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In a binary star system, two stars A and B revolve about their common centre of mass. The masses of star A and star B are 15 * 10³⁰ kg and 45 * 10³⁰ kg respectively. What is the ratio of the area swept per unit time by star A to that swept by star B?
- 1:1
- 1:2
- 1:3
- 1:4
Correct answer: 1:3
Solution
COM condition: m_A * r_A = m_B * r_B => 15 r_A = 45 r_B => r_A/r_B = 3/1 (A is farther from COM since it's lighter). Both stars have same angular velocity omega. Rate of area swept = (1/2)*r²*omega (areal velocity). So ratio of area swept by A to B = r_A²/r_B² = 9/1. But 9:1 is not an option. Let me reconsider: perhaps it means angular momentum / (2*mass)? Areal velocity per star = L/(2m) where L is angular momentum. L_A = m_A * r_A * v_A = m_A * r_A² * omega. Areal velocity of A = L_A/(2*m_A) = r_A²*omega/2. Similarly for B. Ratio = r_A²/r_B² = 9. Still 9:1. None of the options match. Let me try the other interpretation: ratio of area in equal time means (1/2)*r_A²*omega*t: (1/2)*r_B²*omega*t = r_A²:r_B² = 9:1. Since 9:1 is not an option and 1:3 looks most likely given the mass ratio, perhaps the question means rate of area swept considering total area of orbit: pi*r_A² vs pi*r_B² per period T (same T) -> same 9:1 ratio. With mass ratio 1:3 and r_A:r_B = 3:1, ratio = 9:1. Answer should be 9:1 but closest option is 1:3 - this question may have errors in the original.
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