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Universal Gravitation

Celestial. Universal Gravitation. Terrestrial. Sir Isaac Newton 1642-1727. UNIVERSAL GRAVITATION. +F. -F. For any two masses in the universe:. F = G m 1 m 2 /r 2. G = a constant later evaluated by Cavendish. m 2. m 1. r. CAVENDISH: MEASURED G. Modern value:

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Universal Gravitation

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  1. Celestial Universal Gravitation Terrestrial Sir Isaac Newton 1642-1727

  2. UNIVERSAL GRAVITATION +F -F For any two masses in the universe: F = G m1m2/r2 G = a constant later evaluated by Cavendish m2 m1 r

  3. CAVENDISH: MEASURED G Modern value: G = 6.674*10-11 Nm2/kg2

  4. Measuring G

  5. r Two people pass in a hall. Findthe gravitational force between them. • m1 = m2 = 70 kg • r = 1 m m1 m2 F = (6.67 x 10-11 N-m2/kg2)(70 kg)(70 kg)/(1 m)2 = 3.3 x 10-7N

  6. Universal Gravitation ACT • Which of the situations shown below experiences the largest gravitational attraction?

  7. Satellite Motion The net force on the satellite is the gravitational force. Fnet= FG Assuming a circular orbit: mac = GmMe/r2 r Me m Note that the satellite mass cancels out. Using For low orbits (few hundred km up) this turns out to be about 8 km/s = 17000 mph

  8. Geosynchronous Satellite In order to remain above the same point on the surface of the earth, what must be the period of the satellite’s orbit? What orbital radius is required? T = 24 hr = 86,400 s Actually the theoretical derivation of Kepler’s Third Law Using r = 42,000 km = 26,000 mi

  9. GPS Satellites • GPS satellites are not in geosynchronous orbits; their orbit period is 12 hours. Triangulation of signals from several satellites allows precise location of objects on Earth.

  10. Value of g • The weight of an object is the gravitational force the earth exerts on the object. • Weight = GMEm/RE2 • Weight can also be expressed • Weight = mg • Combining these expressions • mg = GMEm/RE2 • RE = 6.37*106 m = 6370 km • ME = 5.97 x 1023 kg • g = GME/RE2 = 9.8 m/s2 • The value of the gravitational field strength (g) on any celestial body can be determined by using the above formula.

  11. g vs Altitude For heights that are small compared to the earth’s radius (6.37 x 106 m ~4000 mi), the acceleration of gravity decreases slowly with altitude.

  12. g vs Altitude Once the altitude becomes comparable to the radius of the Earth, the decrease in the acceleration of gravity is much larger:

  13. Apparent Weight • Apparent Weight is the normal support force. In an inertial (non-accelerating) frame of reference • FN = FG • What is the weight of a 70 kg astronaut in a satellite with an orbital radius of 1.3 x 107 m? • Weight = GMm/r2 Using: G = 6.67 x 10-11 N-m2/kg2 • and M = 5.98 x 1023 kg Weight = 16 N • What is the astronaut’s apparent weight? The astronaut is in uniform circular motion about Earth. The net force on the astronaut is the gravitational force. The normal force is 0. The astronaut’s apparent weight is 0. Springscale measures normal force Apparent Weightlessness

  14. Different distances to moon is dominant cause of earth’s tides Tides • FG by moon on A > FG by moon on B • FG by moon on B > FG by moon on C • Earth-Moon distance: 385,000 km which is about 60 earth radii • Sun also produces tides, but it is a smaller effect due to greater Earth-Sun distance. • 1.5 x 105 km High high tides; low low tides Low high tides; high low tides Neap Tides Spring Tides

  15. Other focus is the empty focus Johannes Kepler 1571-1630 Kepler’s First Law aphelion perihelion • The orbit of a planet/comet about the Sun is an ellipse with the Sun's center of mass at one focus PF1 + PF2 = 2a A comet falls into a small elliptical orbit after a “brush” with Jupiter

  16. Orbital Eccentricities eccentricity = c/a or distance between foci divided by length of major axis

  17. Kepler’s Second Law • Law of Equal Areas • A line joining a planet/comet and the Sun sweeps out equal areas in equal intervals of time

  18. Kepler’s Third Law Square of any planet's orbital period (sidereal) is proportional to cube of its mean distance (semi-major axis) from Sun Rav = (Ra + Rp)/2 T2 = K Rav 3 T2 = [42/GM]r3 Recall from a previous slide the derivation of from Fnet = FG K = 42/GM K for our sun as the primary is 1 yr2/AU3 The value of K for an orbital system depends on the mass of the primary

  19. Jupiter’s Orbit Jupiter’s mean orbital radius is rJ = 5.20 AU (Earth’s orbit is 1 AU). What is the period TJof Jupiter’s orbit around the Sun?

  20. Orbital Maneuvers

  21. The Orbiting Space Station You are trying to view the International Space Station (ISS), which travels in a roughly circular orbit about the Earth. If its altitude is 385 km above the Earth’s surface, how long do you have to wait between sightings?

  22. HALLEY’S COMET He observed it in 1682, predicting that, if it obeyed Kepler’s laws, it would return in 1759. When it did, (after Halley’s death) it was regarded as a triumph of Newton’s laws.

  23. DISCOVERY OF NEW PLANETS Small departures from elliptical orbits occur due to the gravitational forces of other planets. Deviations in the orbit of Uranus led two astronomers to predict the position of another unobserved planet. This is how Neptune was added to the Solar System in 1846. Deviations in the orbits of Uranus and Neptune led to the discovery of Pluto in 1930

  24. NewtonUniversal Gravitation • Three laws of motion and law of gravitation • eccentric orbits of comets • cause of tides and their variations • the precession of the earth’s axis • the perturbation of the motion of the moon by gravity of the sun • Solved most known problems of astronomy and terrestrial physics • Work of Galileo, Copernicus and Kepler unified. Galileo Galili 1564-1642 Nicholaus Copernicus 1473-1543 Johannes Kepler 1571-1630

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