G-1GravityNewton's Universal Law of Gravitation (first stated by Newton): any two masses m1 and m2 exertan attractive gravitational force on each other according to 1 22m mF Gr=This applies to all masses, not just big ones.G = universal constant of gravitation = 6.67 10–11 N m2 / kg2 (G is very small, so it is very difficult to measure!)Don't confuse G with g: "Big G" and "little g" are totally different things.Newton showed that the force of gravity must act according to this rule in order to produce the observed motions of the planets around the sun, of the moon around the earth, and of projectiles near the earth. He then had the great insight to realize that this same force acts between all masses. [That gravity acts between all masses, even small ones, was experimentally verified in 1798 by Cavendish.]Newton couldn't say why gravity acted this way, only how. Einstein (1915) General Theory of Relativity, explained why gravity acted like this.Example: Force of attraction between two humans. 2 people with masses m1 m2 70 kg, distance r = 1 m apart. 11 21 272 2m m6 67 10 70F G 3 3 10 Nr 1--�= = = �( . )( ).This is a very tiny force! It is the weight of a mass of 3.4 10–5 gram. A hair weighs 210–3 grams – the force of gravity between two people talking is about 1/60 the weight of a single hair.Computation of gImportant fact about the gravitational force from spherical masses: a spherical body exerts a gravitational force on surrounding bodies that is the same as if all the sphere's mass were concentrated at its center. This is difficult to prove (Newton worried about this for 20 years.)1/14/2019 University of Colorado at Boulderrm1m2FFG-2We can now compute the acceleration of gravity g ! (Before, g was experimentally determined, and it was a mystery why g was the same for all masses.)Fgrav = m a = m gE2EM mG m gR= (since r = RE is distance from m to center of Earth)m's cancel ! E2EG MgR=If you plug in the numbers for G, ME, and RE, you get g = 9.8 m/s2.Newton's Theory explains why all objects near the Earth's surface fall with the same acceleration(because the m's cancel in grav2G M mF m aR= =.) Newton's theory also makes a quantitative prediction for the value of g, which is correct.Example: g on Planet X. Planet X has the same mass as earth (MX = ME) but has ½ the radius (RX = 0.5 RE). What is gx , the acceleration of gravity on planet X?Planet X is denser than earth, so expect gx larger than g.()( )X E Ex2 22 2EX Eg of earthG M G M G M1g 4 gRR R1 22= = = =123/. Don't need values of G, ME, and RE!Method II, set up a ratio: 1/14/2019 University of Colorado at Bouldersphere, mass Mmass mFgravmass mFgrav (same as with sphere)point mass MrREmass MEmass m, dropped near surfaceEarthG-3X22X2x X EX EEE E X2EG MRg M R1 2 4 g 4 gG Mg M RR,/� �� �� �� �= = = � = =� �/� �� �� �� �_________________ * __________________At height h above the surface of the earth, g is less, since we are further from the surface, further from the earth's center. r = RE + h E E2 2EG M G Mgr R h( )= =+The space shuttle orbits earth at an altitude of about 200 mi 1.6 km/mi 320 km. Earth's radius is RE = 6380 km. So the space shuttle is only about 5% further from the earth's center than we are. If r is 5% larger, then r2 is about 10% larger, andEgrav2EM mF on mass m in shuttle) G about 10% less than on earth's surfaceR h(( )= @+Astronauts on the shuttle experience almost the same Fgrav as when on earth. So why do we say the astronauts are weightless??"Weightless" does not mean "no weight"."Weightless" means "freefall" means the only force acting is gravity.If you fall down an airless elevator shaft, you will feel exactly like the astronauts. You will be weightless, you will be in free-fall.An astronaut falls toward the earth, as she moves forward, just as a bullet fired horizontally from a gun falls toward earth.OrbitsConsider a planet like Earth, but with no air. Fire projectiles horizontally from a mountain top, with faster and faster initial speeds.1/14/2019 University of Colorado at BoulderhearthvEarthFgravFgravNastronautG-4The orbit of a satellite around the earth, or of a planet around the sun obeys Kepler's 3 Laws. Kepler, German (1571-1630). Before Newton. Using observational data from Danish astronomer Tycho Brahe ("Bra-hay"), Kepler discovered that the orbits of the planets obey 3 rules.KI : A planet's orbit is an ellipse with the Sun at one focus. KII : A line drawn from planet P to sun S sweeps out equalareas in equal times.KIII: For planets around the sun, the period T and the mean distance r from the sun are related by 23Tconstantr=. That is for any two planets A and B, 2 2A B3 3A BT Tr r=. This means that planetsfurther from the sun (larger r) have longer orbital periods (longer T).Kepler's Laws were empirical rules, based on observations of the motions of the planets in the sky. Kepler had no theory to explain these rules. Newton (1642-1727) started with Kepler's Laws and NII (Fnet = ma) and deduced thatS Pgrav2Sun planetSPM mF Gr( )-=. Newton applied similar reasoning to the motion of the Earth-Moon system (and to an Earth-apple system) and deduced that Egrav2Earth-mass mEmM mF Gr( )=.1/14/2019 University of Colorado at Boulderorbits!Planetwould go straight,if no gravitySunPlanetSslowerfastersame time intervals,same areasG-5Newton then made a mental leap, and realized that this law applied to any 2 masses, not just to the Sun-planet, the Earth-moon, and Earth-projectile systems. Starting with Fnet = ma and Fgrav = G Mm / r2, Newton was able to derive Kepler's Laws (and much more!). Newton could explain the motion of everything!Derivation of KIII (for special case of circular orbits). Consider a small mass m in circular orbit about a large mass M, with orbital radius r and period T. We aim to show that T2 / r3 = const.Start with NII: Fnet = m aThe only force acting is gravity, and for circular motion a = v2 / r 2222M m 2 rv MG m G vr r r T/p/� �= � = =/� �/� �[recall the v = dist / time = 2r / T ]2 2 2 22 3M 4 r T 4G constant, independent of mr T r G Mp p� = � = =( Deriving this result for elliptical orbits is much harder, but Newton did it. )An extra result of this calculation is a formula for the speed v of a satellite in circular orbit:GMvr=. For low-earth orbit (few hundred miles up), this orbital speed is
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