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Chapter 13 Outline Gravitation • Newton’s law of gravitation • Weight • Gravitational potential energy • Circular orbits • Kepler’s laws • Black holes Gravitation • Gravity is the main (somewhat understood) driving force in the universe. • All masses attract each other. • Newton’s law of gravitation (1687) 𝐺𝑚1 𝑚2 𝐹g = 𝑟2 Gravitational Force • Acts on both bodies with equal magnitude. • When you jump off a diving board, Earth also accelerates towards you, but your mass is 23 orders of magnitude smaller, so this is negligible. • Spherically symmetric masses look like points from the outside. • Replace sphere with a single point mass at the center. Gravitational Constant, 𝐺 • Sir Henry Cavendish used a torsion balance to measure 𝐺 in 1798. • Very difficult to precisely measure 𝐺 = 6.67 × 10−11 N ∙ m2 /kg 2 Weight • So far, we have used 𝑤 = 𝑚𝑔. • While the definition of weight is the total gravitational force exerted on a body, on the surface of a planet (or moon, comet…) we can neglect the forces from everything but that planet. • At the surface of Earth, the weight of a mass, 𝑚, is: 𝐺𝑚E 𝑚 𝑤 = 𝐹g = 𝑅E2 • So, 𝐺𝑚E 𝑔= 2 𝑅E Mass of the Earth • If we measure the local acceleration due to gravity, we should be able to calculate the mass of the earth: 𝑔𝑅E2 9.81 m/s 2 6.37 × 106 kg 2 𝑚E = ≈ 𝐺 6.67 × 10−11 N ∙ m2 /kg 2 𝑚E ≈ 5.97 × 1024 kg • This matches very well with the accepted value of 𝑚E = 5.972 × 1024 kg Density of the Earth • Knowing the mass and radius of the Earth, we can easily calculate an average density. 𝑀E 𝜌E−ave = 4 3 = 5.51 g/cm3 3𝜋𝑅E • But, the rocks we find on the surface generally have densities of about 2 to 3.3 g/cm3 . • The Earth must not have uniform density. • The density near the core is larger than near the surface. Exam #3 Results • Average 10.4 • Standard deviation 3.1 Exam #3 Grades After T-score Adjustment 45 40 35 Frequency 30 25 20 15 10 5 0 50 62 68 72 75 78 82 Grades 85 88 92 95 100 More Gravitational Potential Energy • Previously, we found that the gravitational potential energy was given by: 𝑈g = 𝑚𝑔ℎ • But, this assumed that 𝑔 was constant. • On the surface of a planet, this is a good approximation. • In space, it is not! • Recall our definition of potential energy: ∆𝑈 = −𝑊 = − 𝑭 ∙ 𝑑𝒍 Gravitational Potential Energy • As before, we need to set a zero for the potential energy. • Now, the convenient zero point is infinity. • The potential decreases as the masses get closer together. • Keep in mind that this is the potential energy of the system. 𝐺𝑀𝑚 𝑈g = − 𝑟 Gravitational Potential Energy on Earth • Near the surface of Earth, this expression should reduce to the familiar ∆𝑈 = 𝑚𝑔∆ℎ. • Moving from position 𝑟1 to 𝑟2 , ∆ℎ = 𝑟2 − 𝑟1 1 1 𝐺𝑀𝑚 𝑟2 − 𝑟1 ∆𝑈 = −𝐺𝑀𝑚 − = 𝑟2 𝑟1 𝑟1 𝑟2 • For Earth, 𝑀 = 𝑚E , 𝑟1 𝑟2 ≈ 𝑅E2 , and 𝑔 = 𝐺𝑚E 2 , 𝑅𝐸 so 𝐺𝑚𝐸 ∆𝑈 = 𝑚 2 𝑟2 − 𝑟1 = 𝑚𝑔∆ℎ 𝑅𝐸 Escape Velocity • If you throw a ball up into the air, what happens? • It reaches some maximum height, and then comes back down. • We find this height using conservation of energy. • What if you threw it with an extremely large initial velocity? • If the initial kinetic energy is greater than or equal to the magnitude of the potential energy (with respect to infinity), the ball will not fall down. • When equal, 𝐺𝑀𝑚 1 𝑈1 + 𝐾1 = − + 𝑚𝑣 2 = 0 𝑟 2 𝑣escape = 2𝐺𝑀 𝑟 Motion of Satellites • If a projectile is launch horizontally from a great height, where will it land? • Depends on the initial velocity. • For large enough 𝑣0 , the curvature of the Earth is important • Ignoring air resistance, it may never land. • In an orbit the object keeps falling towards the Earth, but never reaches it. • Closed orbit – ellipse • Open orbit – doesn’t return Circular Orbits • First, we will consider the simplest orbit, a circular orbit. • Satellite mass 𝑚; planet mass 𝑀 • Uniform circular motion: 𝑣2 𝑎rad = 𝑟 • Force provided by gravity: 𝐺𝑀𝑚 𝐹g = 2 𝑟 • Newton’s second law 𝐺𝑀𝑚 𝑣2 =𝑚 2 𝑟 𝑟 Circular Orbital Speed • Solving for the speed, 𝑣circ = 𝐺𝑀 𝑟 • Two things to note: • The mass of the satellite does not matter. • The speed and the radius are linked. (Larger radius, smaller speed) Circular Orbital Period, 𝑇 • The time it takes for a full orbit, or the period, 𝑇, can be found from the speed. 2𝜋𝑟 𝑣circ = 𝑇circ 𝑟 𝑇circ = 2𝜋𝑟 𝐺𝑀 𝑇circ = 2𝜋𝑟 3/2 𝐺𝑀 Total Energy in Circular Orbit • The total energy (potential and kinetic) of the satellite is: 𝐺𝑀𝑚 1 𝐸 =𝑈+𝐾 =− + 𝑚𝑣 2 𝑟 2 • Using 𝑣circ = 𝐺𝑀 , 𝑟 𝐺𝑀𝑚 1 𝐺𝑀 𝐸=− + 𝑚 𝑟 2 𝑟 𝐺𝑀𝑚 𝐸=− 2𝑟 • The total energy is negative (bound system) and equal to one half the potential energy. • This is an example of the virial theorem. Kepler’s Laws • In the 16th century, Nicolaus Copernicus deduced that the Earth is a planet, and, like all planets, it orbits the Sun. • In the early 17th century, Johannes Kepler used Tycho Brahe’s observations of planetary motions to determine three empirical laws of planetary orbits. 1. Each planet moves in an elliptical orbit, with the sun at one focus of the ellipse. 2. A line from the sun to a given planet sweeps out equal areas in equal times. 3. The periods of the planets are proportional to the 32 powers of the major axis lengths of their orbits. • Newton later discovered the reasons for these laws. Kepler’s First Law • Elliptical orbits • Focal points, 𝑆 and 𝑆′ • Semi-major axis, 𝑎 • Eccentricity, 𝑒, between zero (circle) and one. • The sum of the distances 𝑆𝑃 and 𝑆 ′ 𝑃 is constant. • In our solar system, most orbits are fairly circular • Earth: 𝑒 = 0.017 • Most eccentric is Mercury: 𝑒 = 0.206 Kepler’s Second Law • The rate at which area is swept out is called the sector velocity. 𝑑𝐴 1 2 𝑑𝜃 = 𝑟 𝑑𝑡 2 𝑑𝑡 • Triangle area: 𝐴 = 12𝑏ℎ • The perpendicular component of 𝒗 is 𝑣 sin 𝜙. • The rate at which 𝜃 is changing is =𝜔= 𝑣 sin 𝜙 r 𝑑𝐴 1 = 𝑟𝑣 sin 𝜙 𝑑𝑡 2 𝑑𝜃 𝑑𝑡 Kepler’s Second Law • Recall that angular momentum is 𝑳 = 𝒓 × 𝑚𝒗. • 𝐿 = 𝑚𝑟𝑣 sin 𝜙 • So, the sector velocity is simply 𝑑𝐴 𝐿 = 𝑑𝑡 2𝑚 • Kepler’s second law is a natural consequence of the conservation of angular momentum. • There can be no torque because the force is always central (on a line between the masses.) Kepler’s Third Law • For a circular orbit, we already showed that the period was proportional to the 32 power of the radius. • This is also true of an elliptical orbit, but the radius is replaced by the semi-major axis, 𝑎. 𝑇= 2𝜋𝑎3/2 𝐺𝑀 Escape Velocity Revisited • We already calculated the escape velocity at the surface of a planet (or any spherical object). 𝑣escape = 2𝐺𝑀 𝑟 • If the sphere becomes more and more dense, the mass will remain the same, but the radius decreases, and the escape velocity increases. • At what point does this become an issue? Black Holes • The maximum speed for anything is the speed of light, 𝑐 = 3.00 × 108 m/s. • If the escape velocity exceeds this, even light cannot escape. • This is a black hole. Schwartzschild Radius • For a given mass, what will the radius have to be to form a black hole? 𝑐= 2𝐺𝑀 𝑅S 2𝐺𝑀 𝑅S = 2 𝑐 • This is the Schwartzschild radius, and it defines the event horizon. • Note: To properly derive this, we need to use general relativity, but the result is the same. What if the Sun collapsed into a black hole? • What if the Sun collapsed into a black hole? • Would the Earth’s orbit be affected? • What would the radius of the event horizon be? 2𝐺𝑚Sun 𝑅S = ≈ 3 km 2 𝑐 • The Sun’s radius now is 6.96 × 108 m.