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Andrew [12]
3 years ago
12

Consider a neutron star of radius 10 km that spins with a period of 0.8 seconds. Imagine a person is standing at the equator of

this neutron star. Calculate the centripetal acceleration of this person
Physics
1 answer:
Arada [10]3 years ago
8 0

Answer:

a = 616850.28 m/s²

Explanation:

Given that,

The radius of the neutron star, r = 10 Km

                                                     = 10,000 m

The time period of the neutron star, T = 0.8 s

The centripetal acceleration is given by the formula,

                                  a = v²/r

The linear velocity is given by the relation,

                                    v = rω

The time taken to complete one complete rotation is given by the relation

                                   T = 2π /ω

Where,

                                    ω = 2π / T

Substituting v and ω into the equation for centripetal acceleration. It becomes

                                    a = 4π²r/T²

Substituting the given values in the above equation

                                      a = 4π² x 10000 / 0.8²

                                         = 616850.28 m/s²

Hence, the centripetal acceleration of this person is, a = 616850.28 m/s²

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