If a neutron star has a diameter of 10 km and rotates 642 times a second what is the speed at the neutrons star equator as a percentage of the speed of light?
2007-04-24 18:43:38 · 6 answers · asked by Cloud 7 1 in Science & Mathematics ➔ Astronomy & Space
Much better. The speed at the equator is given by the angular velocity times the radius. So
V = 2π*642 Rad/sec * 5,000m
V = 20169024 m/s which is
20169024/300000000 = .067 or about 6.7% of the speed of light.
Doug
2007-04-24 18:51:20 · answer #1 · answered by doug_donaghue 7 · 0⤊ 0⤋
2 pi r is the circumference
2 pi 5 = 10pi
10pi * 642 = 20 169.0248 km/s or 20 169 024.8 m/s
speed of light = 299 792 458 m / s
20 169 024.8/299 792 458 = 0.06728
6.728% of the speed of light
2007-04-24 18:52:29 · answer #2 · answered by James H 2 · 0⤊ 0⤋
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2016-12-04 20:07:59 · answer #3 · answered by philipp 4 · 0⤊ 0⤋
I honestly figured out this answer to be sure and the first two answers here were right.
6.7% (or .067)
2007-04-24 19:36:00 · answer #4 · answered by Edward 5 · 0⤊ 0⤋
doug dude got it spot on give him the best.
2007-04-24 23:41:15 · answer #5 · answered by Eddyking4 2 · 0⤊ 0⤋
C'mon - did you REALLY need help with this question??
2007-04-24 21:05:23 · answer #6 · answered by Anonymous · 0⤊ 1⤋