Applying Newton's second law, F= = ma yields F.for each star.macGMMMv2(2r)2Solving for the mass, we have4v2 rM =We can write r in terms of the period T by considering the time interval and distance of one complete cycle.The distance traveled in one orbit is the circumference of the stars' common orbit, so 2nr = vT. Therefore,M= 4vr - (av) ) = 2v°T4v?rGSubstituting the values, we have10° m/s) (I86400 s/d)M =N. m?/kg?)11T(6.67x10x 1032 kg.This is equivalent to|x 1032kg)solar masses.(1.99x1030kg) Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midwaybetween them. This statement implies that the masses of the two stars are equal (see figure below). Assumethe orbital speed of each star is v| = 225 km/s and the orbital period of each is 11.6 days. Find the mass Mof each star. (For comparison, the mass of our Sun is 1.99 x 1030 kg.)MXCMMPart 1 of 3 - ConceptualizeFrom the given data, it is difficult to estimate a reasonable answer to this problem without working throughthe details and actually solving it. A reasonable guess might be that each star has a mass equal to or slightlylarger than our Sun because fourteen days is short compared to the periods of all the Sun's planets.Part 2 of 3 - CategorizeThe only force acting on each star is the central gravitational force of attraction which results in a centripetalacceleration. When we solve Newton's second law, we can find the unknown mass in terms of the variablesgiven in the problem.

Since, You have only asked bout the calculation part M=2v3TπG Here, "v" is the velocity that is…

Applying Newton's second law, f = ma yields F = ma. for each star. Mv2 (2r)2 GMM Solving for the mass, we have 4v2 r M = We can write r in terms of the period T by considering the time interval and distance of one complete cycle. The distance traveled in one orbit is the circumference of the stars' common orbit, so 2ër = vT. Therefore, 4v² - ()) - 2vT M = Substituting the values, we have |× 10³ m/s)' ([ ]a) (86400 s/d) M = N. m?/kg?) -11 1(6.67x10° x 1032 kg. This is equivalent to 1032kg) solar masses. (1.99×1030 kg)

Applying Newton's second law, f = ma yields F = ma. for each star. Mv2 (2r)2 GMM Solving for the mass, we have 4v2 r M = We can write r in terms of the period T by considering the time interval and distance of one complete cycle. The distance traveled in one orbit is the circumference of the stars' common orbit, so 2ër = vT. Therefore, 4v² - ()) - 2vT M = Substituting the values, we have |× 10³ m/s)' ([ ]a) (86400 s/d) M = N. m?/kg?) -11 1(6.67x10° x 1032 kg. This is equivalent to 1032kg) solar masses. (1.99×1030 kg)

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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