Problem 204 - Binary Stars 

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Solutions: A: m1/m2 = r2/r1 B: T = 2*pi*(r1 + r2)*sqrt(r1/(G*m2)), for both stars together Supporting calcs & reasoning: For part A, center of orbits, must be the common center of mass between the two stars, such that as they orbit this center, the center of mass remains stationary in its own reference frame. Thus, m1*r1 = m2*r2, and we solve it for m1/m2 = r2/r1. For part B, The force causing centripetal acceleration of star 1, equals the force of gravity from star 2 and vice-versa. The two binary stars have the same orbital period (and same angular speed ω), in order to conserve angular momentum and center of mass. Thus: m1*r1*ω^2 = G*m1*m2/(r1+r2)^2 Solve for omega: ω= sqrt(G*m2/r1)/(r1 + r2) Translate omega into period T: T = 2*pi/ω T = 2*pi*(r1 + r2)*sqrt(r1/(G*m2))

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The ratio of the masses = 2 / their orbital periods are the same. r2 looks to be twice r1, which means mass 1 has twice the mass for the barycenter to be located where it is. both planets have equal, but opposite momentum, as they are continually exchanging momentum as they orbit. since mass1 has twice the mass, its tangential velocity is half of the tangential velocity of mass2. the circular distance of the orbit of mass 1 is half the circular distance of mass 2, but its moving half as fast, so the periods are the same.

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Just finished lecture two . Will try it once I finish entirely of 8.01 . Till then , Have a good day 🌟

To me the answer seems to be T= {2π(r1+r2)√r1}/(√Gm2) we also get T = {2π(r1+r2)√r2}/(√Gm1) which are equal since m1r1=m2r2, and it makes sense because the Time period will be the same for both stars

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a) m1 * r1 = m2 * r2 or m1 / m2 = r2 / r1 b) T² = 4 * pi² * (r1 + r2)³ / ( G * (m1 + m2)) or T = sqrt [ 4 * pi² * (r1 + r2)³ / ( G * (m1 + m2)) ] , both stars have the same period.

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a) m1/m2=r2/r1. b) The orbital periods are equal. But they cannot be determined because the data on the masses is insufficient.

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If to binary star systems were circular with radius of 1km in square space given at the rotational axis and gravitational electromagnetic pull of them how long would it take the two star systems to make a full 360 degrees rotational orbit around the planetary star system of stars if the stars net weight and mass were 1 trillion metric kilo tons how much gravitational pull would you need given upon the orbiting binary star systems in orbit around the planetary alignment and grid i know this answer just testing to see if anyone else out there can do the problem in their head like me

a) m1/m2 = r2/r1 b) T = 2pi * sqrt[ (r1+r2)^3 / G / (m1+m2) ] for both masses

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Is it realistic to assume that the angular velocities are equal? This can actually be derived... In CoM frame: r_cm=0=(m1r1+m2r2)/M r1=(m2/M)d, r2=(m1/M)d d=r1+r2 => g1=Gm1/d²=ω2²r2 g2=Gm2/d²=ω1²r1 => Gm1/d²=ω2²(m1/M)d, Gm2/d²=ω1²(m2/M)d => GM/d³=ω2², GM/d³=ω1² => GM/d³=ω2²=ω1²=ω² => μ=m1/m2 =ω2²(m1/M)d/ω1²(m2/M)d =ω2²r2/ω1²r1 =r2/r1 which is part (a) Period is T=2π/ω=2π√(d³/GM) =2π√((r1+r2)³/G(m1+m2)) WHICH IS PART (B). QED.

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a) m1/m2 = r2/r1 b) T = sqrt(4pi(r2+r1)^3/G(m1+m2))

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SOLUTION PART (a): m1/m2 = r2/r1 SOLUTION PART (b): T^2= 4π^2(r1+r2)^3/[G(m1+m2)] Solution to a) is readily obtained from m1 r1 = m2 r2 and b) goes like this [Verbatim form ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-HJXiz5phCZQ.html (8.01 Exams 2 & Solutions)] Let ω be the angular velocity, which must be the same for both stars. The centripetal acceleration must equal that of gravity: star 1: ω^2r1=m2 G/(r1+r2)^2 star 2: ω^2r2=m1 G/(r1+r2)^2 Add these two equations and substitute ω=2π/T to find 4π^2(r1+r2)/T^2=(m1+m2)G/(r1+r2)^2, from where T^2 can be solved.

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Determine the electronic potential at surface of gold where nucleus radius is 6.6×10¹⁵m. Z= 79 Proton = 1.6×10^-19C

Ans(i):- m1/m2= r2/r1 Ans(ii):- T= 2π(r1+r2)×√(r2/Gm1) Note- r2/m1 = r1/m2 = constant

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