/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 When a large star becomes a supe... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(20 \mathrm{~km}\) (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 and (b) what is the magnitude of the particle's centripetal acceleration? (c) If the neutron star rotates faster, do the answers to (a) and (b) increase, decrease, or remain the same?

Short Answer

Expert verified
(a) 125,664 m/s; (b) ~789,569,000 m/s^2; (c) Both increase.

Step by step solution

01

Identify Given Values

The problem gives us the radius of the neutron star, which is approximately 20 km, or 20,000 meters. It also states that the neutron star rotates once every second.
02

Calculate Circumference of the Neutron Star

The circumference of a circle, which in this case is the path traveled by a particle on the equator, is calculated using the formula \(C = 2\pi r\). For a radius \(r = 20,000\) meters, the circumference is:\[C = 2 \pi \times 20,000 = 40,000 \pi \approx 125,664 \, \text{m}\]
03

Determine Speed of a Particle on the Equator

The speed \(v\) of a particle on the equator can be found using the formula \(v = \frac{C}{T}\), where \(T\) is the period of rotation. Given that \(T = 1\) second, the speed is:\[v = \frac{125,664}{1} = 125,664 \, \text{m/s}\]
04

Compute Centripetal Acceleration

Centripetal acceleration \(a_c\) is given by the formula \(a_c = \frac{v^2}{r}\). Using \(v = 125,664\,\text{m/s}\) and \(r = 20,000\,\text{m}\), the acceleration is:\[a_c = \frac{(125,664)^2}{20,000} \approx 789,568,972 \, \text{m/s}^2\]
05

Analyze Effect of Faster Rotation

If the neutron star rotates faster, both the speed \(v\) and the centripetal acceleration \(a_c\) increase. This is because the rotational period \(T\) would decrease, increasing \(v = \frac{C}{T}\), and \(a_c = \frac{v^2}{r}\) directly depends on \(v^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

supernova
A supernova is a colossal explosion that happens at the end of a massive star’s lifecycle. When a star exhausts its nuclear fuel, its core can no longer withstand gravitational forces. As a result, the core collapses inwards, and the outer layers are expelled, creating a supernova.
This phenomenon is crucial in terms of cosmic recycling. It spreads elements throughout space, which can later be part of new stars, planets, or even life forms. About one in every ten of these explosions leads to the creation of a neutron star.
Neutron stars are incredibly dense remnants of the core, often having a radius of just about 20 kilometers. Their density is such that a sugar-cube-sized amount of neutron-star material would weigh about as much as Mount Everest! They can also rotate extremely fast, sometimes making several rotations per second.
centripetal acceleration
Centripetal acceleration is the acceleration experienced by an object moving in a circular path, directed towards the center of the circle. This type of acceleration keeps the object moving along its circular path rather than flying off in a straight line.
When considering a particle at the equator of a rotating neutron star, the centripetal acceleration is crucial to understanding the forces acting upon it. The formula for centripetal acceleration is:
\[a_c = \frac{v^2}{r}\]
where \(v\) is the speed, and \(r\) is the radius of the circular path.
In the case of the neutron star exercise, the particle is moving at an incredible speed due to the star's fast rotation. Thus, it experiences a significant centripetal acceleration because both \(v\) and \(v^2\) are large, due to the rapid rotational motion.
rotational dynamics
Rotational dynamics is an area of physics that deals with the motion of rotating bodies and the forces acting on them. It includes concepts such as rotational inertia, angular velocity, and torque.
In the context of a neutron star, rotational dynamics helps understanding how such a star can spin with tremendous speed. Key factors include:
  • Angular velocity, which refers to the speed at which the star rotates.
  • Moment of inertia, which depends on the mass distribution within the star. For neutron stars, a smaller radius results in a smaller moment of inertia.
The exercise shows that as a neutron star rotates faster, its increased angular velocity directly affects the speed of particles on its equator and thereby influences the centripetal acceleration. Rotational dynamics helps to explain how neutron stars can sustain such rapid rotations without tearing themselves apart, thanks to their intense gravitational forces holding them together.
particle motion on equator
The motion of a particle on the equator of a rotating object, such as a neutron star, involves both linear and circular aspects. As the star spins, so too does the particle at its equator.
In the given exercise, we determine the speed, or linear velocity, of such a particle. Linear velocity on the equator is derived from the circumference of the star's rotational path and the rotational period.
Keep in mind the relationship:
  • Linear velocity \(v\) is determined by \(v = \frac{C}{T}\), where \(C\) is the circumference and \(T\) is the period of rotation.
The particle's velocity and the star's rotation are interdependent. If the star spins more rapidly, the speed at which the particle is "flung" around increases. Consequently, this increase affects the particle's centripetal acceleration and impacts the dynamics of motion, further showcasing the intricate balance in a neutron star's rotational behavior.

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