Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Title here
Document related concepts
Transcript
Universal Gravitation Celestial Terrestrial Sir Isaac Newton 1642-1727 UNIVERSAL GRAVITATION For any two masses in the universe: F = G m1m2/r2 G = a constant later evaluated by Cavendish -F +F m2 m1 r CAVENDISH: MEASURED G Modern value: G = 6.674*10-11 Nm2/kg2 Measuring G Two people pass in a hall. Find the gravitational force between them. m1 = m2 = 70 kg r=1m m1 m2 r m1 m2 F G 2 r F = (6.67 x 10-11 N-m2/kg2)(70 kg)(70 kg)/(1 m)2 = 3.3 x 10-7N Universal Gravitation ACT Which of the situations shown below experiences the largest gravitational attraction? Satellite Motion The net force on the satellite is the gravitational force. r Fnet= FG Assuming a circular orbit: mac = GmMe/r2 Me m 2 mM e mv G 2 r r GM e v r Note that the satellite mass cancels out. Using M e 5.97 1023 kg For low orbits (few hundred km up) this turns out to be about 8 km/s = 17000 mph 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 Fnet FG 2 mM e mv G 2 r r 4 2 r 2 GM e 2 2 rT r GM eT r 4 2 2 3 Using M e 5.97 1023 kg r = 42,000 km = 26,000 mi 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. 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. 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. GM E g ( h) ( RE h) 2 GM E 2 g (0) 9.83 m/s RE 2 g vs Altitude Once the altitude becomes comparable to the radius of the Earth, the decrease in the acceleration of gravity is much larger: GM E g (r ) 2 r 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. Tides FG by moon on A > FG by moon on B FG by moon on B > FG by moon on C Different distances to moon is dominant cause of earth’s tides 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 Spring Tides Neap Tides 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 A comet falls into a small elliptical orbit after a “brush” with Jupiter Johannes Kepler 1571-1630 eccentricity Planet Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto Orbital Eccentricities Eccentricity Notes 0.206 0.007 0.017 0.093 0.048 0.056 0.470 0.009 0.249 Too few observations for Kepler to study Nearly circular orbit Small eccentricity Largest eccentricity among planets Kepler could study Slow moving in the sky Slow moving in the sky Not discovered until 1781 Not discovered until 1846 Not discovered until 1930 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 and the Sun sweeps out equal areas in equal intervals of time vp Ra va R p 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 Recall from a previous slide the derivation of from Fnet = FG T2 = [42/GM]r3 K = 42/GM Planet T (yr) R (AU) T2 R3 Mercury 0.24 0.39 0.06 0.06 Venus 0.62 0.72 0.39 0.37 Earth 1.00 1.00 1.00 1.00 Mars Jupiter Saturn 1.88 11.9 29.5 1.52 5.20 9.54 3.53 3.51 142 141 870 868 Jupiter’s Orbit Jupiter’s mean orbital radius is rJ = 5.20 AU (Earth’s orbit is 1 AU). What is the period TJ of Jupiter’s orbit around the Sun? 2 TJ rJ T r so TE rE 2 3 3 rJ TJ TE rE 3/2 (5.20 AU) (1.0 yr) (1.00 AU) 3/2 11.9 yr Orbital Maneuvers 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? 2 r vT v M E m mv 2 Fg G 2 r r 2 r T GM E 4 2 r 2 r T2 ( RE h)3 4 2 3 r3 T r 2 2 GM E GM E GM E (6375 km 385 km)3 T 2 5,528 s=92.1 min (6.67 1011 N m2 /kg 2 )(5.98 1024 kg) 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. HALLEY’S COMET DISCOVERY OF NEW PLANETS Deviations in the orbits of Uranus and Neptune led to the discovery of Pluto in 1930 Newton Universal 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 Nicholaus Copernicus Johannes Kepler 1564-1642 1473-1543 1571-1630
