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Solution_Set_04

# Solution_Set_04 - SOLUTION SET 4 Physics 2022 1 d = 1.496 x...

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SOLUTION SET 4 Physics 2022 1. d = 1.496 x 10 8 km / 9.46 x 10 12 km/ly = 1.58 x 10 -5 ly No too small 2. d pluto = 39.5 AU / 206,265 AU/pc = 1.92 x 10 -4 pc d prox_c = 1.30 pc d prox_c / d pluto = 1.30 / (1.92 x 10 -4 ) = 6790 3a. d = (2 x 10 6 AU) / (206,265 AU/pc) = 9.7 pc 3b. p = 1 / d = 1 / 9.7 pc = 0.10 arcsec YES 4. p = 0.153 arcsec d = 1 / p = 1 / 0.153 = 6.54 pc 5a. v T = 4.74 / p = 4.74 (8.67 arcsec/yr) / (0.255 arcsec) = 161 km/s 5b. v = sqrt [ v T 2 + v r 2 ] = sqrt [ (161) 2 + (246) 2 ] = 294 km/s 5c. Because the velocity is positive, it is moving away from the Sun. 6a. v T = 4.74 d d = (40 km/s) / [ 4.74 x 0.08] = 106 pc 6b. v T = 4.74 d = (4.74) (10.358 arcsec/yr) (106 pc) = 5.2 x 10 3 km/s The star needs a tangential velocity of about 5200 km/s to exhibit a proper motion equal to that of Barnard’s star. 7. v 2 = v T 2 + v r 2 v T 2 = (120) 2 – (48) 2 v T = 110 km/s 8a.     = v / c v = [ (656.41 – 656.28) / 656.28 ] (3 x 10 5 km/s) = 59 km/s 8b. Redshifted (i.e., positive), therefore it is receding from us.

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9. b A = b B d A = 12 pc d B = 48 pc L A / L B = (d A / d B ) 2 (b A / b B ) = (12 / 48) 2 (1 / 1) = 0.062 (B has greater luminosity) (Square the ratio of the distances – Inverse Square Law) 10. L C = L D d C = 60 pc d D = 12 pc L C / L D = (d C / d D ) 2 (b C / b D ) = (60 / 12) 2 (b C / b D ) = (1 / 1) b C / b
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Solution_Set_04 - SOLUTION SET 4 Physics 2022 1 d = 1.496 x...

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