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# Physics 201: Lecture 24, Pg 1 Chapter 13 The beautiful rings of Saturn consist of countless centimeter-sized ice crystals, all orbiting the planet under.

## Presentation on theme: "Physics 201: Lecture 24, Pg 1 Chapter 13 The beautiful rings of Saturn consist of countless centimeter-sized ice crystals, all orbiting the planet under."— Presentation transcript:

Physics 201: Lecture 24, Pg 1 Chapter 13 The beautiful rings of Saturn consist of countless centimeter-sized ice crystals, all orbiting the planet under the influence of gravity. Goal: To use Newton’s theory of gravity to understand the motion of satellites and planets.

Physics 201: Lecture 24, Pg 2 Who discovered the basic laws of planetary orbits? A. Newton B. Kepler C. Faraday D. Einstein E. Copernicus

Physics 201: Lecture 24, Pg 3 What is geometric shape of a planetary or satellite orbit? A. Circle B. Hyperbola C. Sphere D. Parabola E. Ellipse

Physics 201: Lecture 24, Pg 4 The value of g at the height of the space shuttle’s orbit is A. 9.8 m/s 2. B. slightly less than 9.8 m/s 2. C. much less than 9.8 m/s 2. D. exactly zero.

Physics 201: Lecture 24, Pg 5 Newton proposed that every object in the universe attracts every other object. The Law: Any pair of objects in the Universe attract each other with a force that is proportional to the products of their masses and inversely proportional to the square of their distance Newton’s Universal “Law” of Gravity

Physics 201: Lecture 24, Pg 6 Newton’s Universal “Law” of Gravity “Big” G is the Universal Gravitational Constant G = 6.673 x 10 -11 Nm 2 / kg 2 Two 1 Kg particles, 1 meter away  F g = 6.67 x 10 -11 N (About 39 orders of magnitude weaker than the electromagnetic force.) The force points along the line connecting the two objects.

Physics 201: Lecture 24, Pg 7 Newton’s Universal “Law” of Gravity You have three 1 kg masses, 1 at the origin, 2 at (1m, 0) and 3 at (0, 1m). What would be the Net gravitational force on object 1 by objects 2 & 3? G = 6.673 x 10 -11 Nm 2 / kg 2 Two 1 Kg particles, 1 meter away  F g = 6.67 x 10 -11 N Each single force points along the line connecting two objects. F g =9.4 x 10 -11 N along diagonal

Physics 201: Lecture 24, Pg 8 Motion Under Gravitational Force l Equation of Motion l If M >> m then there are two classic solutions that depend on the initial r and v l Elliptical and hyperbolic paths

Physics 201: Lecture 24, Pg 9 Anatomy of an ellipse l a: Semimajor axis l b: Semiminor axis l c: Semi-focal length a 2 =b 2 +c 2 l Eccentricity e = c/a If e = 0, then a circle Most planetary orbits are close to circular.

Physics 201: Lecture 24, Pg 10 A little history l Kepler’s laws, as we call them today, state that 1. Planets move in elliptical orbits, with the sun at one focus of the ellipse. 2. A line drawn between the sun and a planet sweeps out equal areas during equal intervals of time. 3. The square of a planet’s orbital period is proportional to the cube of the semimajor-axis length. Focus

Physics 201: Lecture 24, Pg 11 Kepler’s 2 nd Law l The radius vector from the sun to a planet sweeps out equal areas in equal time intervals (consequence of angular momentum conservation). l Why?

Physics 201: Lecture 24, Pg 12 Conservation of angular momentum l Area of a parallelogram = base x height l L is constant (no torque)

Physics 201: Lecture 24, Pg 13 Kepler’s 3 rd Law Square of the orbital period of any planet is proportional to the cube of semimajor axis a.

Physics 201: Lecture 24, Pg 14 Suppose an object of mass m is on the surface of a planet of mass M and radius R. The local gravitational force may be written as Little g and Big G where we have used a local constant acceleration: On Earth, near sea level, it can be shown that g surface ≈ 9.8 m/s 2.

Physics 201: Lecture 24, Pg 15 Cavendish’s Experiment F = m 1 g = G m 1 m 1 / r 2 g = G m 2 / r 2 If we know big G, little g and r then will can find the mass of the Earth!!!

Physics 201: Lecture 24, Pg 16 Orbiting satellites v t = (gr) ½ Net Force: ma = mg = mv t 2 / r gr = v t 2 v t = (gr) ½ The only difference is that g is less because you are further from the Earth’s center!

Physics 201: Lecture 24, Pg 17 Geostationary orbit

Physics 201: Lecture 24, Pg 18 Geostationary orbit l The radius of the Earth is ~6000 km but at 36000 km you are ~42000 km from the center of the earth. l F gravity is proportional to 1/r 2 and so little g is now ~10 m/s 2 / 50 n v T = (0.20 * 42000000) ½ m/s = 3000 m/s At 3000 m/s, period T = 2  r / v T = 2  42000000 / 3000 sec = = 90000 sec = 90000 s/ 3600 s/hr = 24 hrs l Orbit affected by the moon and also the Earth’s mass is inhomogeneous (not perfectly geostationary) l Great for communication satellites (1 st pointed out by Arthur C. Clarke)

Physics 201: Lecture 24, Pg 19 Tuesday l All of Chapter 13

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