The Crab Nebula is a cloud of glowing gas about 10...

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 $e x$ plosion, seen on earth in 1054 A.D. Energy is released by the Crab Nebula at a rate of about $5 \times 10^{31} \mathrm{~W},$ about $10^{5}$ 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 $0.0331 \mathrm{~s},$ and this period is increasing by $4.22 \times 10^{-13} \mathrm{~s}$ 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 $\left(3000 \mathrm{~kg} / \mathrm{m}^{3}\right)$ and to the density of an atomic nucleus (about $\left.10^{17} \mathrm{~kg} / \mathrm{m}^{3}\right) .$ Justify the statement that a neutron star is essentially a large atomic nucleus.

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 $e x$ plosion, seen on earth in 1054 A.D. Energy is released by the Crab Nebula at a rate of about $5 \times 10^{31} \mathrm{~W},$ about $10^{5}$ 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 $0.0331 \mathrm{~s},$ and this period is increasing by $4.22 \times 10^{-13} \mathrm{~s}$ 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 $\left(3000 \mathrm{~kg} / \mathrm{m}^{3}\right)$ and to the density of an atomic nucleus (about $\left.10^{17} \mathrm{~kg} / \mathrm{m}^{3}\right) .$ Justify the statement that a neutron star is essentially a large atomic nucleus.

Key Concepts

Rotational Kinetic Energy

Rotational kinetic energy is the energy stored in an object due to its rotation. It is given by the formula (1/2) I ?², where I is the moment of inertia and ? is the angular velocity. This concept is central when analyzing how energy is stored and transformed in rotating celestial bodies like neutron stars.

Moment of Inertia

The moment of inertia measures an object's resistance to changes in its rotational motion and depends on how the object's mass is distributed relative to its rotation axis. It plays a crucial role in linking the rotational kinetic energy to the angular velocity and is essential for understanding energy loss in rotating astronomical objects.

Spin-down Mechanism

The spin-down mechanism refers to the gradual reduction in the angular velocity of a rotating object due to energy losses, which may be radiated away or converted into other forms. In pulsars and neutron stars, this process helps explain how the decrease in rotational speed corresponds to the release of energy observed in surrounding nebulae.

Uniform Solid Sphere Model

The uniform solid sphere model is an idealization used to simplify the calculation of physical properties such as moment of inertia, radius, and density for spherical objects. By assuming a constant density throughout the object, this model provides a useful approximation for calculating parameters of celestial bodies like neutron stars.

Angular and Linear Velocity Relationship

The relationship between angular and linear velocity (v = ?R) allows one to determine the speed at a point on the surface of a rotating object from its angular velocity and radius. This conversion is vital for assessing whether the speeds involved are relativistic or comparable to the speed of light in astrophysical contexts.

Density Calculation and Comparison

Calculating density by dividing mass by volume offers insight into the compactness of an object. For celestial bodies such as neutron stars, comparing the calculated densities to those of ordinary matter or atomic nuclei highlights their extremely dense nature, underscoring the connection between these stars and nuclear matter.

Transcript

00:01 This problem is asking us about a neutron star that is spinning in the crab nebula.

00:06 It is a period of 0 .0331 seconds per rotation.

00:12 That period is changing at a constant rate of 4 .22 times 10 and the minus 13th seconds per second.

00:25 And it is releasing energy from that slowing down of its motion is releasing 5 times 10 to 30 first watts.

00:37 So from that, we are supposed to first calculate the moment of inertia of the neutron star, using that moment of inertia and the mass, which is given to us at 2 .78 times 10 to 30 kilograms.

00:50 We are supposed to calculate the radius of the neutron star.

00:55 And then from there, we can calculate the velocity of a point in its surface at the equator and the density of the neutron star.

01:07 So first, to calculate the inertia, we can use our energy equation.

01:12 E equals 1 .5 i.

01:17 Omega squared.

01:19 But in this case what we have is the power and we have a change in the period so we can instead use this power the power is equal to the derivative of the energy with respect to time and that will of course be equal to the derivative of our energy equation d ddt of one -half i omega squared and we can rewrite omega as a function of the period, which is 2 pi over t, and that'll be squared.

02:10 And so, oops, i forgot the derivative here, dt.

02:15 And so all we need to do is take this derivative.

02:18 One half, an i are going to be the same.

02:22 They're constant, so i'll just put an i over 2 out front times.

02:28 I'll use the chain rule here.

02:29 So first i'll take the derivative of the whole thing.

02:32 So that's 2 times 2 pi over t.

02:40 And then i will take the derivative of the inside.

02:44 And i'll put it out here.

02:47 And using the quotient rule, first we'll take the derivative of the top, multiply it by the bottom.

02:53 But the derivative of the top is 0, so that'll just be and then plus the, no, i'm sorry, minus the derivative of the bottom with respect to the top, which will give us dt over dt times 2 pi, and then that'll all be over t squared.

03:33 Okay, so now we can simplify this down a little.

03:38 If we multiply these things together, get rid of the zero, then we will, will be left with minus i times two pi squared d t d t all over t to the third and then all we need to do is solve for all this and we will get our moment of inertia so i will just multiply both sides by t over t to the third and divide both sides by this whole thing which will give us i is equal to 2 pi squared.

04:28 I just put the negative up front and do that correctly.

04:34 Times d t times p for the power...

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