When a large star becomes a supernova, its core may be compressed so tightly tha

Question

When a large star becomes a supernova, its core may be compressed so tightly that it becomes a neutron star, with a radius of about 23 km (about the size of the San Francisco area). If a neutron star rotates once every second, what is the speed of a particle on the star's equator?
What is the particle's centripetal acceleration in meters per second squared?
What is the particle's centripetal acceleration in g-units (ratio to g)?
If the neutron star rotates even faster, what happens, i.e. what would be the particle's centripetal acceleration in g-units if the period was 0.5 s?

Explanation / Answer

(1) v = 2 R / T

        = 2 * 3.14 * 22 x 103 / 1.0

        = 1.38 x 105 m/s

(2) a = v2 / R

        = (1.38 x 105)2 / 22 x 1000

        = 867644.8 m/s2

(3) a = 88535.18 g

(4) acceleration and velocity increases.

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