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PHYS 608

by: Jamaal McGlynn

PHYS 608 PHYS 608

Jamaal McGlynn
GPA 3.78

M. Bromley

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M. Bromley
Class Notes
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This 12 page Class Notes was uploaded by Jamaal McGlynn on Tuesday October 20, 2015. The Class Notes belongs to PHYS 608 at San Diego State University taught by M. Bromley in Fall. Since its upload, it has received 20 views. For similar materials see /class/225317/phys-608-san-diego-state-university in Physics 2 at San Diego State University.


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Date Created: 10/20/15
Lecture 12 Outline In and Out Burger o The Kepler Problem Section 38 o Hohmann Transfer Parabolic motion 6 1 Hyperbolic motion 6 gt 1 Intro to Scattering Section 310 Hohmann Transfer 0 Requires two velocity Changes for planet transfer 0 Which is the most energy ef cient way to do so Hohmann Transfer II o For both circles and ellipses we have E k2a o For circular path of the planets we get simple velocities k 1 k k k E mv2 thus 211 and v2 2T0 2 7 0 m r l 77173 a Transfer ellipse has at 7 1 7 2 With conserved energy k k k l k 2 mv2 k 2T2 6 271 Up gt 01 and va lt 02 mm 7 1 7 2 mrg 7 1 7 2 0 Whilst most energy ef cient also the slowest transit o and return must wait a While for the planets to be re aligned Parabolic Motion in time for V7 c7 Parabolic Motion in time for F7 c7 2 t e3 f0 d0 WW2 90 1 l ecos9 0 2 0 eg For Parabola s e 1 use 1 3086 2 C082 3 0 4 6 3 tang 1 2 t 4mk20 sec 2 d6 2mk20 51 da 0 Above used a tan so da sec2 d6 0 and also 8602 1 tar o Sowehave 7Tlt6lt7T o Where 6 7T gt t oo7 oo o Where 6 0 gt t 07 apsidal Stealing Gravity from a kid planet o But for interplanetary travel steal energy from a planet 0 Let U and v be the velocities relative to P inertial frame a then we must have from an attractive force 27f gt a Conservation of energy says that Planet s z7f lt Intergalactic Planetary o Voyager 2 path departing 1977 a this procedure is also known as gravity assist Central Force Scattering 0 Consider effective one body scattering 0 Where problem has rotational symmetry about axis a De ne cross section 09 of units of area for scattering in given direction as 0Qd 2 0Q27T sin dG the of particles scattered into solid angle d9 per unit time DIVIDED by the incident intensity I which is of particles crossing area per unit time Impact Parameter Given the impact parameter s and initial speed 210 imagine in the lab frame particle passing by with no central force Then angular momentum must be 6 mvos 5V 2mE Given known E7 5 then we know 6 and thus we also know the scattering angle 9 as well Solid Angle Element d9 0 The 7 of particles scattered into d9 27T sin dG lying between 9 and 9 d is equal to the of particles with s to s d5 ie 27TIsds 27w sin 9d9 0 Given that can consider s E differential cross section E d 09 sin 9 0 eg for repulsive scattering Solid Angle Expression 0 Want to nd an expression for 95 take a so we go back to the equation of orbital motion 7 d 7 67 60 2mE 2mV 1 T0 7 2 T2 T2 72 o with 7 0 007 60 7T and for 67 mm 6min 7T III 00 dr II 7T 6min 7a 2 2mE2mVi mzn 7quot E2 62 T2 a rewrite using 5 5V 2mE AND 27T II II 9 s7r 2mm 5d 0 1 M 2 E Hyperbolic Orbit Equation 0 Coulomb Force f7 ZZ e27 2 thus k ZZ e2 o For E gt 0 we have hyperbolic motion eccentricity 12E 2 1 2E6 1 2E5 2 6 Mk2 771ZZ 822 ZZ e2 o Eqn of orbit Choose 6 7T ie 6 0 at periapsis 1 mk m 6 2 COSlt6 1 ZZ m 2 e ecos6 1 rhyperbola g 0 Now nd 11 from when rhyp gt 00 with 6 7T II 1 6cos6 1 0 thus cosIIl e


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