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1. The document discusses orbital parameters of a binary star system. It provides the orbital period, radii, and velocities of the two stars. 2. Doppler shift formulas are used to calculate the orbital velocities of the stars based on differences between maximum and minimum observed wavelengths from the stars. 3. Gravitational force equations are used to calculate the masses of the stars based on their orbital radii and velocities.
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Solution 3 i ph o 38
- 1. Question “Pink” 1.1 Period = 3.0 days = s106.2 5 × . (0.4) Period = ω π2 (0.2) ⇒ 15 srad104.2 −− ×=ω . (0.2) 1.2 Calling the minima in the diagram 1, 90.001 ==αII and 63.002 == βII , we have: α 1 1 4 1 2 2 1 2 1 0 = += T T R R I I (0.4) α β = − −= 4 1 2 2 1 2 1 2 11 T T R R I I (0.4) (or equivalent relations) From above, one finds: 6.1 1 2 1 2 1 =⇒ − = R R R R β α (0.2+0.2) and 4.1 1 1 2 14 2 1 =⇒ − − = T T T T α β (0.2+0.2) 2.1) Doppler-Shift formula: c v ≅ ∆ 0λ λ (or equivalent relation) (0.4) Maximum and minimum wavelengths: 7.5897max,1 =λ Å , 1.5894min,1 =λ Å 0.5899max,2 =λ Å , 8.5892min,2 =λ Å Difference between maximum and minimum wavelengths: 6.31 =∆λ Å , 2.62 =∆λ Å (All 0.6) Using the Doppler relation and noting that the shift is due to twice the orbital speed: (Factor of two 0.4) 0 1 1 2λ λ∆ = cv 4 102.9 ×= m/s (0.2) 0 2 2 2λ λ∆ = cv 5 106.1 ×= m/s (0.2)
- 2. The student can use the wavelength of central line and maximum (or minimum) wavelengths. Marking scheme is given in the Excel file. 2.2) As the center of mass is not moving with respect to us: (0.5) 1 2 2 1 v v m m = = 7.1 (0.2) 2.3) Writing ω i i v r = for 2,1=i , we have (0.4) 9 1 108.3 ×=r m, (0.2) 9 2 105.6 ×=r m (0.2) 2.4) 10 21 100.1 ×=+= rrr m (0.2) 3.1) The gravitational force is equal to mass times the centrifugal acceleration 2 2 2 2 1 2 1 12 21 r v m r v m r mm G == (0.7) Therefore, = = 1 2 1 2 2 2 2 2 2 1 rG vr m rG vr m (0.1) ⇒ ×= ×= kg103 kg106 30 2 30 1 m m (0.2 + 0.2)
- 3. 4.1) As it is clear from the diagram, with one significant digit, 4=α . (0.6) 4.2) As we have found in the previous section: 4 = Sun i Suni M M LL (0.2) So, Watt103 28 1 ×=L (0.2) Watt104 27 2 ×=L (0.2) 4.3) The total power of the system is distributed on a sphere with radius d to produce 0I , that is: 2 21 0 4 d LL I π + = (0.5) 0 21 4 I LL d π + =⇒ = 18 101× m (0.2) = 100 ly. (0.2) 4.4) d r =≅ θθ tan = 8 101 − × rad. (0.2 + 0.2) 4.5) A typical optical wavelength is 0λ . Using uncertainty relation: ≅= r d D 0λ 50 m. (0.2 + 0.2)