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10/10/2013PHY 113 C Fall 2013-- Lecture 131 PHY 113 C General Physics I 11 AM - 12:15 PM MWF Olin 101 Plan for Lecture 13: Chapter 13 – Fundamental gravitational.

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Presentation on theme: "10/10/2013PHY 113 C Fall 2013-- Lecture 131 PHY 113 C General Physics I 11 AM - 12:15 PM MWF Olin 101 Plan for Lecture 13: Chapter 13 – Fundamental gravitational."— Presentation transcript:

1 10/10/2013PHY 113 C Fall Lecture 131 PHY 113 C General Physics I 11 AM - 12:15 PM MWF Olin 101 Plan for Lecture 13: Chapter 13 – Fundamental gravitational force law and planetary motion 1.Gravitational force law; relationship with g near Earth’s surface 2.Orbital motion of planets, satellites, etc. 3.Energy associated with gravitation and planetary motion

2 10/10/2013 PHY 113 C Fall Lecture 132

3 10/10/2013PHY 113 C Fall Lecture 133 Question from Webassign Assignment #11

4 10/10/2013PHY 113 C Fall Lecture 134 X 11 33 22

5 10/10/2013PHY 113 C Fall Lecture 135 X 11 33 22

6 10/10/2013PHY 113 C Fall Lecture 136 Question from Webassign Assignment #11 X

7 10/10/2013PHY 113 C Fall Lecture 137 Question from Webassign Assignment #11 iclicker question: A. K i =K f ? B. K i =K f ?

8 10/10/2013PHY 113 C Fall Lecture 138 Universal law of gravitation  Newton (with help from Galileo, Kepler, etc.) 1687

9 10/10/2013PHY 113 C Fall Lecture 139 Newton’s law of gravitation: m 2 attracts m 1 according to: x y m1m1 m2m2 r2r2 r1r1 r2r1r2r1 G=6.67 x N m 2 /kg 2

10 F 12 10/10/2013PHY 113 C Fall Lecture 1310 Vector nature of Gravitational law: m1m1 m2m2 m3m3 x y d d F 13

11 10/10/2013PHY 113 C Fall Lecture 1311 Gravitational force of the Earth RERE m Note: Earth’s gravity acts as a point mass located at the Earth’s center.

12 10/10/2013PHY 113 C Fall Lecture 1312 Question: Suppose you are flying in an airplane at an altitude of 35000ft~11km above the Earth’s surface. What is the acceleration due to Earth’s gravity? a/g = 0.997

13 10/10/2013PHY 113 C Fall Lecture 1313 Attraction of moon to the Earth: Acceleration of moon toward the Earth: F = M M a  a = 1.99x20 20 N/7.36x10 22 kg = m/s 2 R EM

14 10/10/2013PHY 113 C Fall Lecture 1314

15 10/10/2013PHY 113 C Fall Lecture 1315 Gravity on the surface of the moon Gravity on the surface of mars

16 10/10/2013PHY 113 C Fall Lecture 1316 iclicker question: In estimating the gravitational acceleration near the surfaces of the Earth, Moon, or Mars, we used the full mass of the planet or moon, ignoring the shape of its distribution. This is a reasonable approximation because: A.The special form of the gravitational force law makes this mathematically correct. B.Most of the mass of the planets/moon is actually concentrated near the center of the planet/moon. C.It is a very crude approximation.

17 10/10/2013PHY 113 C Fall Lecture 1317 Stable circular orbit of two gravitationally attracted objects (such as the moon and the Earth) R EM F a v

18 10/10/2013PHY 113 C Fall Lecture 1318 iclicker question: In the previous discussion, we saw how the moon orbits the Earth in a stable circular orbit because of the radial gravitational attraction of the moon and Newton’s second law: F=ma, where a is the centripetal acceleration of the moon in its circular orbit. Is this the same mechanism which stabilizes airplane travel? Assume that a typical cruising altitude of an airplane is 11 km above the Earth’s surface and that the Earth’s radius is 6370 km. (a) Yes (b) No

19 10/10/2013PHY 113 C Fall Lecture 1319 Stable (??) circular orbit of two gravitationally attracted objects (such as the airplane and the Earth) R Ea F a v

20 10/10/2013PHY 113 C Fall Lecture 1320 Newton’s law of gravitation: Earth’s gravity: Stable circular orbits of gravitational attracted objects: RERE m R EM F a v M Summary:

21 10/10/2013PHY 113 C Fall Lecture 1321 More details If we examine the circular orbit more carefully, we find that the correct analysis is that the stable circular orbit of two gravitationally attracted masses is about their center of mass. m1m1 R2R2 R1R1 m2m2

22 10/10/2013PHY 113 C Fall Lecture 1322 m1m1 R2R2 R1R1 m2m2 Radial forces on m 1 : T 2 ?

23 10/10/2013PHY 113 C Fall Lecture 1323 iclicker question: What is the relationship between the periods T 1 and T 2 of the two gravitationally attracted objects rotating about their center of mass? (Assume that m 1 < m 2.) (A) T 1 =T 2 (B) T 1 T 2 m1m1 R2R2 R1R1 m2m2

24 10/10/2013PHY 113 C Fall Lecture 1324 m1m1 R2R2 R1R1 m2m2

25 10/10/2013PHY 113 C Fall Lecture 1325 iclicker questions: How is it possible that all of these relations are equal? A.Magic. B.Trick. C.Algebra.

26 10/10/2013PHY 113 C Fall Lecture 1326 m1m1 R2R2 R1R1 m2m2 For m 2 >>m 1  R 2 << R 1 : m2m2 m1m1 R1R1 Larger mass

27 10/10/2013PHY 113 C Fall Lecture 1327 What is the physical basis for stable circular orbits? 1.Newton’s second law? 2.Conservation of angular momentum? L = (const) Note: Gravitational forces exert no torque

28 10/10/2013PHY 113 C Fall Lecture 1328 m1m1 R2R2 R1R1 m2m2 v1v1 v2v2 L 1 =m 1 v 1 R 1 L 2 =m 2 v 2 R 2 L = L 1 + L 2 Question: How are the magnitudes of L 1 and L 2 related? Note: More generally, stable orbits can be elliptical.

29 10/10/2013PHY 113 C Fall Lecture 1329 Satellites orbiting earth (approximately circular orbits): R E ~ 6370 km Examples: Satellite h (km)T (hours)v (mi/h) Geosynchronous35790 ~ NOAA polar orbitor 800 ~ Hubble 600 ~ Inter. space station * 390 ~

30 10/10/2013PHY 113 C Fall Lecture 1330 Planets in our solar system – orbiting the sun PlanetMass (kg)Distance to Sun (m) Period of orbit (years) Mercury3.30x x Venus4.87x x Earth5.97x x Mars6.42x x Jupter1.90x x Saturn5.68x x

31 10/10/2013PHY 113 C Fall Lecture 1331 m1m1 R2R2 R1R1 m2m2 v1v1 v2v2 Review: Circular orbital motion about center of mass CM

32 10/10/2013PHY 113 C Fall Lecture 1332 m1m1 R1R1 m2m2 v1v1 Review: Circular orbital motion about center of mass

33 10/10/2013PHY 113 C Fall Lecture 1333 Example: Satellite in circular Earth orbit

34 10/10/2013PHY 113 C Fall Lecture 1334 Work of gravity: riri rfrf

35 10/10/2013PHY 113 C Fall Lecture 1335 Gravitational potential energy Example:

36 10/10/2013PHY 113 C Fall Lecture 1336 h U=mgh iclicker exercise: We previously have said that the gravitational potential of an object of mass m at a height h is U=mgh. How is this related to A.No relation B.They are approximately equal C.They are exactly equal

37 10/10/2013PHY 113 C Fall Lecture 1337 h g  Near the surface of the Earth U=mgh is a good approximation to the gravitation potential.

38 10/10/2013PHY 113 C Fall Lecture 1338 iclicker exercise: How much energy kinetic energy must be provided to an object of mass m=1000kg, initially on the Earth’s surface to outer space? A.This is rocket science and not a fair question. B.It is not possible to escape the Earth’s gravitational field. C.We can estimate the energy by simple conservation of energy concepts.

39 10/10/2013PHY 113 C Fall Lecture 1339 Total energy of a satellite in a circular Earth orbit

40 10/10/2013PHY 113 C Fall Lecture 1340 Total energy of a satellite in a circular Earth orbit iclicker question: Compared to the energy needed to escape the Earth’s gravitational field does it take A.more B.less energy to launch a satellite to orbit the Earth?

41 10/10/2013PHY 113 C Fall Lecture 1341

42 10/10/2013PHY 113 C Fall Lecture 1342 Energy involved with changing orbits h’

43 10/10/2013PHY 113 C Fall Lecture 1343 Stable Elliptical Orbits


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