- #1
chaose
- 10
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A triple star system consists of two stars each of mass m revoling in the same circular orbit of radius r around a central star of mass M. The two orbiting stars are always at opposite ends of a diameter of the orbit. Derive an expression for the period of the revolution of the stars.
I don't understand this problem that well. Any hints?
I thought that if the two stars are at opposite ends of the orbit, won't the gravitational forces from one cancel out the forces from the other star?
Then won't the period just be Kepler's Third law?
T^2 = (r^3)*(4pi^2)/(GM), G = universal gravitational constant, M = mass of central star, T = period of revolution, r = radius of the orbit, pi = 3.14159...
I don't understand this problem that well. Any hints?
I thought that if the two stars are at opposite ends of the orbit, won't the gravitational forces from one cancel out the forces from the other star?
Then won't the period just be Kepler's Third law?
T^2 = (r^3)*(4pi^2)/(GM), G = universal gravitational constant, M = mass of central star, T = period of revolution, r = radius of the orbit, pi = 3.14159...