Presentation on theme: "Ch 8.1 Motion in the Heavens and on Earth"— Presentation transcript:
1Ch 8.1 Motion in the Heavens and on Earth Thursday December 9, 2010Ch 8.1Motion in the Heavens and on Earth
2Newton’s Law of Universal Gravitation If the force of gravity is being exerted on objects on Earth, what is the origin of that force?Newton’s realization was that the force must come from the Earth.He further realized that this force must be what keeps the Moon in its orbit.
3Gravity Near the Earth’s Surface; Geophysical Applications The acceleration due to gravity varies over the Earth’s surface due to altitude, local geology, and the shape of the Earth, which is not quite spherical.
4SatellitesSatellites are routinely put into orbit around the Earth. The tangential speed must be high enough so that the satellite does not return to Earth, but not so high that it escapes Earth’s gravity altogether.
5SatellitesThe satellite is kept in orbit by its speed – it is continually falling, but the Earth curves from underneath it.
6Satellites and “Weightlessness” Objects in orbit are said to experience weightlessness. They do have a gravitational force acting on them, though!The satellite and all its contents are in free fall, so there is no normal force. This is what leads to the experience of weightlessness.
7Kepler’s Laws and Newton's Synthesis Kepler’s laws describe planetary motion.1st law - The orbit of each planet is an ellipse, with the Sun at one focus.
8Kepler’s Laws and Newton's Synthesis 2nd law - An imaginary line drawn from each planet to the Sun sweeps out equal areas in equal times.
9Kepler’s Laws and Newton's Synthesis 3rd law - The ratio of the square of a planet’s orbital period is proportional to the cube of its mean distance from the Sun.
10Kepler’s 3rd law (Ta/Tb)2 = (ra/rb)3 T= period, r=radius (distance) Look at example problem on pg 180 to see an application of Kepler’s 3rd lawThen do practice problems 1 and 2 (p ) for homework
11Newton’s Law of Gravitation Thursday December 16, 2010Newton’s Law of Gravitation
12Newton’s Law of Universal Gravitation Earth’s gravitational force on you is one-half of a Third Law pairSuch a disparity in masses = reaction force undetectableFor bodies more equal in mass it can be significant.
13Newton’s Law of Universal Gravitation Gravitational force must be proportional to both masses.Observing planetary orbits, Newton concluded the gravitational force decreases as the inverse of the square of the distance between the masses.Law of Universal Gravitation :where
14Practice ProblemTwo bowling balls each have a mass of 6.8 kg. They are located next to each other with their centers 21.8 cm apart. What gravitational force do they exert on each other?
15Solution F=Gm1m2/r2 F=6.67E-11 Nm2/kg2 ((6.8kg) 2/(0.218 m) 2)
16Gravity Near the Earth’s Surface; Geophysical Applications Relate the gravitational constant to the local acceleration of gravity. On the surface of the Earth:Solving for g gives:
17Knowing the radius of the Earth, the mass of the Earth can be calculated:
18Kepler’s Laws and Newton's Synthesis Kepler’s laws can be derived from Newton’s lawsIrregularities in planetary motion led to the discovery of Neptune, and planets outside our solar system.
19Using Newton’s Laws and Universal Gravitation together Newton’s second law and Kepler’s third law work together to provide information about planetary motion and gravitational attraction amongst them.Ms = mass of sunMp= mass of planetr = radius of the planets orbit.
20Newton’s version of Kepler’s 3rd When using Newton’s second law…Fg= mp*ac.ac= 4p2r/T2 (see p164)Fg = mp4p2r/T2 set this equal to NLGrav and solve for T2T2= (4p2/Gms)r3(Kepler’s 3rd law…”the square of the period is proportional to the cube of the distance that separates the masses”)
22Ch 8.2 Using the Law of Universal Gravitation Friday December 17, 2010Ch 8.2Using the Law of Universal Gravitation
23Motion of Planets and Satellites Newton discovered1. projectile launched at a horizontal speed of 8 km/s and a height of 150 km above earth’s surface will fall to earth at same rate and curvature of the earth(very little air resistance at 150km)2. it will stay 150 km above the surface and travel at a constant speed of 8 km/s. We us this information to put objects into orbit around the earth.
24Newton and Kepler Objects in orbit exhibit uniform circular motion: ac = v2/rNewton’s second law then becomes: F=mv2/rCombining this with Newton’s Law of Gravitation givesGMeM/r2 = mv2/rSolving for speed:v = GMe/rSolving for time (period) givesT = 2p * r3/GMe
25HINT***When solving problems where you need the radius of an object in orbit*****Add the height above earth’s surface to the radius of the earth (6.38 X 106 m).Sometimes, due to sig figs, the height above Earth’s surface is so small compared to rE that it is the same as using rE.
26Practice ProblemA satellite orbits earth 3.25 X 105 m above earth’s surface. What is its speed in orbit and its period?
27Find the velocity first 1. To find the radius add 3.25 X 105 to Earth’s radius which is 6.38 X 106 m.r = 6.71 X 1062. V = GMe/r3. V = 7.71 X 103 m/s
28Find the Period T = 2p * r3/GMe OR use v=d/t and remember d = 2pr. We get T = 2pr/v*You will get the same answer for both equations. Write this equation in your notes because you will probably make less math mistakes if you use it instead.T = 5547 seconds
30Weight and Weightlessness By using Newton’s law of universal gravitation and his second law motion we were able to calculate acceleration due to gravity.g=GMe/Re2As you move further from the center of earth, gravity is reduced.
31Gravitational FieldCommon forces that we see acting in day to day life are contact forces. Gravity is not a contact force, it is a long distance force. This was a concept that Newton initially struggled with.
32Gravitational FieldLater Michael Faraday developed the idea that anything with a mass is surrounded by a gravitational field. This results in a force of attraction. To find the strength of the gravitational field you use the equation:g=F/m
33Gravitational Vs. Inertial Mass Inertial Mass – the ratio of the net force exerted on an object and its acceleration. This is measured by applying force to an object and measuring its acceleration.F = m*aGravitational Mass – determines the size of the gravitational attraction between two objectsgravitational mass = r2Fgravity/Gm
34HomeworkPg. 192, Section 8.2 Section Review and page 196, 52-54