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Thursday December 9, 2010 Ch 8.1 Motion in the Heavens and on Earth.

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Presentation on theme: "Thursday December 9, 2010 Ch 8.1 Motion in the Heavens and on Earth."— Presentation transcript:

1 Thursday December 9, 2010 Ch 8.1 Motion in the Heavens and on Earth

2 Newton’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.

3 Gravity 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.

4 Satellites Satellites 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.

5 Satellites The satellite is kept in orbit by its speed – it is continually falling, but the Earth curves from underneath it.

6 Satellites 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.

7 Kepler’s Laws and Newton's Synthesis Kepler’s laws describe planetary motion. 1 st law - The orbit of each planet is an ellipse, with the Sun at one focus.

8 Kepler’s Laws and Newton's Synthesis 2 nd law - An imaginary line drawn from each planet to the Sun sweeps out equal areas in equal times.

9 Kepler’s Laws and Newton's Synthesis 3 rd law - The ratio of the square of a planet’s orbital period is proportional to the cube of its mean distance from the Sun.

10 Kepler’s 3 rd law (T a /T b ) 2 = (r a /r b ) 3 T= period, r=radius (distance) Look at example problem on pg 180 to see an application of Kepler’s 3 rd law Then do practice problems 1 and 2 (p ) for homework

11 Thursday December 16, 2010 Newton’s Law of Gravitation

12 Newton’s Law of Universal Gravitation Earth’s gravitational force on you is one-half of a Third Law pair Such a disparity in masses = reaction force undetectable For bodies more equal in mass it can be significant.

13 Newton’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

14 Practice Problem Two 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?

15 Solution F=Gm 1 m 2 /r 2 F=6.67E-11 Nm 2 /kg 2 ((6.8kg) 2 /(0.218 m) 2 ) F=6.5E-8 N

16 Gravity 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:

17 Knowing the radius of the Earth, the mass of the Earth can be calculated:

18 Kepler’s Laws and Newton's Synthesis Kepler’s laws can be derived from Newton’s laws Irregularities in planetary motion led to the discovery of Neptune, and planets outside our solar system.

19 Using 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 sun Mp= mass of planet r = radius of the planets orbit.

20 Newton’s version of Kepler’s 3 rd When using Newton’s second law… F g = m p *a c. a c = 4  2 r/T 2 (see p164) F g = m p 4  2 r/T 2 set this equal to NLGrav and solve for T 2 T 2 = (4  2 /Gm s )r 3 (Kepler’s 3rd law…”the square of the period is proportional to the cube of the distance that separates the masses”)

21 Homework P195/ 34, 38, 40

22 Friday December 17, 2010 Ch 8.2 Using the Law of Universal Gravitation

23 Motion of Planets and Satellites Newton discovered 1. 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.

24 Newton and Kepler Objects in orbit exhibit uniform circular motion: a c = v 2 /r Newton’s second law then becomes: F=mv 2 /r Combining this with Newton’s Law of Gravitation gives GMeM/r 2 = mv 2 /r Solving for speed: v =  GMe/r Solving for time (period) gives T = 2  *  r 3 /GMe

25 HINT ***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 10 6 m). Sometimes, due to sig figs, the height above Earth’s surface is so small compared to r E that it is the same as using r E.

26 Practice Problem A satellite orbits earth 3.25 X 10 5 m above earth’s surface. What is its speed in orbit and its period?

27 Find the velocity first 1. To find the radius add 3.25 X 10 5 to Earth’s radius which is 6.38 X 10 6 m. r = 6.71 X V =  GMe/r 3. V = 7.71 X 10 3 m/s

28 Find the Period T = 2   r 3 /GMe OR use v=d/t and remember d = 2  r. We get T = 2  r/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

29 Homework Page 196: 50, 51

30 Weight 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/Re 2 As you move further from the center of earth, gravity is reduced.

31 Gravitational Field Common 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.

32 Gravitational Field Later 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

33 Gravitational 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*a Gravitational Mass – determines the size of the gravitational attraction between two objects gravitational mass = r 2 Fgravity/Gm

34 Homework Pg. 192, Section 8.2 Section Review and page 196, 52-54


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