Universal Gravitation Sir Isaac Newton 1642-1727 Terrestrial Celestial F = G m1m

Universal Gravitation Sir Isaac Newton 1642-1727 Terrestrial Celestial F = G m1m

Nugent, Mary, Features Reporter has reference to this Academic Journal, PHwiki organized this Journal Universal Gravitation Sir Isaac Newton 1642-1727 Terrestrial Celestial F = G m1m2/r2 For any two masses in the universe: G = a constant later evaluated by Cavendish r m1 m2 UNIVERSAL GRAVITATION CAVENDISH: MEASURED G Modern value: G = 6.67410-11 Nm2/kg2

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Two people pass in a hall. Find the gravitational as long as ce between them. m1 = m2 = 70 kg r = 1 m F = (6.67 x 10-11 N-m2/kg2)(70 kg)(70 kg)/(1 m)2 F = 3.3 x 10-7 N Earth-Moon Force Mass of Earth: 5.97 x 1024 kg Mass of Moon: 7.35 x 1022 kg Earth-Moon Distance: 3.84 x 108 m What is the as long as ce between the earth in addition to the moon F = (6.67 x 10-11 N m2/ kg2 )(5.97x1024kg)(7.35×1022)/(3.84×108)2 1.98 x 1020 N Practice What is the gravitational as long as ce of attraction between a 100 kg football player on the earth in addition to the earth

Definition of Weight The weight of an object is the gravitational as long as ce the earth exerts on the object. Weight = GMEm/RE2 Weight can also be expressed Weight = mg Combining these expressions mg = GMEm/RE2 RE = 6.37106 m = 6370 km ME = 5.97 x 1024 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 as long as mula. Apparent Weight is the normal support as long as ce. 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 in addition to M = 5.98 x 1024 kg Weight = 165 N What is the astronauts apparent weight The astronaut is in uni as long as m circular motion about Earth. The net as long as ce on the astronaut is the gravitational as long as ce. The normal as long as ce is 0. The astronauts apparent weight is 0. Apparent Weight Apparent Weightlessness Spring scale measures normal as long as ce 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 108 km Different distances to moon is dominant cause of earths tides High high tides; low low tides Low high tides; high low tides Spring Tides Neap Tides

Tide Animation http://www.youtube.com/watchv=Ead8d9wVDTQ Satellite Motion The net as long as ce on the satellite is the gravitational as long as ce. Fnet= FG Assuming a circular orbit: mac = GmMe/r2 For low orbits (few hundred km up) this turns out to be about 8 km/s = 17000 mph Note that the satellite mass cancels out. m Me r Using TRMM Tropical Rainfall Measuring Mission The TRMM orbit is circular in addition to is at an altitude of 218 nautical miles (350 km) in addition to an inclination of 35 degrees to the Equator. The spacecraft takes about 91 minutes to complete one orbit around the Earth. This orbit allows as long as as much coverage of the tropics in addition to extraction of rainfall data over the 24-hour day (16 orbits) as possible.

Geosynchronous Satellite r = 42,000 km = 26,000 mi 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 Using Actually the theoretical derivation of Keplers Third Law A Colorful Character Highly accurate data Gave his data to Kepler Copper/silver nose Lost nose in a duel Keplers First Law 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 perihelion aphelion A comet falls into a small elliptical orbit after a brush with Jupiter

Orbital Eccentricities eccentricity = c/a or distance between foci divided by length of major axis Keplers Second Law Law of Equal Areas A line joining a planet/comet in addition to the Sun sweeps out equal areas in equal intervals of time Keplers Third Law T2 = K Rav 3 T2 = [42/GM]r3 Square of any planet’s orbital period (sidereal) is proportional to cube of its mean distance (semi-major axis) from Sun Recall from a previous slide the derivation of Rav = (Ra + Rp)/2 from Fnet = FG K as long as our sun as the primary is 1 yr2/AU3 K = 42/GM The value of K as long as an orbital system depends on the mass of the primary

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. DISCOVERY OF NEW PLANETS Small departures from elliptical orbits occur due to the gravitational as long as ces 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 in addition to Neptune led to the discovery of Pluto in 1930 Newton Universal Gravitation Three laws of motion in addition to law of gravitation eccentric orbits of comets cause of tides in addition to 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 in addition to terrestrial physics Work of Galileo, Copernicus in addition to Kepler unified. Galileo Galili 1564-1642 Nicholaus Copernicus 1473-1543 Johannes Kepler 1571-1630