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Chapter 11: Problem 4

Neutron stars consist only of neutrons and have unbelievably high densities. A typical mass and radius for a neutron star might be \(2.7 \times 10^{28}\) \(\mathrm{kg}\) and \(1.2 \times 10^{3}\) \(\mathrm{m} .\) (a) Find the density of such a star. (b) If a dime \(\left(V=2.0 \times 10^{-7} \mathrm{m}^{3}\right)\) were made from this material, how much would it weigh (in pounds)?

Short Answer

Expert verified
The density of the neutron star is approximately \(3.73 \times 10^{18}\) kg/m\(^3\). The dime would weigh about 1.645 trillion pounds.

Step by step solution

01

Calculate Volume of the Neutron Star

To calculate the density, we need the volume of the neutron star. Since it is spherical, use the formula for the volume of a sphere: \[ V = \frac{4}{3} \pi r^3 \] where \( r = 1.2 \times 10^{3} \) m is the radius. First, calculate \( r^3 \): \( 1.2^3 = 1.728 \), and then substitute into the formula.\[ V = \frac{4}{3} \pi (1.728 \times 10^{9}) \] Calculate \( V \): approximately 7.24 \( \times 10^{9} \) m\(^3\).
02

Calculate Density of the Neutron Star

Density \( \rho \) is defined as mass divided by volume: \[ \rho = \frac{m}{V} \] Substitute the values for mass \( m = 2.7 \times 10^{28} \) kg and volume \( V \approx 7.24 \times 10^{9} \) m\(^3\): \[ \rho = \frac{2.7 \times 10^{28}}{7.24 \times 10^{9}} \approx 3.73 \times 10^{18} \text{ kg/m}^3 \]
03

Calculate Mass of the Neutron Star Material in a Dime

To find out how much a dime made of the neutron star material would weigh, calculate its mass using the density: \[ m = \rho \times V \] The volume of the dime \( V = 2.0 \times 10^{-7} \) m\(^3\). Substitute: \[ m = 3.73 \times 10^{18} \times 2.0 \times 10^{-7} \approx 7.46 \times 10^{11} \text{ kg} \]
04

Convert Mass from Kilograms to Pounds

To convert the mass from kilograms to pounds, use the conversion factor \(1 \text{ kg} = 2.20462 \text{ lbs}\). Calculate: \[ 7.46 \times 10^{11} \times 2.20462 \approx 1.645 \times 10^{12} \text{ lbs} \] Thus, the weight of the dime would be approximately 1.645 trillion pounds.

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Key Concepts

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

Density Calculation
Understanding density is crucial when dealing with astronomical objects like neutron stars. Density is determined by how much mass is contained within a certain volume. The formula for density \( \rho \) is given by:\[ \rho = \frac{m}{V} \]where \( m \) represents the mass and \( V \) represents the volume of the object.

In the case of a neutron star, with a mass of \( 2.7 \times 10^{28} \text{ kg} \) and a volume of approximately \( 7.24 \times 10^{9} \text{ m}^3 \), we calculate the density by dividing the mass by the volume:- Substitute the values into the density formula.- Calculate to find the density \( \rho \approx 3.73 \times 10^{18} \text{ kg/m}^3 \).

This extraordinarily high density shows how tightly packed matter is in a neutron star, which is several times denser than any found material on Earth.
Volume of a Sphere
To comprehend the size of spherical objects like neutron stars, we use the formula for the volume of a sphere:\[ V = \frac{4}{3} \pi r^3 \]Here, \( r \) is the radius of the sphere. Given a neutron star with radius \( 1.2 \times 10^{3} \text{ m} \):
  • First, calculate \( r^3 \) by raising the radius to the third power.
  • Then substitute \( r^3 \) into the volume formula.
Compute this to find that the neutron star's volume is approximately \( 7.24 \times 10^{9} \text{ m}^3 \).

This large volume, when paired with the star's massive density, illustrates how immense yet compact these celestial bodies can be. Despite their small relative volume compared to other types of stars, neutron stars pack a punch with their extreme mass.
Mass Conversion to Weight
Converting mass to weight is necessary when we want to know how heavy an object would be on Earth, typically expressed in pounds. Mass is different from weight, as mass measures the amount of matter, while weight considers the force of gravity on that mass. To convert mass in kilograms to weight in pounds, use the conversion factor:\[ 1 \text{ kg} = 2.20462 \text{ lbs} \]

Given a dime with mass \( 7.46 \times 10^{11} \text{ kg} \) crafted from neutron star material:
  • Multiply the mass by 2.20462.
  • Calculate to get the weight in pounds, approximately \( 1.645 \times 10^{12} \) pounds.
This conversion highlights how such dense material, even in small quantities like a dime, results in remarkable weight, showcasing the extreme nature of neutron star matter.

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Most popular questions from this chapter

(a) The volume flow rate in an artery supplying the brain is \(3.6 \times 10^{-6} \mathrm{m}^{3} / \mathrm{s} .\) If the radius of the artery is \(5.2\) \(\mathrm{mm},\) determine the average blood speed. (b) Find the average blood speed at a constriction in the artery if the constriction reduces the radius by a factor of \(3 .\) Assume that the volume flow rate is the same as that in part (a).

In the human body, blood vessels can dilate, or increase their radii, in response to various stimuli, so that the volume flow rate of the blood increases. Assume that the pressure at either end of a blood vessel, the length of the vessel, and the viscosity of the blood remain the same, and determine the factor \(R_{\text {dilated }} / R_{\text {normal }}\) by which the radius of a vessel must change in order to double the volume flow rate of the blood through the vessel.

Water flows straight down from an open faucet. The cross-sectional area of the faucet is \(1.8 \times 10^{-4} \mathrm{m}^{2},\) and the speed of the water is \(0.85 \mathrm{m} / \mathrm{s}\) as it leaves the faucet. Ignoring air resistance, find the cross-sectional area of the water stream at a point \(0.10 \mathrm{m}\) below the faucet.

Three fire hoses are connected to a fire hydrant. Each hose has a radius of \(0.020 \mathrm{m} .\) Water enters the hydrant through an underground pipe of radius \(0.080 \mathrm{m} .\) In this pipe the water has a speed of \(3.0 \mathrm{m} / \mathrm{s}\) (a) How many kilograms of water are poured onto a fire in one hour by all three hoses? (b) Find the water speed in each hose.

A small crack occurs at the base of a 15.0 -m-high dam. The effective crack area through which water leaves is \(1.30 \times 10^{-3} \mathrm{m}^{2}\) (a) Ignoring viscous losses, what is the speed of water flowing through the crack? (b) How many cubic meters of water per second leave the dam?
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