Class Note for PTYS 206 with Professor Yelle at UA
Class Note for PTYS 206 with Professor Yelle at UA
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Date Created: 02/06/15
P Y82 62 28 Feb 2 8 List of Symbols F fOrce V a acceleratin not semi major axis in this lecture vz velocity 392 M mass of Sun 2 m mass of planet V a general distance 39 rradius of Circle semimajor axis of orbit radius of Earth Newton devised a uniform and systematic method for describing motion which we today refer to as the Science of Mechanics It remains the basic description of motion requiring correction only at very high velocities and very small distances Newton summarized his theory in 3 laws 1 An object remains at rest or continues in uniform motion unless acted upon by a force 2 Force is equal to mass x acceleration Fma 3 For every action there is an equal and opposite reaction Cambridge was closed because of the Plague As the story goes Newton was sitting under the apple tree outside his farmhouse shown right and while watching the apples fall he realized that the force that made the apples fall also made the planets orbit the sun Using his newly invented Calculus Newton was able to show that Kepler s 3 laws of planetary motion followed directly from this hypothesis Falling Apples and Orbiting Planets w iii 1 f Fk l g 39t quot9 3 39 1 1 A a J 3 hi3 395 JV i Jupi l er39 What do these have in common Newton s cannonball From Principia Gravitational Force Units ACC ord39ing to Newton s 2nd law Forcemass acceleration The units must also match Units of kilograms Units of acceleration meterssecz Unit of force must be ki lograms meterssec2 kg m 3392 shorthand We de ne a new unit to make notation more simple Lets call it a Newton From the definition we can See that 1 Newton 1 kg m s2 From now on we measure force in Newtons What are the off Newtonis law of gravitation F GMmd Let s solve for G multiply by d2 divide by Mm G Fd2Mm Examine the units Fd27Mm has units of N m2kg2 or N mz39kgZ Or expressing Newtons in kg m and s 391 N 1 kg m Fd2Mm has units of N m2 kfg mS39Zm2k39g39239 5392 kgquot G has units of m3 s2 Numer39ijcazlly G 63967xr703911 m3 32 kg1 E W s u The separation d is the distance between the centers of the objects Newton Explains Galileo Newton s 2nd Law F ma Newton s law of gravity F G Mmd52 The separation d is the distance between the 7 V falling body and the center F G MmR of the Earth dR Set forces equal ma 39 GMmR2 Cancel m39on both sides of a GMRz the equation The acceleration does not depend on m Bodies fall at the same rate regardless of mass Planetary motion more complicated but governed by the same laws First we need to consider the acceleration of orbiting bodies Acceleration is any cha ge in speed ofdirec ti39on of motion Circular motion is accelerated motion because direction is changing Forcircu ar Cenm e al motion for Real Life Example A Circular Race Track Orbiting Planets Continued So orbiting planets are accelerating This Gravitational Constant G A Fundamental ConstanL must be caused by a 0 eumverse I ve39oc39ty V Pl force Lets assume MESStm that the force IS graVIty II Gravitational Force We should be able to FGMmr2 calculate the force and l acceleration usmg I Newton s second law Mass M I and NeWtonS IaW Of Newton s 2nd Law Fma Ill graVIty 39 Orbits come in a variety of shapes eccentricities In order to keep the math simple we will consider in this lecture only circular orbits All of our results also apply to elliptical orbits but we will not derive them that way Step 1 Calculate the Velocity We take as given that acceleration and velocity 39in circular motion are related by a v2r According to Newton s 2 nd law F ma var According to Newton s law of gravity F GM mrz Equati39ng the expressions for force We have mVZr r GMmr2 Solving for v2 gives v2 GMr Step The Velocity related to the semimajoraxis and period The velocity is related to the semimajor axis and the perio in a simple way velocity distancetime distance 2m Where rsvernimajor axis radius of circle time Period P v 2mrPv distancetime Step Relate the Period to the Orbital Radius We have V2 GiMr And V 2nr So it follows that hr2 GMr 4 39n 2lr239I32 GMr How Does This Relate to Kepler s Third Law We have 4712 r2 F 2 Mrquot Multiply both sides by r 4752r F 2 GM Multiply both sides y p2 4zn2r3 GM P2 Divide both Sides by GM4n 2 Newton s form of Kepler s Third Law ave r3 GM4n P2 K epler s third law was a3P 2J where asemimajior axis not acceleration Sincetoday we are using rsem39i major this equation is the same as Kepler s 3rd if GM4752 1 AU3year2 Let s check Do Newton and Kepler Agree We Want to kn M if GM47121 1 A39U39quot year2 Plug in G 67 x10391 1 kgquot M20gtlt1030 kg GM4712 34x1018 m3 Recall 1 AU 1011 m and 1 year 31 x1 07 S 80 391 AU3year2 31 5x1011 m39331x1 07 s 1 AU3year2 3439x1078 m 318392 WOW 3M7quot M39 r73 wW A PQQFMM 6 Gm 3 r3 ya MMA i W 2F am Pr HM M V7 r m4 1 2 Cw M m 6W5 11 st AugWMZ 1L 4W CL 2 4 WM 3 ME my i Amigym 47 55 099 v1 251quot 2424ij MugHM M if m 223 cf 227 P Mg wag X61 ch f h x 6 Pl V1 W3 mm P 5 s P 25 2 W Z EKmOw 4 r 339 quotf M 2M9 F AW 5 3 k e l fixw f 1 L33 HM u V gg ymra m f 4455 X r 393 t M ya 5 6 39i I H l 5 may Same 9quot ME m Q z r3 Mg W my T 22gtquot U 5E 33 MWE EEQMMyWW V quot a j z Qa X SEQ 154 g it up l f V lwf f l 51m Ra E 2 E f I s L 41 Ag MD ME X zg
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