This preview shows page 1. Sign up to view the full content.
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 = L/2 R = L/(2cos ), where = 30 . 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. Newtons 2 nd law yields M(v 2 /R) = F 1 M(v 2 /(L/(2cos ))) = F 12 cos + F 13 cos = 2(Gm 2 /L 2 )cos 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 R/v = 2 (L/(2cos ))/v = 8.16 10 5 s...
View Full Document
This note was uploaded on 12/10/2011 for the course PHY 2048 taught by Professor Field during the Spring '08 term at University of Florida.
- Spring '08