# Lec09 - G M/r 2 = v 2/r v = 2 π r P thus G M/r 2 = 2 π...

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Assignments • For Monday, 26 Sep: – Read Ch. 5 and Do Online Quiz #3 • For Friday, 30 Sep: – Do Written HW #2

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Outing to Mount Cuba Observatory • Would be optional , and note: – space limited to ca. 44 – ca. 12 miles north of Newark -- car pool. .. – can’t guarantee clear skies – but if cloudy, still can see scopes; use planetarium • Straw poll: – How many seriously interested? – How many can drive? – Is a Sunday night best?? e.g. Sunday, 13 Nov http://www.physics.udel.edu/MCAO/
Deriving Kepler’s 3rd law For circular orbits around large mass: Newton. centrifugal gravity = acceleration ; speed = circum/period

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Unformatted text preview: G M/r 2 = v 2 /r ; v = 2 π r / P thus G M/r 2 = ( 2 π a/P) 2 /r = 4 π 2 r/P 2 so M = (4 π 2 /G) r 3 /P 2 constant Must apply to Earth’s orbit around sun, so if: M in Msun; a = r in AU ; P in years, then (4 π 2 /G)=1 M = a 3 /P 2 Circular Orbital motion but V=2 π r/P r 3 /P 2 =(G/4 π 2 ) M M/M sun = (r/au) 3 (P/yr) 2 _______ V 2 = GM/r with M=M sun with and r=a => Kepler’s 3 rd law: (a/au) 3 = (P/yr) 2 Pop Quiz 1. A planet orbits a star of mass 2 M sun at a distance of 2 AU. In earth years, about how often would beings on this planet celebrate birthdays? a. 1 b. 1/2 c. 4 d. 2...
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## This note was uploaded on 02/01/2012 for the course PHYS 133 taught by Professor Staff during the Fall '08 term at University of Delaware.

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Lec09 - G M/r 2 = v 2/r v = 2 π r P thus G M/r 2 = 2 π...

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