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Certain stars are believed to be rotating at about 1 rot/s. If such a star has a diameter of 40 km, what is the linear speed of a point on the equator of the star? |
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Answer» Data : d = 40 km, f = 1 rot/s ∴ r = \(\frac{d}2\) = \(\frac{40\,km}2\) = 20 km = 2 x 104 m Linear speed, v = ωr = (2πf)r = (2 × 3.142 × 1)(2 × 104) = 6.284 × 2 × 104 = 1.257 × 10 m/s (or 125.6 km/s) |
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