Gravity James H Dann, Ph.D. Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required)
To access a customizable version of this book, as well as other interactive content, visit www.ck12.org AUTHOR James H Dann, Ph.D. CONTRIBUTORS Chris Addiego Antonio De Jesus López CK-12 Foundation is a non-profit organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform. Copyright 2014 CK-12 Foundation, www.ck12.org The names CK-12 and CK12 and associated logos and the terms FlexBook and FlexBook Platform (collectively CK-12 Marks ) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the CC License ), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 26, 2014
www.ck12.org Chapter 1. Gravity CHAPTER 1 Gravity CHAPTER OUTLINE 1.1 Universal Law of Gravity 1.2 Gravity and Space Problems Introduction The Big Idea Over 2500 years ago Aristotle proposed two laws of physics governing motion. One for Earthly bodies (objects on Earth) that states objects naturally go in straight lines and one for Heavenly bodies (objects in space) that states objects naturally go in circles. This idea held steady for 2,000 years until Isaac Newton, in a triumph of brilliance, declared that there is one law of physics that governs motion and he unified earthly bodies and heavenly bodies with the The Universal Law of Gravitation. Newton used the universal law of gravity to prove that the moon is accelerating towards Earth just as a body on Earth is falling towards it. The acceleration of both the moon and the object on Earth is calculated using the same formula. This theory is so well regarded that it has been used to hypothesize the existence of black holes and dark matter, later verified by observation. The search to unify the laws of physics into one theory continues today and is often referred to as the yet undiscovered Grand Unified Theory (GUT). 1
1.1. Universal Law of Gravity www.ck12.org 1.1 Universal Law of Gravity Use Newton s Universal Law of Gravity equation to solve problems. Explain how the acceleration of gravity on Earth is calculated. Students will learn to use Newton s Universal Law of Gravity equation to solve problems. They will also learn how the acceleration of gravity on Earth is calculated and a bit about gravitational fields in general. Key Equations F G = Gm 1m 2 ; the force of gravity between an object with mass m r 2 1 and another object of mass m 2 and a distance between them of r. G = 6.67 10 11 Nm 2 /kg 2 ; the universal constant of gravity g = Gm ; gravitational field strength or gravitational acceleration of a planet with mass m and radius r. Note r 2 that this is not really a separate equation but comes from Newton s second law and the law of universal gravitation. Some data needed for the problems: The radius of Earth is 6.4 10 6 m The mass of Earth is about 6.0 10 24 kg The mass of Sun is about 2.0 10 30 kg The Earth-Sun distance is about 1.5 10 11 m The Earth-Moon distance is about 3.8 10 8 m Guidance When using the Universal Law of Gravity formula and the constant G above, make sure to use units of meters and kilograms. The direction of the force of gravity is in a straight line between two objects. It is always attractive. Newton invented calculus in order to prove that for a spherical object (like Earth) one can assume all of its mass is at the center of the sphere (thus in his formula, one can use the radius of Earth for the distance between a falling rock and Earth). Newton s Laws apply to all forces; but when he developed them only one was known: gravity. Newton s major insight and one of the greatest in the history of science was that the same force that causes objects to fall when released is also responsible for keeping the planets in orbit. Universal Gravity Any two objects in the universe, with masses m 1 and m 2 with their centers of mass at a distance r apart will experience a force of mutual attraction along the line joining their centers of mass equal to: F G = Gm 1m 2 r 2 Universal Gravitation, 2
www.ck12.org Chapter 1. Gravity where G is the Gravitational constant: G = 6.67300 10 11 m 3 kg 1 s 2 Here is an illustration of this law for two objects, for instance the earth and the sun: Gravity on the Earth s Surface On the surface of a planet such as earth the r in formula [3] is very close to the radius of the planet, since a planet s center of mass is usually at its center. It also does not vary by much: for instance, the earth s radius is about 6,000 km, while the heights we consider for this book are on the order of at most a few kilometers so we can say that for objects near the surface of the earth, the r in formula [3] is constant and equal to the earth s radius. This allows us to say that gravity is more or less constant on the surface of the earth. Here s an illustration: For any object a height h above the surface of the earth, the force of gravity may be expressed as: Now we make the approximation that F G = Gm earthm ob j (r earth + h) 2 r earth + h r earth 3
1.1. Universal Law of Gravity www.ck12.org then, we can rewrite the force of gravity equation as F G = Gm earth rearth 2 m ob j = m ob j g } {{ } g earth We can do this because the quantity in braces only has constants; we can combine them and call their product g. Remember, this is an approximation that holds only when the r in formula [3] is more or less constant. We call the quantity mg an object s weight. Unlike an object s mass, an object s weight can change and depends on the gravitational force it experiences. In fact, an object s weight is the magnitude of the gravitational force on it. To find the weight of an object on another planet, star, or moon, use the appropriate values in the formula for the force of gravity. Simulation Gravity and Orbits (PhET Simulation) Time for Practice 4 1. Which is greater the gravitational force that the Sun exerts on the moon, or the force the Earth exerts on the moon? Does the moon orbit the Earth or the Sun? Explain. 2. Use Newton s Law of Universal Gravitation to explain why even though Jupiter has 300 times the mass of the earth, on the surface of Jupiter you d weigh only 3 times what you weigh on earth. What other factor has to be considered here? 3. Prove g is approximately 10 m/s 2 on Earth by following these steps: a. Calculate the force of gravity between a falling object (for example an apple) and that of Earth. Use the symbol m o to represent the mass of the falling object. b. Now divide that force by the object s mass to find the acceleration g of the object. c. Calculate the force of gravity between the Sun and the Earth. (sun mass = 2.0 10 30 kg; average distance from sun to earth = 150 million km)
www.ck12.org Chapter 1. Gravity 4. Calculate the gravitational force that your pencil or pen pulls on you. Use the center of your chest as the center of mass (and thus the mark for the distance measurement) and estimate all masses and distances. a. If there were no other forces present, what would your acceleration be towards your pencil? Is this a large or small acceleration? b. Why, in fact, doesn t your pencil accelerate towards you? 5. Mo and Jo have been traveling through the galaxy for eons when they arrive at the planet Remulak. Wanting to measure the gravitational field strength of the planet they drop Mo s lava lamp from the top deck of their spacecraft, collecting the velocity-time data shown below. TABLE 1.1: velocity (m/s) time (s) 0 0 3.4 1.0 7.0 2.0 9.8 3.0 14.0 4.0 17.1 5.0 (a) Plot a velocity-time graph using the axes above. Put numbers, labels and units on your axes. Then draw a best-fit line (use a ruler) and use that line to find the gravitational field strength of Remulak. Explain below how you did that. (b) Mo and Jo go exploring and drop a rock into a deep canyon it hits the ground in 8.4 s. How deep is the canyon? (c) If the rock has a mass of 25 g and makes a hole in the ground 1.3 cm deep, what force does the ground exert to bring it to a stop? (d) Mo and Jo observe the shadows of their lava lamps at different positions on the planet and determine (a la Eratosthenes, the Greek astronomer, around 200 B.C.) that the radius of Remulak is 4500 km. Use that and your result for g to find the mass of Remulak. 5
1.1. Universal Law of Gravity www.ck12.org Answer to 5 5a. 3.4 m/s 2 5b. 120 m 5c. 800 N 5d. 10 23 kg 6
www.ck12.org Chapter 1. Gravity 1.2 Gravity and Space Problems Apply the universal law of gravity equation in order to determine the parameters of orbits. Students will learn to apply the universal law of gravity equation in order to determine the parameters of orbits. Key Equations v = 2πr T for a particle travels a distance 2πr in an amount of time T F G = Gm 1m 2 r 2 a c = v2 r Guidance When can Gravity Act as a Centripetal Force? We saw last chapter that the force of Gravity causes an attraction between two objects of mass m 1 and m 2 at a distance r of F G = Gm 1m 2 r 2. [4] By Newton s Third Law, both objects experience the force: equal in magnitude and opposite in direction, and both will move as an effect of it. If one of the objects is much lighter than the other (like the earth is to the sun, or a satellite is to earth) we can approximate the situation by saying that the heavier mass (the sun) does not move, since its acceleration will be far smaller due to its large mass. Then, if the lighter mass remains at a relatively constant absolute distance from the heavier one (remember, centripetal force needs to be constant in magnitude), we can say that the lighter mass experiences an effectively centripetal force and thus has a centripetal acceleration. Math of Centripetal Gravity Gravity is not always a centripetal force. This is a really important point. It only acts as a centripetal force when conditions approximate those listed above very much like it isn t constant near the surface of the earth, but very close to it. If gravity provides centripetal force and acceleration, we can set [2] equal to [4]. It s important to remember that in [2] m refers to the lighter mass, since that is the one traveling. Then, G m 1 m 2 r r 2 = m 1v2 r So, the relationship between velocity and radius for a circular orbit of a light object around an heavy mass (note the mass of the lighter object cancels) is: Gm 2 = v orb 2 r orb 7
1.2. Gravity and Space Problems www.ck12.org Geosynchronous orbit This is the orbit where a satellite completes one orbit of the Earth every 24 hours, staying above the same spot (longitude) on Earth. This is a very important orbit for spy satellites and TV satellites among others. You force the speed of the satellite to be a value such that the satellite makes one rotation every 24 hours. Example 1 Watch this Explanation MEDIA Click image to the left for more content. Time for Practice 1. A digital TV satellite is placed in geosynchronous orbit around Earth, so it is always in the same spot in the sky. a. Using the fact that the satellite will have the same period of revolution as Earth, calculate the radius of its orbit. b. What is the ratio of the radius of this orbit to the radius of the Earth? c. Draw a sketch, to scale, of the Earth and the orbit of this digital TV satellite. d. If the mass of the satellite were to double, would the radius of the satellite s orbit be larger, smaller, or the same? Why? 8 2. A top secret spy satellite is designed to orbit the Earth twice each day (i.e., twice as fast as the Earth s rotation). What is the height of this orbit above the Earth s surface? 3. Two stars with masses 3.00 10 31 kg and 7.00 10 30 kg are orbiting each other under the influence of each other s gravity. We want to send a satellite in between them to study their behavior. However, the satellite needs to be at a point where the gravitational forces from the two stars are equal. The distance between the two stars is 2.0 10 10 m. Find the distance from the more massive star to where the satellite should be placed. ( Hint: Distance from the satellite to one of the stars is the variable.) 4. Calculate the mass of the Earth using only: (i) Newton s Universal Law of Gravity; (ii) the Moon-Earth distance (Appendix B); and (iii) the fact that it takes the Moon 27 days to orbit the Earth.
www.ck12.org Chapter 1. Gravity 5. 6. A space station was established far from the gravitational field of Earth. Extended stays in zero gravity are not healthy for human beings. Thus, for the comfort of the astronauts, the station is rotated so that the astronauts feel there is an internal gravity. The rotation speed is such that the apparent acceleration of gravity is 9.8 m/s 2. The direction of rotation is counter-clockwise. a. If the radius of the station is 80 m, what is its rotational speed, v? b. Draw vectors representing the astronaut s velocity and acceleration. c. Draw a free body diagram for the astronaut. d. Is the astronaut exerting a force on the space station? If so, calculate its magnitude. Her mass m = 65 kg. e. The astronaut drops a ball, which appears to accelerate to the floor, (see picture) at 9.8 m/s 2. a. Draw the velocity and acceleration vectors for the ball while it is in the air. b. What force(s) are acting on the ball while it is in the air? c. Draw the acceleration and velocity vectors after the ball hits the floor and comes to rest. d. What force(s) act on the ball after it hits the ground? Answers to Selected Problems 1. a. 4.23 10 7 m b. 6.6 R e d. The same, the radius is independent of mass 2. 1.9 10 7 m 3. You get two answers for r, one is outside of the two stars one is between them, that s the one you want, 1.32 10 10 m from the larger star. 9
1.2. Gravity and Space Problems www.ck12.org 4.. 5. a. v = 28 m/s b. v down, a right c. f right d. Yes, 640N Summary In this chapter students will learn about the Universal Law of Gravity and how to solve gravitational problems in space. 10