An astronomer is measuring the light emitted by a binary star system. One of the stars appears to be much more massive than the other. The astronomer notices that the smaller star orbits the larger one in a nearly circular orbit at a distance of r 845 106 km. Through careful observation, she measures the smaller star's orbital speed to be v 12.0 km/s. Using Keppler's Laws she calculates the mass of the larger star. What is it? Use G = 6.67 x 10-¹1 N-m²/kg². M = x 1030 kg (± 0.05 × 1030 kg)

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Chapter1: Units, Trigonometry. And Vectors
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An astronomer is measuring the light emitted by a binary star system. One of
the stars appears to be much more massive than the other. The astronomer
notices that the smaller star orbits the larger one in a nearly circular orbit at
fr = 845 ×10º|
a distance of r = 845 x 106 km. Through careful observation, she measures
the smaller star's orbital speed to be v 12.0 km/s. Using Keppler's Laws
she calculates the mass of the larger star. What is it? Use G = 6.67 x 10-¹1
N-m²/kg².
-
M =
30
x1030 kg (± 0.05 × 10³0 kg)
Transcribed Image Text:An astronomer is measuring the light emitted by a binary star system. One of the stars appears to be much more massive than the other. The astronomer notices that the smaller star orbits the larger one in a nearly circular orbit at fr = 845 ×10º| a distance of r = 845 x 106 km. Through careful observation, she measures the smaller star's orbital speed to be v 12.0 km/s. Using Keppler's Laws she calculates the mass of the larger star. What is it? Use G = 6.67 x 10-¹1 N-m²/kg². - M = 30 x1030 kg (± 0.05 × 10³0 kg)
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