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(a) What is the average density of the sun? (b) What is the average density of a neutron star that has the same mass as the sun but a radius of only 20.0 km?

Short Answer

Expert verified
The average density of the sun is about 1408 kg/m³, and that of a neutron star is approximately \( 5.934 \times 10^{17} \) kg/m³.

Step by step solution

01

Formula for Density

Density, \( \rho \), is defined as the mass \( M \) of an object divided by its volume \( V \). The formula is \( \rho = \frac{M}{V} \).
02

Volume of the Sun

The sun is approximately spherical, so we use the formula for the volume of a sphere: \( V = \frac{4}{3} \pi R^3 \). The radius of the sun is given as \( R = 696,340 \) km or \( 6.96340 \times 10^8 \) meters.
03

Calculate the Volume of the Sun

Using the radius of the sun, compute its volume: \[ V_{\text{sun}} = \frac{4}{3} \pi (6.96340 \times 10^8)^3 \approx 1.412 \times 10^{27} \text{ cubic meters}. \]
04

Density Calculation for the Sun

The mass of the sun is \( M_{\text{sun}} = 1.989 \times 10^{30} \) kg. Using the formula for density: \[ \rho_{\text{sun}} = \frac{1.989 \times 10^{30}}{1.412 \times 10^{27}} \approx 1408 \text{ kg/m}^3. \]
05

Volume of the Neutron Star

A neutron star has the same mass as the sun but a much smaller radius. Use the radius of 20.0 km or 20,000 meters to find the volume: \[ V_{\text{neutron}} = \frac{4}{3} \pi (20,000)^3 \approx 3.351 \times 10^{12} \text{ cubic meters}. \]
06

Density Calculation for the Neutron Star

With the mass of the neutron star being identical to the sun, \( M_{\text{neutron}} = 1.989 \times 10^{30} \) kg, the density is: \[ \rho_{\text{neutron}} = \frac{1.989 \times 10^{30}}{3.351 \times 10^{12}} \approx 5.934 \times 10^{17} \text{ kg/m}^3. \]

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Key Concepts

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

Sun
The Sun is the center of our solar system and a giant ball of hot, glowing gases. It is primarily composed of hydrogen, which undergoes nuclear fusion at its core to produce helium. This fusion process releases the energy that we feel as sunlight. The Sun is a massive object with a mass of about \( 1.989 \times 10^{30} \) kilograms. It is considered a nearly perfect sphere, meaning its volume can be calculated using the formula for the volume of a sphere: \( V = \frac{4}{3} \pi R^3 \). Here, \( R \) is the radius of the Sun, approximately 696,340 km or \( 6.96340 \times 10^8 \) meters.

To understand the concept of density in relation to the Sun, let's reflect on the fact that density is the ratio of mass to volume. By inserting the measurements into the equation for density, we calculate the density of the Sun as \( 1408 \text{ kg/m}^3 \). This may be surprising, as it's less dense than many substances on Earth, due to the Sun's vast size spreading its mass over a tremendous volume.

Understanding the density of the Sun helps scientists study its structure and behavior as well as predict its future interactions with the Earth and other bodies in the solar system.
Neutron Star
A neutron star is a fascinating astronomical object. It is the remnant of a massive star that reached the end of its life cycle and exploded in a supernova. What's left behind is a core composed almost entirely of neutrons. This core is unimaginably dense, as its mass rivals that of the Sun, yet it is packed into a volume with a radius of only 20.0 km (20,000 meters).

When calculating the volume of a neutron star using the sphere volume formula \( V = \frac{4}{3} \pi R^3 \), with \( R = 20,000 \) meters, we find it is approximately \( 3.351 \times 10^{12} \) cubic meters. Despite its tiny size compared to the Sun, its mass is the same, resulting in an astounding density of \( 5.934 \times 10^{17} \text{ kg/m}^3 \).

This incredible density is due to the collapse of matter in the original star, compressing the core to the point where protons and electrons have combined into neutrons. Because of these extreme conditions, neutron stars exhibit unique behaviors, such as emitting powerful radiation bursts and having intense gravitational fields.
Volume
Volume is a fundamental concept in physics and mathematics, representing the amount of space an object occupies. For three-dimensional objects, calculating volume involves geometric formulas that are specific to the object's shape. In our discussions related to astronomy, we commonly use the formula for the volume of a sphere, \( V = \frac{4}{3} \pi R^3 \), where \( R \) is the radius.

For large astronomical bodies like the Sun, the volume gives us insight into how its mass is distributed throughout its huge structure. By knowing the radius, the Sun's volume calculates to roughly \( 1.412 \times 10^{27} \) cubic meters. This allows scientists to make calculations related to the Sun’s density and gravitational field, both critical for understanding stellar physics.

In the case of neutron stars, volume helps to illustrate how a massive amount of matter can be crammed into a comparably tiny space, leading to densities that surpass those of any known materials on Earth. The volume calculation of \( 3.351 \times 10^{12} \) cubic meters for a neutron star reinforces the enormity of its density and the significance of volume as a concept in understanding cosmic phenomena.

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Most popular questions from this chapter

A sealed tank containing seawater to a height of 11.0 m also contains air above the water at a gauge pressure of 3.00 atm. Water flows out from the bottom through a small hole. How fast is this water moving?

When an open-faced boat has a mass of 5750 kg, including its cargo and passengers, it floats with the water just up to the top of its gunwales (sides) on a freshwater lake. (a) What is the volume of this boat? (b) The captain decides that it is too dangerous to float with his boat on the verge of sinking, so he decides to throw some cargo overboard so that 20\(\%\) of the boat's volume will be above water. How much mass should he throw out?

A closed container is partially filled with water. Initially, the air above the water is at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{Pa}\right)\) and the gauge pressure at the bottom of the water is 2500 Pa. Then additional air is pumped in, increasing the pressure of the air above the water by 1500 Pa. (a) What is the gauge pressure at the bottom of the water? (b) By how much must the water level in the container be reduced, by drawing some water out through a valve at the bottom of the container, to return the gauge pressure at the bottom of the water to its original value of 2500 Pa? The pressure of the air above the water is maintained at 1500 Pa above atmospheric pressure.

A block of balsa wood placed in one scale pan of an equalarm balance is exactly balanced by a 0.115-kg brass mass in the other scale pan. Find the true mass of the balsa wood if its density is 150 \(\mathrm{kg} / \mathrm{m}^{3} .\) Explain why it is accurate to ignore the buoyancy in air of the brass but not the buoyancy in air of the balsa wood.

A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill \(2200.355-\) L cans per minute. At point 2 in the pipe, the gauge pressure is 152 kPa and the cross-sectional area is 8.00 \(\mathrm{cm}^{2} .\) At point \(1,1.35\) above point \(2,\) the cross-sectional area is 2.00 \(\mathrm{cm}^{2} .\) Find the (a) mass flow rate; (b) volume flow rate; (c) flow speeds at points 1 and \(2 ;\) (d) gauge pressure at point \(1 .\)

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