The Crab Pulsar is a neutron star, rotating with a period of about 33.085 ms. It is estimated to have an equatorial radius of 15 km, about average for a neutron star. The pulsar is slowing in its rotation so that it is expected to come to rest 9.5 × 10¹⁰ s in the future. Assuming it is slowing with a constant angular acceleration, what is the tangential acceleration of an object on the neutron star's equator? A) 3.0 × 10⁻⁵ m/s² B) 2.0 × 10⁻⁹ m/s² C) 3.2 × 10⁻¹⁰ m/s² D) 7.6 × 10⁻⁷ m/s² E) 4.8 × 10⁻⁶ m/s²

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The correct answer is A.The tangential acceleration of an object on the neutron star's equator can be calculated using the formula:$$a_t = rω^2$$where:* $a_t$ is the tangential acceleration* $r$ is the radius of the neutron star* $ω$ is the angular velocity of the neutron starThe angular velocity of the neutron star can be calculated using the formula:$$ω = 2π/T$$where:* $ω$ is the angular velocity* $T$ is the period of rotationThe period of rotation of the neutron star is given as 33.085 ms, which is equal to 0.033085 s.The radius of the neutron star is given as 15 km, which is equal to 15,000 m.Substituting these values into the formula for angular velocity, we get:$$ω = 2π/0.033085 s = 188.49 rad/s$$Substituting the values of $r$ and $ω$ into the formula for tangential acceleration, we get:$$a_t = 15,000 m * (188.49 rad/s)^2 = 5.18 × 10^8 m/s^2$$Therefore, the tangential acceleration of an object on the neutron star's equator is 5.18 × 10^8 m/s^2.

The Crab Pulsar Is a Neutron Star, Rotating with a Period