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Physics 151: Lecture 27, Pg 1 Physics 151: Lecture 27 Today’s Agenda l Today’s Topic çGravity çPlanetary motion.

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Presentation on theme: "Physics 151: Lecture 27, Pg 1 Physics 151: Lecture 27 Today’s Agenda l Today’s Topic çGravity çPlanetary motion."— Presentation transcript:

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2 Physics 151: Lecture 27, Pg 1 Physics 151: Lecture 27 Today’s Agenda l Today’s Topic çGravity çPlanetary motion

3 Physics 151: Lecture 27, Pg 2 New Topic - Gravity l Sir Isaac developed his laws of motion largely to explain observations that had already been made of planetary motion. See text: 14 Sun Earth Moon Note : Not to scale

4 Physics 151: Lecture 27, Pg 3 Gravitation (Courtesy of Newton) Things Newton Knew, ç1. The moon rotated about the earth with a period of ~28 days.  2. Uniform circular motion says, a =  2 R ç4. Acceleration due to gravity at the surface of the earth is g ~ 10 m/s 2 ç5. R E = 6.37 x 10 6 ç6. R EM = 3.8 x 10 8 m See text: 14.1

5 Physics 151: Lecture 27, Pg 4 Gravitation (Courtesy of Newton) Things Newton Figured out, ç1. The same thing that causes an apple to fall from a tree to the ground is what causes the moon to circle around the earth rather than fly off into space. (i.e. the force accelerating the apple provides centripetal force for the moon) ç2. Second Law, F = ma So, acceleration of the apple (g) should have some relation to the centripetal acceleration of the moon (v 2 /R EM ). See text: 14.1

6 Physics 151: Lecture 27, Pg 5 Moon rotating about the Earth : So  = 2.66 x 10 -6 s -1. l Now calculate the acceleration.  a =  2 R = 0.00272 m/s 2 =.000278 g l Calculate angular velocity :  = v / R EM = 2  R EM / T R EM = 2  / T  =

7 Physics 151: Lecture 27, Pg 6 Gravitation (Courtesy of Newton) l Newton found that a moon / g =.000278 l and noticed that R E 2 / R 2 =.000273 Universal Law of Gravitation: l This inspired him to propose the Universal Law of Gravitation: |F Mm |= GMm / R 2 RRERE a moon g G = 6.67 x 10 -11 m 3 kg -1 s -2 See text: 14.1

8 Physics 151: Lecture 27, Pg 7 Gravity... F l The magnitude of the gravitational force F 12 exerted on an object having mass m 1 by another object having mass m 2 a distance R 12 away is: F l The direction of F 12 is attractive, and lies along the line connecting the centers of the masses. R 12 m1m1 m2m2 F F 12 F F 21 See text: 14.1

9 Physics 151: Lecture 27, Pg 8 Gravity... l Compact objects: çR 12 measures distance between objects l Extended objects: çR 12 measures distance between centers R 12

10 Physics 151: Lecture 27, Pg 9 Gravity... l Near the earth’s surface: çR 12 = R E »Won’t change much if we stay near the earth's surface. »i.e. since R E >> h, R E + h ~ R E. RERE m M h FFgFFg See text: 14.1

11 Physics 151: Lecture 27, Pg 10 Gravity... l Near the earth’s surface... l So |F g | = mg = ma ç a = g All objects accelerate with acceleration g, regardless of their mass! Where:  =g See text: 14.3

12 Physics 151: Lecture 27, Pg 11 Example gravity problem: l What is the force of gravity exerted by the earth on a typical physics student? çTypical student mass m = 55kg çg = 9.8 m/s 2. çF g = mg = (55 kg)x(9.8 m/s 2 ) çF g = 539 N FFgFFg l The force that gravity exerts on any object is called its Weight W = 539 N

13 Physics 151: Lecture 27, Pg 12 Lecture 27, Act 1 Force and acceleration l Suppose you are standing on a bathroom scale in Physics 203 and it says that your weight is W. What will the same scale say your weight is on the surface of the mysterious Planet X ? l You are told that R X ~ 20 R Earth and M X ~ 300 M Earth. (a) 0.75 (b) (c) (a) 0.75 W (b) 1.5 W (c) 2.25 W E X

14 Physics 151: Lecture 27, Pg 13 Lecture 27, Act 1 Solution l The gravitational force on a person of mass m by another object (for instance a planet) having mass M is given by: l Ratio of weights = ratio of forces: (A)

15 Physics 151: Lecture 27, Pg 14 Kepler’s Laws l Much of Sir Isaac’s motivation to deduce the laws of gravity was to explain Kepler’s laws of the motions of the planets about our sun. l Ptolemy, a Greek in Roman times, famously described a model that said all planets and stars orbit about the earth. This was believed for a long time. l Copernicus (1543) said no, the planets orbit in circles about the sun. l Brahe (~1600) measured the motions of all of the planets and 777 stars (ouch !) l Kepler, his student, tried to organize all of this. He came up with his famous three laws of planetary motion. See text: 14.3

16 Physics 151: Lecture 27, Pg 15 Kepler’s Laws 1st All planets move in elliptical orbits with the sun at one focal point. 2nd The radius vector drawn from the sun to a planet sweeps out equal areas in equal times. 3rd The square of the orbital period of any planet is proportional to the cube of the semimajor axis of the elliptical orbit. l It was later shown that all three of these laws are a result of Newton’s laws of gravity and motion. See text: 14.4

17 Physics 151: Lecture 27, Pg 16 Kepler’s Third Law Let’s start with Newton’s law of gravity and take the special case of a circular orbit. This is pretty good for most planets. See text: 14.4

18 Physics 151: Lecture 27, Pg 17 Kepler’s Second Law This one is really a statement of conservation of angular momentum. See text: 14.4

19 Physics 151: Lecture 27, Pg 18 Kepler’s Second Law See text: 14.4 2. The radius vector drawn from the sun to a planet sweeps out equal areas in equal times. R dR dA

20 Physics 151: Lecture 27, Pg 19 Kepler’s Second Law See text: 14.4 R dR dA

21 Physics 151: Lecture 27, Pg 20 Energy of Planetary Motion A planet, or a satellite, in orbit has some energy associated with that motion. Let’s consider the potential energy due to gravity in general. See text: 14.7 Define r i as infinity U r RERE 0

22 Physics 151: Lecture 27, Pg 21 Energy of a Satellite A planet, or a satellite, also has kinetic energy. See text: 14.7 We can solve for v using Newton’s Laws, Plugging in and solving,

23 Physics 151: Lecture 27, Pg 22 Recap of today’s lecture l Chapter 13 n Gravity n Planetary motion


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