Unformatted text preview: stars. We label them star 1, 2 and 3. Star 1 rotates in a circle. Therefore, it accelerates toward the center with magnitude v 2 /R, where R is the radius of the circular orbit. The radius of the circle and the edge length of the triangle are related as Rcos θ = L/2 ⇒ R = L/(2cos ), where = 30 º . The centripetal acceleration is provided by two gravitational forces from other 2 stars. The two forces have the same magnitude and each makes an angle = 30 to the radial direction. Newton’s 2 nd law yields M(v 2 /R) = F 1 M(v 2 /(L/(2cos θ ))) = F 12 cos θ + F 13 cos θ = 2(Gm 2 /L 2 )cos θ v = √ (Gm/L) = 1.334 × 10 5 m/s b) What is the period of the circular motion of the stars? The period of the circular motion is T = 2 π R/v = 2 π (L/(2cos θ ))/v = 8.16 × 10 5 s...
View
Full Document
 Spring '08
 Field
 Mass, Rotation, triangle

Click to edit the document details