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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 \((V=\) \(\left.2.0 \times 10^{-7} \mathrm{m}^{3}\right)\) were made from this material, how much would it weigh (in pounds)?

Short Answer

Expert verified
a) The density is approximately \(3.73 \times 10^{18} \mathrm{kg/m}^3\). b) The dime would weigh about \(1.65 \times 10^{12}\) pounds.

Step by step solution

01

Calculate Volume of Neutron Star

The formula for the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \). Given the radius \( r = 1.2 \times 10^3 \mathrm{m} \), substitute this into the formula to find the volume:\[ V = \frac{4}{3} \pi (1.2 \times 10^3)^3 \]Calculating this gives \( V \approx 7.238 \times 10^9 \mathrm{m}^3 \).
02

Calculate Density of Neutron Star

Density \( \rho \) is defined as mass per unit volume. We have the mass \( m = 2.7 \times 10^{28} \mathrm{kg} \) and the volume from Step 1 as \( V \approx 7.238 \times 10^9 \mathrm{m}^3 \). Use the density formula:\[ \rho = \frac{m}{V} = \frac{2.7 \times 10^{28} \mathrm{kg}}{7.238 \times 10^9 \mathrm{m}^3} \] Solving this, \( \rho \approx 3.73 \times 10^{18} \mathrm{kg/m}^3 \).
03

Calculate Mass of Neutron Star Material Dime

Use the density of the neutron star to find the mass of a dime. The volume of the dime is given as \( V = 2.0 \times 10^{-7} \mathrm{m}^3 \).\[ m = \rho \times V = 3.73 \times 10^{18} \mathrm{kg/m}^3 \times 2.0 \times 10^{-7} \mathrm{m}^3 \] This calculation gives \( m \approx 7.46 \times 10^{11} \mathrm{kg} \).
04

Convert Mass of Dime to Pounds

To convert from kilograms to pounds, use the conversion factor: 1 kg = 2.20462 pounds. Multiply the mass obtained by this factor:\[ m \approx 7.46 \times 10^{11} \mathrm{kg} \times 2.20462 \text{ pounds/kg} \] This results in a mass of approximately \( 1.65 \times 10^{12} \text{ pounds} \).

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Key Concepts

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

Volume Calculation
When calculating the volume of a neutron star, we treat it like a sphere due to its nearly perfect spherical shape. Spheres have a specific formula for volume which is:
  • \( V = \frac{4}{3} \pi r^3 \)
This formula helps calculate the volume based on the radius \( r \) of the sphere. For our neutron star example, the given radius is \( 1.2 \times 10^3 \ \mathrm{m} \).
To find out how much space our neutron star occupies, plug in this radius into the sphere volume formula. Doing so, we find the volume of the neutron star to be approximately \( 7.238 \times 10^9 \ \mathrm{m}^3 \). This tells us how large this massive celestial body is. Volume is a crucial component when calculating density as it represents the space an object takes in the universe.
Mass Conversion
Understanding mass conversion is important to express mass in forms we might find more relatable. Our task involved converting the mass of a tiny piece of neutron star material—a small dime's volume worth—into pounds.
Initially, we calculated the mass of this neutron matter sphere using the density formula and found it to be approximately \( 7.46 \times 10^{11} \ \mathrm{kg} \). Since everyday measurements in the US are frequently in pounds, we converted this kilogram value into pounds.
Here's how this works using the conversion factor:
  • 1 kilogram equals approximately 2.20462 pounds.
Therefore, multiplying the mass in kilograms by this conversion factor, we get a staggeringly large number: about \( 1.65 \times 10^{12} \) pounds! This helps put the extreme mass of neutron star material into perspective for tangible understanding.
Density Formula
The density formula gives us a way to measure how much mass is packed into a given volume. This is useful for understanding the properties of neutron stars and other celestial bodies. The formula is as follows:
  • Density \( \rho = \frac{m}{V} \)
This relationship means density is the mass per unit volume.
For our neutron star, with a mass \( m \) of \( 2.7 \times 10^{28} \ \mathrm{kg} \) and a calculated volume \( V \) of \( 7.238 \times 10^9 \ \mathrm{m}^3 \), we found the density to be approximately \( 3.73 \times 10^{18} \ \mathrm{kg/m}^3 \).
Such density values are incredibly large, explaining why neutron stars are unbelievably dense, essentially packing massive amounts of matter into a compact space.
Neutron Star Properties
Neutron stars are remnants of massive stars that have ended their life cycles. Once their nuclear fuel is depleted, these stars explode in a supernova, leaving the core to collapse under gravity into an extremely dense state.
Here are some defining properties of neutron stars:
  • Composition: Primarily made up of neutrons, lending them their name.
  • Mass: Generally between 1.4 and 3 times the mass of our sun, but squeezed into a tiny sphere.
  • Density: As showcased, they have densities that dwarf most known materials, reaching values like \( 3.73 \times 10^{18} \ \mathrm{kg/m}^3 \).
  • Size: Despite their mass, they are only about 10 kilometers in radius.
The study of neutron stars helps physicists understand the properties of matter under extreme conditions—conditions that cannot be replicated on Earth.

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Most popular questions from this chapter

(a) The mass and the radius of the sun are, respectively, \(1.99 \times\) \(10^{30} \mathrm{kg}\) and \(6.96 \times 10^{8} \mathrm{m} .\) What is its density? (b) If a solid object is made from a material that has the same density as the sun, would it sink or float in water? Why? (c) Would a solid object sink or float in water if it were made from a material whose density was the same as that of the planet Saturn (mass \(=5.7 \times 10^{26} \mathrm{kg},\) radius \(\left.=6.0 \times 10^{7} \mathrm{m}\right) ?\) Provide a reason for your answer.

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 hollow cubical box is \(0.30 \mathrm{m}\) on an edge. This box is floating in a lake with one-third of its height beneath the surface. The walls of the box have a negligible thickness. Water from a hose is poured into the open top of the box. What is the depth of the water in the box just at the instant that water from the lake begins to pour into the box from the lake?

An aneurysm is an abnormal enlargement of a blood vessel such as the aorta. Because of the aneurysm, the normal cross-sectional area \(A_{1}\) of the aorta increases to a value of \(A_{2}=1.7 A_{1} .\) The speed of the blood \(\left(\rho=1060 \mathrm{kg} / \mathrm{m}^{3}\right)\) through a normal portion of the aorta is \(v_{1}=0.40 \mathrm{m} / \mathrm{s} .\) Assuming that the aorta is horizontal (the person is lying down), determine the amount by which the pressure \(P_{2}\) in the enlarged region exceeds the pressure \(P_{1}\) in the normal region.

A liquid is flowing through a horizontal pipe whose radius is \(0.0200 \mathrm{m}\). The pipe bends straight upward through a height of \(10.0 \mathrm{m}\) and joins another horizontal pipe whose radius is \(0.0400 \mathrm{m} .\) What volume flow rate will keep the pressures in the two horizontal pipes the same?

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