/*! 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 6 (a) What is the average density ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) What is the average density of the sun? (b) What is the average density of a neutron star hat has the same mass as the sun but a radius of only 20.0 \(\mathrm{km}\) ?

Short Answer

Expert verified
(a) The average density of the Sun is approximately \( 1.41 \times 10^3 \text{ kg/m}^3 \). (b) The average density of the neutron star is approximately \( 5.94 \times 10^{17} \text{ kg/m}^3 \).

Step by step solution

01

Understand the formula for density

The formula for density is given by \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). To find the density of any object, including stars, we need its mass and volume.
02

Calculate the volume of the Sun

The Sun is approximately a sphere, so its volume \( V \) can be calculated using the formula for the volume of a sphere: \( V = \frac{4}{3} \pi r^3 \). The average radius \( r \) of the Sun is approximately \( 6.96 \times 10^5 \) km, or \( 6.96 \times 10^8 \) meters.
03

Calculate the Sun's volume

Plug the Sun's radius into the volume formula: \[ V = \frac{4}{3} \pi (6.96 \times 10^8)^3 \approx 1.41 \times 10^{27} \text{ cubic meters} \].
04

Calculate the density of the Sun

The mass of the Sun is approximately \( 1.99 \times 10^{30} \) kg. Using the density formula: \[ \text{Density of Sun} = \frac{1.99 \times 10^{30}}{1.41 \times 10^{27}} \approx 1.41 \times 10^3 \text{ kg/m}^3 \].
05

Calculate the volume of a neutron star

A neutron star with the same mass as the Sun is also a sphere, but it has a much smaller radius of 20.0 km, or \( 2.0 \times 10^4 \) meters. Use the volume formula: \[ V = \frac{4}{3} \pi (2.0 \times 10^4)^3 \approx 3.35 \times 10^{12} \text{ cubic meters} \].
06

Calculate the density of the neutron star

Using the neutron star's volume and the Sun's mass: \[ \text{Density of Neutron Star} = \frac{1.99 \times 10^{30}}{3.35 \times 10^{12}} \approx 5.94 \times 10^{17} \text{ kg/m}^3 \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Volume of a Sphere
When calculating the volume of a sphere, it's essential to understand the formula used to find it. A sphere is a perfectly round geometric object in three-dimensional space, similar to a ball. The volume of a sphere can be determined using the formula: \[ V = \frac{4}{3} \pi r^3 \] where:\
    \
  • \( V \) is the volume,
  • \
  • \( \pi \) (Pi) is a constant approximately equal to 3.14159,
  • \
  • \( r \) is the radius of the sphere.
  • \
This formula is integral for various applications, from calculating the space inside a ball to understanding the massive size of stars like the Sun. When we apply this volume formula to the Sun with a radius of approximately \( 6.96 \times 10^8 \) meters, we find that its volume is a staggering \( 1.41 \times 10^{27} \) cubic meters. Similarly, if we apply the same formula to a neutron star with a much smaller radius, such as 20 km (\( 2.0 \times 10^4 \) meters), we get a volume of only \( 3.35 \times 10^{12} \) cubic meters. Understanding this formula allows us to better grasp the vast differences in volume between different astrophysical objects.
Neutron Star Density
Neutron stars provide a fascinating example of extreme density in the universe. Despite having a mass similar to other large stars, such as the Sun, neutron stars pack this mass into a much smaller volume. This is why they are much denser than many other astrophysical objects.To understand the density of a neutron star, we use the formula for density: \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \] For instance, a typical neutron star with the same mass as the Sun (approximately \( 1.99 \times 10^{30} \) kg) but an immensely smaller radius of 20 km results in a dramatic increase in density. By finding the volume of such a neutron star using our sphere volume formula, we calculate its density to be \( 5.94 \times 10^{17} \) kg/m³. This incredible density arises because the neutrons are packed closely together, more than any other material in the universe. Neutron star density eclipses even that of white dwarfs and suggests the compactness of nuclear particles in these mysterious remnants of supernova explosions.
Astrophysical Objects
Astrophysical objects like the Sun and neutron stars span a vast range of sizes and densities, offering valuable insights into the diversity of the universe. While the Sun is a typical star, serving as a bright example of hydrogen fusion in action, its relatively low density reflects the large space between particles in a gaseous star.In stark contrast, neutron stars are remnants of supernova explosions. They present an extreme state of matter where gravity has compacted the remains into a high-density object. Comparing the two:
    \
  • The Sun, with a density of approximately \( 1.41 \times 10^3 \) kg/m³, shows a much lower density than might be expected, contrasting sharply with the denser, remnant cores of dead stars.
  • \
  • Neutron stars have densities reaching \( 5.94 \times 10^{17} \) kg/m³, illustrating the incredible transformation that occurs during a star's death.
  • \
Such differences not only highlight the variety of evolutionary paths these objects can take but also remind us of the complexity of forces at work in the universe, from nuclear fusion in stars to gravitational collapse resulting in neutron stars. Understanding these celestial lifecycles is key to appreciating the dynamic processes shaping our cosmos.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Submarines on Europa. Some scientists are eager to send a remote-controlled submarine to Jupiter's moon Europa to search for life in its oceans below an icy crust. Europa's mass has been measured to be \(4.78 \times 10^{22} \mathrm{kg}\) , its diamcter is 3130 \(\mathrm{km}\) , and it has no appreciable atmosphere. Assume that the layer of ice at the surface is not thick enough to exert substantial force on the water. If the windows of the submarine you are designing are 25.0 \(\mathrm{cm}\) square and can stand a maximum inward force of 9750 \(\mathrm{N}\) per window, what is the greatest depth to which this submarine can safely dive?

Hydraulic Lift II. The piston of a hydraulic automobile lift is 0.30 \(\mathrm{m}\) in diameter. What gauge pressure, in pascals, is required to lift a car with a mass of 1200 \(\mathrm{kg}\) ? Also express this pressure in atmospheres.

Untethered helium balloons, floating in a car that has all the windows rolled up and outside air vents closed, move in the direction of the car's acceleration, but loose balloons filled with air move in the opposite direction. To show why, consider only the horizontal forces acting on the balloons. Let \(a\) be the magnitude of the car's forward acceleration. Consider a horizontal tube of air with a cross-sectional area \(A\) that extends from the windshield, where \(x=0\) and \(p=p_{0},\) back along the \(x\) -axis. Now consider a volume element of thickness \(d x\) in this tube. The pressure on its front surface is \(p\) and the pressure on its rear surface is \(p+d p\) . Assume the air has a constant density \(\rho\) . (a) Apply Newton's second law to the volume element to show that \(d p=\rho a d x\) . (b) Integrate the result of part (a) to find the pressure at the front surface in terms of \(a\) and \(x\) . (c) To show that considering \(\rho\) constant is reasonable, calculate the pressure difference in atm for a distance as long as 2.5 \(\mathrm{m}\) and a large acceleration of 5.0 \(\mathrm{m} / \mathrm{s}^{2}\) . (d) Show that the net horizontal force on a balloon of volume \(\dot{V}\) is \(\rho V a\) . (e) For negligible friction forces, show that the acceleration of the balloon (average density \(\rho_{\text { bal }} )\) is \(\left(\rho / \rho_{\text { bal }}\right) a,\) so that the acceleration relative to the car is \(a_{n e 1}=\left[\left(\rho / \rho_{\text { bel }}\right)-1\right] a\) (f) Use the expression for \(a_{\mathrm{rel}}\) in part (e) to explain the movement of the balloons.

A hydrometer consists of a spherical bulb and a cylindrical stem with a cross- sectional area of 0.400 \(\mathrm{cm}^{2}\) (see Fig. 14.13a). The total volume of bulb and stem is \(13.2 \mathrm{cm}^{3} .\) When immersed in water, the hydrometer floats with 8.00 \(\mathrm{cm}\) of the stem above the water surface. When the hydrometer is immersed in an organic fluid, 3.20 \(\mathrm{cm}\) of the stem is above the surface. Find the density of the organic fluid. (Nore: This illustrates the precision of such a hydrometer. Relatively small density differences give rise to relatively large differences in hydrometer readings.)

A uniform lead sphere and a uniform aluminum sphere have the same mass. What is the ratio of the radius of the aluminum sphere to the radius of the lead sphere?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.