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Get Full Access to University Physics - 13 Edition - Chapter 13 - Problem 72p
Get Full Access to University Physics - 13 Edition - Chapter 13 - Problem 72p

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# CP Binary Star—Different Masses. Two stars, with masses M1 ISBN: 9780321675460 31

## Solution for problem 72P Chapter 13

University Physics | 13th Edition

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Problem 72P

CP Binary Star—Different Masses. Two stars, with masses M1 and M2 are in circular orbits around their center of mass. The star with mass M1has an orbit of radius the star with mass has an orbit of radius R2(a) Show that the ratio of the orbital radii of the two stars equals the reciprocal of the ratio of their masses—that is (c) The two stars in a certain binary star system move in circular orbits. The first star, Alpha, has an orbital speed of 36.0 km/s The second star, Beta, has an orbital speed of 12.0 km/s The orbital period is 137 d. What are the masses of each of the two stars? (d) One of the best candidates for a black hole is found in the binary system called A0620-0090. The two objects in the binary system are an orange star, V616 Monocerotis, and a compact object believed to be a black hole (see Fig. 13.27). The orbital period of A0620-0090 is 7.75 hours, the mass of V616 Monocerotis is estimated to be 0.67 times the mass of the sun, and the mass of the black hole is estimated to be 3.8 times the mass of the sun. Assuming that the orbits are circular, find the radius of each object’s orbit and the orbital speed of each object. Compare these answers to the orbital radius and orbital speed of the earth in its orbit around the sun

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a) The radii R and R are measured with respect to the center of mass, and so 1 2 M R = M R 1 1 2 2 R1 M 2 R 2= M 1 b)The forces on each star are equal in magnitude, so the product of the mass and the radial accelerations are equal 4 M 1 1 4 M 2 2 T 2 = T2 1 2 GM M1 2 F g (R +R )2 The equivalent expression for the period are 1 2 2 2 2 4 R1(1 +R2) 2 4 R2(R1+R2) M T2= G and M 1 = G Adding above expressions 2 3 2 4 (R1+R2) (M +1M )T 2 G 2(R +R )/2 T = 1 2 G(M 1M )2 2R c)v = T vT R = 2 3 R = 36×10 m/s×137 days×86,400 seconds/day 2 10 = 6.78 × 10 m 3 R = 12×10 m/s×137 days×86,400 seconds/day 2 10 = 2.26 × 10 m 4 (R +R )3 (M + M ) = T G substitute the values of t and radii 2 10 10 3 (M + M ) = 4 (6.78×10 m+2.26×10 m) [(1.37 days(86,400 seconds /day)] (6.6N.m /kg2 30 = 3.12 × 10 kg M R M = = 3M R 30 29 30 4 M =3.12 × 10 kg, or M = 7.8 × 10kg and M = 2.34 × 10 kg d) use the relation in part (a) and part (b) R = ( M )R M 0.67 = ( 3.8) = 0.176R 3 2 2 R R = (M M )TG/4 substitute M and T values For monocerotis 9 R 1.9 × 10 m 2 v 4.4 × 10 km/s For block hole 8 R 34 × 10 m v = 77 km/s

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