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Get Full Access to University Physics - 13 Edition - Chapter 9 - Problem 98p
Get Full Access to University Physics - 13 Edition - Chapter 9 - Problem 98p

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# CALC Neutron Stars and Supernova Remnants. The Crab Nebula

ISBN: 9780321675460 31

## Solution for problem 98P Chapter 9

University Physics | 13th Edition

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Problem 98P

CALC Neutron Stars and Supernova Remnants.? The Crab Nebula is a cloud of glowing gas about 10 light-years across, located about 6500 light-years from the earth (?Fig. P9.86?). It is the remnant of a star that underwent a ?supernova explosion,? seen on earth in 1054 A.D. Energy is released by the Crab Nebula at a rate of about 5 X 1031 W, about 105 times the rate at which the sun radiates energy. The Crab Nebula obtains its energy from the rotational kinetic energy of a rapidly spinning ?neutron star? at its center. This object rotates once every 0.0331 s, and this period is increasing by 4.22 X 10-13 s for each second of time that elapses. (a) If the rate at which energy is lost by the neutron star is equal to the rate at which energy is released by the nebula, find the moment of inertia of the neutron star. (b) Theories of supernovae predict that the neutron star in the Crab Nebula has a mass about 1.4 times that of the sun. Modeling the neutron star as a solid uniform sphere, calculate its radius in kilometers. (c) What is the linear speed of a point on the equator of the neutron star? Compare to the speed of light. (d) Assume that the neutron star is uniform and calculate its density. Compare to the density of ordinary rock (3000 kg/m3) and to the density of an atomic nucleus (about 1017 kg/m3). Justify the statement that a neutron star is essentially a large atomic nucleus.

Step-by-Step Solution:

Solution 98P CALC Neutron Stars and Supernova Remnants. The Crab Nebula is a cloud of glowing gas about 10 light-years across, located about 6500 light-years from the earth (Fig. P9.86). It is the remnant of a star that underwent a supernova explosion, seen on earth in 1054 A.D. Energy is released by the Crab Nebula at a rate of about 5 X 1031 W, about 105 times the rate at which the sun radiates energy. The Crab Nebula obtains its energy from the rotational kinetic energy of a rapidly spinning neutron star at its center. This object rotates once every 0.0331 s, and this period is increasing by 4.22 X 10-13 s for each second of time that elapses. (a) If the rate at which energy is lost by the neutron star is equal to the rate at which energy is released by the nebula, find the moment of inertia of the neutron star. (b) Theories of supernovae predict that the neutron star in the Crab Nebula has a mass about 1.4 times that of the sun. Modeling the neutron star as a solid uniform sphere, calculate its radius in kilometers. (c) What is the linear speed of a point on the equator of the neutron star Compare to the speed of light. (d) Assume that the neutron star is uniform and calculate its density. Compare to the density of ordinary rock (3000 kg/m3) and to the density of an atomic nucleus (about 1017 kg/m3). Justify the statement that a neutron star is essentially a large atomic nucleus. Solution 98 P Step 1: Data given Rate of energy stored E = 5 × 10 W31 Time period of rotation is T = 00331 s 13 Rate of increase of time period dT/dt = 4.22 × 10 s 3 Density of rock = 3r00 kg /m Density of atomic nucleus = 10 kg /m 17 3 Step 1 : Kinetic energy of the rotation is given by 2 K = 1/2I But = 2/T 2 Hence we have K = (1/2) × I × (2/T) Differentiating with respect to time we have 42 dT dK/dt = I × T 3 × dt The rate of loss of energy is given by 4 I dT E = ( 3 ) × dt T Rearranging the equation to find moment of inertia we have 3 I = ET2 (1/dT ) 4 dt Substituting values we get 31 3 5×10 W×(0.0331s) 13 I = 42 × (1/4.22 × 10 s) 27 12 1 I = 4.59 × 10 × 2.36 × 10 s 38 2 I = 1.09 × 10 kg m 38 2 Hence we have the moment of inertia as 1.09 × 10 kg m

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