Physics 12. Unit 5 Circular Motion and Gravitation Part 2
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1 Physics 12 Unit 5 Circular Motion and Gravitation Part 2
2 1. Newton s law of gravitation We have seen in Physics 11 that the force acting on an object due to gravity is given by a well known formula: F g = mg The quantity m is the inertial mass of the object. The quantity g is called the gravitational field strength, and its value is approximately 9.8 m/s 2 when close to the surface of the Earth. This formula tells us that the heavier an object is, the stronger the downward pull it will experience from the Earth. But it works only for the case of the gravity on Earth. Unit 5 - Circular Motion and Gravitation (Part 2) 2
3 In fact, gravitational attraction exists between any pair of objects. It was Newton who came up with the following conclusions based on extensive experimental observations: 1. Mass attracts mass, and the strength of attraction is directly proportional to the size of each mass: F g M 1 M 2 2. The size of force is inversely proportional to the square of the distance between the centers of mass of the two objects: F g 1 R 2 Unit 5 - Circular Motion and Gravitation (Part 2) 3
4 Combining these two Newton derived the following equation, called Newton s law of gravitation, that describes the gravitational attraction between two masses: F g = GM 1M 2 R 2 The coefficient G is the universal gravitational constant with the value of Nm 2 /kg 2. This value is the same for all kinds of masses! Note that this force is always attractive, meaning that it is pointing toward one another. Unit 5 - Circular Motion and Gravitation (Part 2) 4
5 Example: Determine the force of attraction between a 35 kg dog and a 7.6 kg cat, watching each other from a distance of 4.8 m. Unit 5 - Circular Motion and Gravitation (Part 2) 5
6 Example: Given the mass of Earth is kg. a. Find the weight of a 50 kg person on Earth using F g = mg. b. Find the weight of the same person on Earth using the Newton s law of gravitation. c. Find the weight of this person at an altitude of 170 km. Given the radius of Earth is m. Unit 5 - Circular Motion and Gravitation (Part 2) 6
7 The Cavendish experiment This is the experiment performed by a British scientist Henry Cavendish in to determine the density of the Earth. Using his results, scientists were able to determine the value of the universal gravitational constant to be Nm 2 /kg 2. This value is within 1% of the most recent value of G! Unit 5 - Circular Motion and Gravitation (Part 2) 7
8 2. Gravity in space Recall that the well-known formula for gravity, F = mg, where the gravitational acceleration is g = 9.8 m/s 2, works only on Earth s surface. This quantity tells how strongly the gravitational field due to the Earth is acting on every kg of an object when it is on Earth. This field strength would be different if an object is placed anywhere else. To find the field strength, we connect the two equations for F g together: F g = mg = GMm R 2 Unit 5 - Circular Motion and Gravitation (Part 2) 8
9 Cancelling the mass term of the object (i.e., m) gives: g = GM R 2 The gravitation field strength depends on two factors: (1) The mass M This is the mass of the object that exerts the gravitational field. (2) The distance R This is the distance between the centers of mass of the two objects. Unit 5 - Circular Motion and Gravitation (Part 2) 9
10 Example: Find the gravitational field strength at the following altitudes: 1 m, 1 km, 10 km, 100 km. Discuss the trend. Unit 5 - Circular Motion and Gravitation (Part 2) 10
11 Note that gravitational attraction between any pair of objects. Therefore, when an object is placed among many others, not only does it experience the gravity from the closest neighbor, but all the others, either big or small. The net gravitational force is determined by the vector sum of these individual forces. Unit 5 - Circular Motion and Gravitation (Part 2) 11
12 Example: Determine the net force acting on Planet B by the two other planets, as illustrated below. Unit 5 - Circular Motion and Gravitation (Part 2) 12
13 Example: Consider the following two-planet system: The above situation is now changed as follows; find the new gravitational force in each case. 1. The mass m is doubled. 2. The mass M is tripled and the mass m is halved. 3. The mass M is one-tenth, and the distance R is tripled. Unit 5 - Circular Motion and Gravitation (Part 2) 13
14 3. Centripetal motion and gravity Consider a satellite that goes around the Earth. It travels a straight path when it is moving. In the meantime, it experiences the gravitational force exerted by the Earth. The result is that it falls constantly. However, remember that Earth is a nearly spherical object, and the gravitational force is always pointing towards its center. Consequently, the satellite is moving under the influence of a centripetal force, and its path is circular (in approximation). That means, F c = F g Unit 5 - Circular Motion and Gravitation (Part 2) 14
15 A stable path of a satellite looks like the following: Unit 5 - Circular Motion and Gravitation (Part 2) 15
16 We can work out the speed at which a satellite is moving stably in an orbit as follows: Since F c = F g, we have Cancelling m and R gives mv 2 R = GMm R 2 v 2 = GM R Hence v = GM R Unit 5 - Circular Motion and Gravitation (Part 2) 16
17 This velocity is called the first cosmic velocity. For the Earth, its value is about v = = m/s At the altitude of medium Earth orbit (about km), the first cosmic velocity is reduced to m/s. (You can try to verify this value yourself.) Unit 5 - Circular Motion and Gravitation (Part 2) 17
18 Example: Determine the stable parking orbit velocity for a surveying satellite located 230 km above the Moon s surface. If that orbital radius is reduced by one-tenth, by what factor would the orbiting speed increase? Unit 5 - Circular Motion and Gravitation (Part 2) 18
19 4. Kepler s laws of planetary motion About 80 years before Newton published his work on classical mechanics in his famous book, Philosophiæ Naturalis Principia Mathematica, the German astronomer Johannes Kepler ( ) had proposed three empirical laws that describe the motion of planets around the Sun. These three laws were developed based on the data collected by another scientist Tycho Brahe ( ). Unit 5 - Circular Motion and Gravitation (Part 2) 19
20 The Kepler s first law (the law of orbit) This law states that the path of each planet around the Sun is an ellipse with the Sun at one focus. F 1 P + F 2 P = constant Perihelion Aphelion Foci Unit 5 - Circular Motion and Gravitation (Part 2) 20
21 The Kepler s second law (the law of area) The line joining the planet to the Sun sweeps out equal areas in equal intervals of time. This observation implies that the planet is fastest when closest to the Sun (i.e., at perihelion, 1 2) while it is slowest when farthest from the Sun (i.e., at aphelion, 3 4). Unit 5 - Circular Motion and Gravitation (Part 2) 21
22 The Kepler s third law (the law of period) It states that for a given central mass, if R is the average orbital radius, and T is the orbital period, then T 2 R 3 It means that T 2 R 3 = constant This constant, called the Kepler s constant, is the same for all planets in the Solar system (where the central mass is the mass of the Sun). Unit 5 - Circular Motion and Gravitation (Part 2) 22
23 These three laws can be derived based on Newtonian mechanics using the concept of conservation of angular momentum. We will skip this part, and will only see how the third law can be developed. Newton s synthesis Assume that the orbit of a planet of mass m around another one of mass M is very close to a circular path. In this case, F g = F c. Hence, Simplifying: GMm R 2 = 4π2 Rm T 2 T 2 R 3 = 4π2 GM Unit 5 - Circular Motion and Gravitation (Part 2) 23
24 Recall that the Kepler s constant is the ratio of T 2 to R 3. Thus, K = 4π2 GM The observed Kepler s constants for the planets in the Solar system: Planet Kepler s constant Planet Kepler s constant Mercury Jupiter Venus Saturn Earth Uranus Mars Neptune Unit 5 - Circular Motion and Gravitation (Part 2) 24
25 Based on the Kepler s third law, the orbital period of a satellite depends only on the mass of the planet and the orbital radius of the satellite. It is therefore possible for a certain satellite, when launched to an appropriate altitude, to orbit around the Earth with the same period as the rotation of the Earth. Such an orbit is called a geosynchronous orbit or geostationary orbit. Unit 5 - Circular Motion and Gravitation (Part 2) 25
26 Note that R 3 = GMT2 4π 2 It gives the geosynchronous orbital radius: R = 3 GMT 2 4π 2 Unit 5 - Circular Motion and Gravitation (Part 2) 26
27 Example: Find the orbital radius and the speed of a satellite that is geostationary above Earth s equator. Solution: Given M = kg and T = s, R = π 2 = m Since the radius of Earth is m, the altitude of this satellite is: h = = m It is about 36,000 km above the Earth s surface. Unit 5 - Circular Motion and Gravitation (Part 2) 27
28 Its orbital speed can be calculated by v = GM R Since R = m, v = = Its speed is km/s which is about 40% of the first cosmic velocity of Earth. Unit 5 - Circular Motion and Gravitation (Part 2) 28
29 5. Gravitational potential energy In physics 11, we learned that in order to move an object up by an altitude of h m, we need to do a work that corresponds to the change of its potential energy: W = E p = mg h This equation works well near the Earth s surface, but is not valid when applied to space objects because g is no longer a constant. Recall that the Earth s gravitational field strength is given by g = GM R 2 Unit 5 - Circular Motion and Gravitation (Part 2) 29
30 Substituting this into the potential energy expression yields: E p = m GM R 2 h = GMm R Note that gravitational force is always attractive; therefore to move one mass away from the other, energy is required. When two masses are infinitely far away, they don t feel one another, and E p = 0. Because of this, a negative sign is added to the expression above to give the gravitational potential energy for two objects separated by a distance R: E p = GMm R Unit 5 - Circular Motion and Gravitation (Part 2) 30
31 The gravitational potential energy formula can also be derived using calculus. Consider the following F g versus R graph: The work done required to move an object from R to is the area under the curve. W = E p = න R F g dr = GMm r ቤ = E p E p R = GMm R Since E p = 0, E p R < 0 R Unit 5 - Circular Motion and Gravitation (Part 2) 31
32 This formula is used to find the potential energy of an object of mass m in space at a position R relative to: 1. The central planetary mass M. 2. E p = 0 at R =. Unit 5 - Circular Motion and Gravitation (Part 2) 32
33 Example: A 2500 kg satellite is in orbit m above the Earth s surface. What is the gravitational potential energy of the satellite due to the gravitational force due to the Earth? What is the total energy of the satellite? Unit 5 - Circular Motion and Gravitation (Part 2) 33
34 Example: Find the potential energy (relative to infinity) of a 50 kg person flying at an altitude of 10 km above Earth s surface. If the same person is now in a space shuttle orbiting at an altitude of 250 km, what would be his new potential energy relative to infinity? Unit 5 - Circular Motion and Gravitation (Part 2) 34
35 If an object is moved from one altitude to another under the influence of the Earth s gravitational field: The necessary work done will be given by W = E p,f E p,i = GMm R f GMm R i = GMm 1 R i 1 R f Unit 5 - Circular Motion and Gravitation (Part 2) 35
36 Example: Determine the work done to move a 35,000 kg cargo of space junk from an altitude of 430 km above the Moon s surface to a radial distance of 2800 km away. Unit 5 - Circular Motion and Gravitation (Part 2) 36
37 Example: The International Space Station drops a 250 kg waste shuttle from an altitude of 350 km. At what speed would it impact Earth if there were no air resistance? (Assume it starts at rest.) Unit 5 - Circular Motion and Gravitation (Part 2) 37
38 6. Conservation of energy in space Just like all the systems that we have studied, systems in space obey the law of conservation of energy because energy can neither be created or destroyed. Objects in space are under the combined influence of the gravitational fields of planets, galaxies, etc. Therefore, they have gravitational potential energy E p. In the meantime, they may have kinetic energy E k if they are moving relative to stars, planets or galaxies. It is assumed that space is empty, and therefore it is unlikely to have heat loss due to friction. Unit 5 - Circular Motion and Gravitation (Part 2) 38
39 Because of these, the conservation of energy yields the following relationship for space objects: E t = E p + E k Explicitly, it is written as E t = GMm R mv2 Unit 5 - Circular Motion and Gravitation (Part 2) 39
40 Example: A 12,000 kg spaceship is m from the center of a planet that has a mass of kg. As it falls back to the planet s surface, the spaceship gains J of kinetic energy. What is the radius of the planet? Unit 5 - Circular Motion and Gravitation (Part 2) 40
41 Due to the conservation of energy, when an object moves away from a planet s gravitational field, its kinetic energy is converted to its potential energy. If the object has the kinetic energy sufficient to compensate for the cost of potential energy, it will be able to overcome the gravitational attraction fully and escape from the planet. In this case, E k = E p. That means 1 2 mv2 = GMm R The minimum velocity required is called the escape velocity (or second cosmic velocity): v = 2GM R Unit 5 - Circular Motion and Gravitation (Part 2) 41
42 Relationship between first and second cosmic velocities: Unit 5 - Circular Motion and Gravitation (Part 2) 42
43 For Earth, the escape velocity is v = = m/s That means, for an object to escape from Earth, it has to be launched at 11.2 km/s at least! Note that this value is independent of the mass of the object. Unit 5 - Circular Motion and Gravitation (Part 2) 43
44 Example: Determine the escape velocity for any spacecraft launched from the surface of Mars, which has a planetary mass of kg and a radius of m. Unit 5 - Circular Motion and Gravitation (Part 2) 44
45 When a satellite is rotating around a planet in a stable orbit, its total energy is given by E t = E p + E k The kinetic energy can be determined as follows: mv 2 R F c = F g = GMm R 2 mv 2 = GMm R E k = 1 2 mv2 = 1 GMm 2 R Unit 5 - Circular Motion and Gravitation (Part 2) 45
46 The potential energy is given by definition: Hence, the total energy is E p = GMm R E t = 1 2 GMm R GMm R = 1 2 GMm R That means the total energy possessed by any object in a stable orbit is always equal to half of the potential energy contained in that object. Unit 5 - Circular Motion and Gravitation (Part 2) 46
47 Unit 5 - Circular Motion and Gravitation (Part 2) 47
48 Unit 5 - Circular Motion and Gravitation (Part 2) 48
49 Unit 5 - Circular Motion and Gravitation (Part 2) 49
50 Unit 5 - Circular Motion and Gravitation (Part 2) 50
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