PHY 102 1st EditionOutline of Last LectureI. Bullet Problema. Kinematicsb. Conservation of EnergyII. Bullet in block of wooda. Conservation of EnergyIII. Newton’s Applea. Derivations of formulasb. Acceleration toward earthIV. Newton Derives Kepler’s Third LawV. Dark MatterVI. Relationship of Weight v. MassOutline of Current Lecture I. Gravitya. EquationsII. Orbital VelocityIII. Escape VelocityIV. ExamplesV. Sample ProblemsCurrent LectureThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.I. Gravitya. Weight = m * force due to gravityb. F = GMm/r2i. Force = Universal Gravitational Constant * Mass of the earth * mass of theobject pulled by earth/distance between the two objects 2c. G does not equal g (force due to gravity)d. g = GM/r2 because masses cancel out in the equatione. acceleration due to gravity is independent of the massII. Orbital Velocitya. Fg = GMm/r2b. Fc = mv2/rc. In orbital motion, the Force due to gravity = forcec which is mv2/ri. So GM/r2 = v2/rii. When reduced, vorbit = √GM / rIII. Escape Velocitya. Kinetic energy + potential energyi. This is equal to ½ mv2 – GMm/rii. Etotal = 0 for v = 0 at infinityiii. ½ mv2 = GMm/rb. Vescape = √2GM rIV. Examplesa. Force between two objects: Fg = GMm/r2b. Orbital Velocity: vorbit = √GMrc. Escape velocity: vescape = √2 GMrd. –G = 6.67x10-11Nm2/kg2e. –M = 5.97x1024kgf. –r = 6,371,000mV. Samplesa. What is the gravitational force between a proton and a n electron in a Hydrogen atom?mp = 1.673x10-27kgme = 9.109x10-31kg-ro = 5.29x10-11m-G = 6.67x10–11Nm2/kg2F = Gmemp/ro2Plug & Chug from this pointb. What is the orbital velocity of the International Space Station (ISS) whose altitudeis 330km? Take the radius of Earth to be 6378.1km. The mass of Earth is
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