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 39 km (about the size of the San Francisco area). If a neutron star rotates once every second,

A) what is the speed of a particle on the star's equator?

B) What is the particle's centripetal acceleration in meters per second squared?

C) What is the particle's centripetal acceleration in g-units (ratio to g)?

D) 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

A) speed of a particle on the star's equator = 2 * 3.14 * 39000/1

                                                                        = 244920 m/sec

                                                                          = 244.92 km/sec

B)   particle's centripetal acceleration = 244920 * 244920/39000

                                                              = 1538097.6 m/sec2

C) particle's centripetal acceleration in g-units = 1538097.6 /9.8

                                                                            = 156948.73   g units

D) particle's centripetal acceleration in g-units if the period was 0.5 s = 4 * 156948.73

                                                                                                                = 627794.92    g units

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