/*! 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 26 Consider a pulsar, a collapsed s... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider a pulsar, a collapsed star of extremely high density, with a mass \(M\) equal to that of the \(\operatorname{Sun}\left(1.98 \times 10^{30} \mathrm{~kg}\right)\), a radius \(R\) of only \(12 \mathrm{~km}\), and a rotational period \(T\) of \(0.041 \mathrm{~s}\). By what percentage does the free-fall acceleration \(g\) differ from the gravitational acceleration \(a_{g}\) at the equator of this spherical star?

Short Answer

Expert verified
The free-fall acceleration differs from the gravitational acceleration by approximately 0.011%. The gravitational pull is thereby reduced by this margin due to rotational effects.

Step by step solution

01

Calculate the gravitational acceleration

The gravitational acceleration at the surface of the pulsar can be determined using the formula for gravitational acceleration: \[ a_g = \frac{GM}{R^2} \] where \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \), \( M \) is the mass of the star, and \( R \) is the radius of the star. Plug in the given values: \[ a_g = \frac{6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \times 1.98 \times 10^{30} \, \text{kg}}{(12000 \, \text{m})^2} \].
02

Compute the centripetal acceleration

The centripetal acceleration at the equator due to the rotation of the pulsar is given by: \[ a_c = \frac{4\pi^2R}{T^2} \] where \( R = 12000 \, \text{m} \) and \( T = 0.041 \, \text{s} \). Substitute these values to find \( a_c \): \[ a_c = \frac{4\pi^2 \times 12000}{0.041^2} \].
03

Determine the free-fall acceleration

The effective free-fall acceleration \( g \) at the equator takes into account both gravitational and centripetal forces: \[ g = a_g - a_c \]. Use the values of \( a_g \) and \( a_c \) calculated in the previous steps to find \( g \).
04

Calculate the percentage difference

The percentage difference between \( g \) and \( a_g \) is given by: \[ \text{Percentage Difference} = \left(\frac{a_g - g}{a_g}\right) \times 100\% \]. Substitute the values of \( g \) and \( a_g \) to find this percentage.

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Key Concepts

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

Gravitational Acceleration
Gravitational acceleration refers to the acceleration of an object due to the force of gravity from a star like a pulsar. In our exercise, the gravity is pulling the object towards the pulsar's surface. It is determined by the formula \[ a_g = \frac{GM}{R^2} \] where:
  • \(G\) is the gravitational constant \(6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2\).
  • \(M\) represents the mass of the pulsar, which in this case, equals that of the Sun \(1.98 \times 10^{30} \, \text{kg}\).
  • \(R\) is the radius of the pulsar \(12,000 \, \text{m}\).
Plug these values into the formula to compute the gravitational acceleration at the pulsar's surface. The result quantifies the force exerted by gravity alone, unaffected by any other forces or motions.
Centripetal Acceleration
When an object rotates, centripetal acceleration keeps it moving in a circular path. For a pulsar rotating at the equator, this acceleration stems from the rotation of the star itself. The centripetal acceleration can be calculated using:\[ a_c = \frac{4\pi^2R}{T^2} \]Here:
  • \(R\) is the radius \(12,000 \, \text{m}\),
  • \(T\) is the rotational period \(0.041 \, \text{s}\),
  • \(\pi\) is the mathematical constant approximately equal to 3.14159.
By substituting these values, we determine the centripetal acceleration due to the star's rotation. This force acts outwardly on the pulsar's equator and must be considered when evaluating the total acceleration experienced there.
Free-Fall Acceleration
Free-fall acceleration reflects the net acceleration experienced at the pulsar's equator. It considers both gravitational effects and the additional force of centripetal acceleration as the star rotates. We find it using:\[ g = a_g - a_c \]This calculation subtracts the outward centripetal acceleration from the gravitational acceleration, yielding the effective force pulling matter toward the pulsar's surface without other influences like air resistance. It's vital for understanding how forces balance at the star's rapidly spinning equator.
Percentage Difference
The percentage difference measures how much the free-fall acceleration diverges from the calculated gravitational acceleration. This involves comparing how rotational effects alter the gravitational pull experienced at the pulsar's surface. The calculation is:\[ \text{Percentage Difference} = \left(\frac{a_g - g}{a_g}\right) \times 100\% \]Substituting the values from previous calculations allows us to see this discrepancy. It highlights the pulsar's rapid rotation and shows the importance of considering other forces beyond mere gravitational pull in such environments. This concept is crucial for understanding how dynamic motions influence the forces we measure, especially in celestial mechanics.

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