10/21/2003 PHY Lecture 14 1
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1 Announcements. Second exam scheduled for Oct. 8 th -- practice exams now available -- Thursday review of Chapters Today s lecture Universal law of gravitation Gravity near the planet s surface Gravitational potential energy Planetary and satelite motion 0//003 PHY 3 -- Lecture 4
2 Newton s law of gravitation: m attracts m according to: y F Gm m r rˆ m r r r m G6.67 x 0 - N m /kg r xample: m F x m 70 kg; N r m : 8 N 0//003 PHY 3 -- Lecture 4
3 Vector nature of Gravitational law: y d m 3 m m d x m ( m i j) F G + m 3 d 0//003 PHY 3 -- Lecture 4 3
4 Gravitational force of the arth m GM F m GM g ( ) 4 m/s 9.8m/s 0//003 PHY 3 -- Lecture 4 4
5 Question: Suppose you are flying in an airplane at an altitude of 35000ft~km above the arth s surface. What is the acceleration due to arth s gravity? F GM m ( + h) GM a ( + h) ma (( ) 0 ) 4 m/s 9.796m/s a/g //003 PHY 3 -- Lecture 4 5
6 Attraction of moon to the arth: F GM M M N ( ) M N Acceleration of moon toward the arth: F M M a a.99x0 0 N/7.36x0 kg m/s 0//003 PHY 3 -- Lecture 4 6
7 Stable circular orbit of two gravitationally attracted objects (such as the moon and the arth) M F a v a T v π M v ω GM M π T 3 M GM M M ( ) π s 7.4 days //003 PHY 3 -- Lecture 4 7
8 Peer instruction question In the previous discussion, we saw how the moon orbits the arth in a stable circular orbit because of the radial gravitational attraction of the moon and Newton s second law: Fma, 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 km above the arth s surface and that the arth s radius is 6370 km. (a) Yes (b) No 0//003 PHY 3 -- Lecture 4 8
9 Stable (??) circular orbit of two gravitationally attracted objects (such as the airplane and the arth) v GM a a F a a v ω v T π a GM 3 a a π T a 6 3 ( ) π s.4 hours v 7.9 km/s 8000 mi/h!! 4 0//003 PHY 3 -- Lecture 4 9
10 Newton s law of gravitation: arth s gravity: GM m F GM g m/s 9.8m/s 6 ( ) Stable circular orbits of gravitational attracted objects: m F Gm 4 m r rˆ M F a v M M 0//003 PHY 3 -- Lecture 4 0
11 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. x 0//003 PHY 3 -- Lecture 4
12 adial forces on m : m m F v T r π ( + ) π T Gm m ( + ) Gm m a r m v T? Tangential forces? 0//003 PHY 3 -- Lecture 4
13 Peer instruction question What is the relationship between the periods T and T of the two gravitationally attracted objects rotating about their center of mass? (Assume that m < m.) (A) T T (B) T <T (C) T >T m m 0//003 PHY 3 -- Lecture 4 3
14 0//003 PHY 3 -- Lecture 4 4 m m ( ) ( ) ( ) 3 m m G T T m Gm m m v m m m π ω ω ω ω
15 What is the physical basis for stable circular orbits? dp. Newton s second law? F ma dt. Conservation of mechanical energy? K + U (const) 3. Conservation of linear momentum? p (const) 4. Torqued motion? τ I α? 5. Conservation of angular momentum? L (const) 0//003 PHY 3 -- Lecture 4 5
16 dl τ 0 dt L (const) v m L m v L m v m v Question: How are the magnitudes of L and L related? 0//003 PHY 3 -- Lecture 4 6
17 The potential energy associated with the gravitational force. r r Gm m r Gmm Gmm U ( r) dr dr r r r r ref 0//003 PHY 3 -- Lecture 4 7
18 Total mechanical energy for circular orbits: (assume M >> m) U(J) mv + U ( r) mv GMm r r GMm U ( r) r v GM r h/ GMm r 0//003 PHY 3 -- Lecture 4 8
19 Peer instruction question What is wrong with the previous analysis? A. Nothing is wrong. (The description of circular motion due to gravitational attraction is complete.) B. depends on r and therefore must not be constant. C. can only be constant if r is constant, but it is not obvious why r is constant. D. Conservation of angular moment will come to the rescue. 0//003 PHY 3 -- Lecture 4 9
20 Angular momentum: L r x mv For circular orbit: M F v L M M M v L v M K M M L M M v M M M a M M 0//003 PHY 3 -- Lecture 4 0
21 L mr for elliptical orbit r for circular orbit U(r) 0//003 PHY 3 -- Lecture 4
22 Circular orbit: y x r 0 + r y 0 x lliptical orbit: y x a + b y x 0//003 PHY 3 -- Lecture 4
23 Satellites orbiting earth (approximately circular orbits): T π ( + h / ) 5058( + h / ) s GM ~ 6370 km xamples: 600 Satellite h (km) Geosynchronous NOAA polar orbitor 800 Hubble Inter. space station * 390 T (hours) ~4 ~.7 ~.6 ~.5 v (mi/h) * Link: 0//003 PHY 3 -- Lecture 4 3
24 Sample question: Suppose that the space shuttle (m0 5 kg) was initially in the same orbit as the International space station (h i 390km) and the engines are fired to give it exactly the amount of energy W to raise it to the same orbit as the Hubble space telescope (h f 600km). What is the energy W? You can show that the energy of a satellite of mass m in a circular orbit of height h above the arth s surface is given by: GM m K + U mech ( + h) W GM m ( ) 89, 000J h ( ) f h i + + 0//003 PHY 3 -- Lecture 4 4
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