Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 5 - Circular Motion; Gravitation - General Problems: 75

Answer

(a) The gravitational force provides the centripetal force for each star to continue moving in a circle around the midpoint between the two stars. (b) The mass of each star is $9.6 \times 10^{29} ~kg$

Work Step by Step

(a) The gravitational force provides the centripetal force for each star to continue moving in a circle around the midpoint between the two stars. Let $r$ be half the distance between the stars: $v = \frac{2\pi r}{T}$ First, we need to find $T$: $T = (12.6 ~yrs)(365 ~d/yr)(24 ~h/d)(3600 ~s/h)$ $T = 397353600 ~s$ We can use the expression for $v$ and the value for $T$ in the force equation. Let's consider the circular motion of one of the stars: $\frac{Gm^2}{(2r)^2} = \frac{mv^2}{r}$ $m = \frac{4v^2r}{G} = \frac{16\pi^2 r^3}{GT^2}$ $m = \frac{(16\pi^2)(4.0 \times 10^{11} ~m)^3}{(6.67 \times 10^{-11} ~N\cdot m^2/kg^2)(397353600 ~s)^2}$ $m = 9.6 \times 10^{29} ~kg$ The mass of each star is $9.6 \times 10^{29} ~kg$
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