 2 shows that even while the PE and KE change continuously in an elliptical orbit, the total energy TE stays constant. Because there are no external forces acting on the system consisting of the satellite and the body it orbits, nor any internal dissipative forces, its total mechanical energy must be conserved. Any increase in kinetic energy is matched by an equivalent loss in potential energy, and vice versa. In a circular orbit, a satellite s speed is constant and its distance from the central body remains the same, as shown in Concept 3. This means that both its kinetic and potential energies are constant. Orbital energy Satellites have kinetic and potential energy Since PE = 0 at infinite distance, PE always negative The total energy of a satellite increases with the radius (in the case of circular orbits) or the semimajor axis (in the case of elliptical orbits). Moving a satellite into a larger orbit requires energy; the source of that energy for a satellite might be the chemical energy present in its rocket fuel. Equations used to determine the potential and kinetic energies and the total energy of a satellite in circular orbit are shown in Equation 1. These equations can be used to determine the energy required to boost a satellite from one circular orbit to another with a different radius. The KE equation can be derived from the equation for the velocity of a satellite. The PE equation holds true for any two bodies, and can be derived by calculating the work done by gravity as the satellite moves in from infinity. The equations have an interesting relationship: The kinetic energy of the satellite equals one-half the absolute value of the potential energy. This means that when the radius of a satellite s orbit increases, the total energy of the satellite increases. Its kinetic energy decreases since it is moving more slowly in its higher orbit, but the potential energy increases twice as much as this decrease in KE. For a given orbit: Total energy is constant Equation 2 shows the total energy equation for a satellite in an elliptical orbit. This equation uses the value of the semimajor axis a instead of the radius r. For a given circular orbit Both PE, KE constant Total energy increases with radius Copyright Kinetic Books Co. Chapter

19 Energy in circular orbits G = gravitational constant M = mass of planet m = mass of satellite r = orbit radius Energy in elliptical orbits E tot = total energy a = semimajor axis Sample problem: energy and orbital radius A 175 kg satellite in circular orbit fires its rockets twice to move to a new circular orbit with a larger radius. How much energy does the satellite expend to attain the wider orbit? The illustration shows a satellite in a circular orbit performing a two-stage maneuver called a Hohmann transfer that boosts it to a circular orbit with a larger radius. The diagram is not drawn to scale. We assume that the satellite s mass stays effectively constant. 364 Copyright Kinetic Books Co. Chapter 13

20 Variables first orbital radius second orbital radius mass of Earth mass of satellite r i = m r f = m M = kg m = 175 kg gravitational constant G = N m 2 /kg 2 total energy of satellite E What is the strategy? 1. Write an equation for the change in the total energy of the satellite. 2. Substitute the total energy expressions for the satellite in each orbit. 3. Enter the given values and calculate the work performed. Physics principles and equations The work done by the satellite s rocket engine equals its change in total energy. The equation for total energy of a satellite in a circular orbit is, Step-by-step solution The change in the satellite s energy is equal to its final energy minus its initial energy. We write this as an equation, then substitute the energy expressions for the satellite in each orbit and evaluate the result. Step Reason 1. E = E f E i change in energy 2. substitute energy expressions 3. substitute values for radii and simplify 4. enter remaining values 5. E = J = J evaluate change in total energy Copyright Kinetic Books Co. Chapter

21 Interactive checkpoint: Kepler s third law and energy A planet orbits a star with a period of s and a semimajor axis of m. What is the mass of the star? The total energy of the planet in this orbit is J. What is the mass of the planet? Answer: M = kg m = kg Sample problem: energy of a rocket to the Moon A 20,000 kg rocket is in a circular orbit above the Earth having a radius 6.58x10 6 m. How much energy will it take to boost the rocket to the Moon on the orbital path shown? Upon firing its engines, the rocket will follow an elliptical orbital path to the Moon. This problem simplifies the actual orbital mechanics required to fire a rocket to the Moon. For instance, the question ignores the gravitational influence of other bodies, such as the Moon and the Sun. However, it provides a good starting point for determining the energy. Variables energy of circular orbit energy of elliptical orbit E c E e radius of circular orbit Earth-Moon distance semimajor axis of ellipse mass of Earth mass of rocket What is the strategy? 1. Determine the energy of the circular orbit around the Earth. 2. Using the dimensions of the elliptical transfer orbit to the Moon, determine the semimajor axis of the elliptical orbit. A diagram useful for accomplishing this is supplied below. 3. Use the semimajor axis to determine the energy of the elliptical transfer orbit. 4. The energy required by the rocket is the energy of the elliptical orbit minus the energy of the circular orbit. Physics principles and equations The total energy of a circular orbit is r min = m r max = m a M = kg m = kg gravitational constant G = N m 2 /kg Copyright Kinetic Books Co. Chapter 13