University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

272

Pages:

5
Author:

Jabin F.

# Binary Star Problem -> Two stars revolving in circles under their mutual gravitational attraction. → line joining passes through centre of circle (always). → Separation btw two stars r₀ = r₁ + r₂, where r₁, r₂ are radii of their orbits. Now radial EOM for m₁: → F₁x = m₁a₁x → (Gm₁m₂)/(r₀³) = m₁(r₁ω²) → -(Gm₁m₂)/(r₀²) = m₁(-r₁ω²) for m₂: → F₂x = m₂a₂x → -(Gm₁m₂)/(r₀³) = m₂(-r₂ω²) by adding (1) and (2) we get (Gm₁m₂)/(r₀³) = (r₁ + r₂)ω² = (r₀)ω² → ω² = G(m₁ + m₂)/(r₀³) → ω = √[G(m₁ + m₂)/(r₀³)] In binary star problem, any question can be solved using eq. (1) & (2). → ratio of linear speed / kinetic energy / angular momentum. → ratio of linear speed to tangential speed. ω₁m₁ ω₀r₀ m₂ m₁ g₁ g₂ P-133 A binary star system consists of two stars S₁ and S₂ with masses m₁ and 3m resp. separated by a distance r. If both S₁ and S₂ individually follow circular orbits around the centre of the mass with instantaneous speeds v₁ and v₂ resp. the ratio of speeds is: Solution: P-134 Q-25 A particle moves in 3rd space in a central potential V(r) = kr⁻³, where k is a constant. The angular frequency ω for a circular orbit depends on its radius. R as → Solution: V(r) = kr⁻³ → F = -dV/dr = -3kr⁻⁴ radial EOM → -3kr⁻⁴ = ma = m(-rω²) → ω² = 2k/mr⁴ → ω = 2√(k/m) * r⁻² → ω ∝ R⁻²