Problem 1 It is estimated that perhaps as many as 85% of all stars are in binary

Question

Problem 1 It is estimated that perhaps as many as 85% of all stars are in binary systems, which consist of two stars bound to each other by their mutual gravity and orbiting around the system’s center of mass. For this problem, consider the mathematically simplest (but physically unlikely) case in which the two stars have the same mass. The system’s center of mass is then located halfway between the two. Take the distance from the center of each star to the center of mass to be R = 1.5 × 1010m, and the mass of each star to be M1 = M2 M = 8.96 × 1030kg.

(a)Derive a formula, based on this information, for the angular speed of the stars (it is the same for both), as they revolve around the center of mass. Hint: think of the force of gravity between the stars, and of what it takes to get an object to move in a circle!

(b)Using the formula derived in part (a), give a numerical value for the angular speed of the stars.

(c)Find the period of the motion, both in seconds and in days.

Explanation / Answer

A)

Gravitational force n one star = F = GMM/(2R)^2 = GM^2/(4R^2) ------------(1)

B)

By Newton’s law

Fnet = ma

F= Fc

GM^2/(4R^2) = Mv^2/R

Gives,

v= sqrt(RGM)/2R

Angular speed = w= v/R = sqrt(RGM)/2R^2---------(1)

B)

Plugging values in above eqn (1),

w = sqrt(RGM)/2R^2 = sqrt(1.5*10^10*6.67*10^-11*8.96*10^30)/(2*(1.5*10^10)^2) = 6.65*10^-6 rad/s

C)

T = 2R/v = 2R/sqrt(RGM)/2R = 4R^2/sqrt(RGM)

Plugging values,

T = (4*3.14*(1.5*10^10)^2)/sqrt(1.5*10^10*6.67*10^-11*8.96*10^30)= 9.44*10^5 s = 10.92 days

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