/*! 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 35 The sun emits energy in the form... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 32: Problem 35

The sun emits energy in the form of electromagnetic waves at a rate of \(3.9 \times 10^{26} \mathrm{~W}\). This energy is produced by nuclear reactions deep in the sun's interior. (a) Find the intensity of electromagnetic radiation and the radiation pressure on an absorbing object at the surface of the sun (radius \(\left.r=R=6.96 \times 10^{5} \mathrm{~km}\right)\) and at \(r=R / 2,\) in the sun's interior. Ignore any scattering of the waves as they move radially outward from the center of the sun. Compare to the values given in Section 32.4 for sunlight just before it enters the earth's atmosphere. (b) The gas pressure at the sun's surface is about \(1.0 \times 10^{4} \mathrm{~Pa} ;\) at \(r=R / 2,\) the gas pressure is calculated from solar models to be about \(4.7 \times 10^{13} \mathrm{~Pa}\). Comparing with your results in part (a), would you expect that radiation pressure is an important factor in determining the structure of the sun? Why or why not?

Short Answer

Expert verified
The specific numerical answers depend on the calculations and resulting comparisons. However, in general, if the calculated radiation pressures are significantly smaller than the gas pressures given in the exercise, then it can be concluded that radiation pressure plays a less significant role in the determination of the Sun's structure.

Step by step solution

01

Calculation of intensity and radiation pressure at the sun's surface

The first step is to calculate the intensity of radiation at the surface of the Sun. Using the formula for intensity, we substitute for \(Power = 3.9 \times 10^{26} W\) and \(Area = 4\pi (6.96 \times 10^{8} km)^{2}\). Now, the area must be in metres, so we multiply the radius in km by \(10^{3}\). Then we calculate the radiation pressure by substituting the calculated intensity into the formula for radiation pressure. Here, we use \(c = 3 \times 10^{8} ms^{-1}\).
02

Calculation of intensity and radiation pressure at \(r=R/2\)

Repeat the same procedure as in Step 1, but now with \(r = R / 2\), hence \(Area = 4\pi (6.96 \times 10^{8}/2 km)^{2}\). Make sure to convert radii from km to m again.
03

Comparing with Section 32.4 values

This step involves a comparison of the calculated values with those provided in Section 32.4 (not given in this exercise). One way to approach this could be checking if the numbers for intensity and pressure match, or if they are off by a constant factor.
04

Comparing with given gas pressure values

Compare the radiation pressure values on the Sun's surface and at half of its radius with the gas pressures given in part (b) of the exercise (i.e., \(1.0 \times 10^{4} Pa\) at the Sun's surface and \(4.7 \times 10^{13} Pa\) at \(r = R/2\) respectively). This will involve critical analysis of the values and their implications.
05

Analysis of radiation pressure's importance in determining solar structure

Now draw conclusions derived from the comparisons made in above steps. The point here is to understand whether radiation pressure is of significant magnitude as compared to gas pressure, and hence, would play a vital role in determining the Sun's structure.

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 91影视!

Key Concepts

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

Intensity of Electromagnetic Radiation
The intensity of electromagnetic radiation is a measure of how much energy passes through a certain area in a certain amount of time. Essentially, it tells us how much energy you would feel if you were standing in the path of the radiation. Imagine the intensity as the density of energy: the higher the intensity, the more 'packed' the area is with energy.

For the sun, which is essentially a massive energy-emitting body, the intensity can be calculated using the formula: \[Intensity = \frac{Power}{Area}\]
where the power is the total energy emitted by the sun per second, also known as luminosity, and the area is the sphere surrounding the sun where the energy is spread out. As the distance from the sun increases, the same amount of energy is spread over a larger area, resulting in a lower intensity. This is why the intensity of sunlight is less on Earth than it is on the sun's surface.

Understanding the intensity is crucial, not just for studying the sun, but for many applications ranging from calculating the potential for solar power on Earth to understanding how different frequencies of light are absorbed or transmitted through various materials.
Nuclear Reactions in the Sun
The sun is like a natural nuclear reactor, where nuclear reactions occur at its core. These reactions are mostly hydrogen nuclei fusing to form helium, which releases a massive amount of energy. The process is known as nuclear fusion. It occurs under extreme conditions of high temperature and pressure at the sun鈥檚 core and is responsible for producing almost all the sun's energy.

The formula that describes the energy released in these reactions is Einstein鈥檚 famous equation: \[E = mc^{2}\]
where 'E' represents energy, 'm' is the mass of the hydrogen that's converted to energy, and 'c' is the speed of light. Here's a fascinating fact: every second, the sun converts about 4 million tons of its own mass into energy! Yet, it's so large that it has been doing this for billions of years and will continue to do so for billions more. The energy produced by these nuclear reactions then makes its way through various layers of the sun before reaching the surface and radiating out into space as sunlight.
Solar Structure
The sun, a glowing sphere of gas, is structured in layers, much like an onion. Each layer has distinct characteristics and plays a role in the sun's overall functionality. Starting from the inside out, the core is where nuclear fusion takes place. Surrounding the core, we have the radiative zone, where energy travels outward slowly in the form of electromagnetic radiation.

Next is the convective zone, where the energy transfer occurs by convection (hot plasma rising, cooling and sinking back down). The visible surface, or photosphere, is where the light we see is emitted. Above that is the chromosphere and finally, the outermost layer, the corona, which extends millions of kilometers into space and is visible during a total solar eclipse.

Understanding the sun鈥檚 structure helps us grasp why the sun's surface and the region halfway towards the center have very different properties. For instance, the pressure and temperature change drastically between these regions, which affects both the gas and radiation pressures. It's the balance of these pressures that supports the sun's structure and prevents it from collapsing under its own gravity or exploding outwards.

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

Interplanetary space contains many small particles referred to as interplanetary dust. Radiation pressure from the sun sets a lower limit on the size of such dust particles. To see the origin of this limit, consider a spherical dust particle of radius \(R\) and mass density \(\rho\). (a) Write an expression for the gravitational force exerted on this particle by the sun (mass \(M\) ) when the particle is a distance \(r\) from the sun. (b) Let \(L\) represent the luminosity of the sun, equal to the rate at which it emits energy in electromagnetic radiation. Find the force exerted on the (totally absorbing) particle due to solar radiation pressure, remembering that the intensity of the sun's radiation also depends on the distance \(r\). The relevant area is the cross-sectional area of the particle, not the total surface area of the particle. As part of your answer, explain why this is so. (c) The mass density of a typical interplanetary dust particle is about \(3000 \mathrm{~kg} / \mathrm{m}^{3} .\) Find the particle radius \(R\) such that the gravitational and radiation forces acting on the particle are equal in magnitude. The luminosity of the sun is \(3.9 \times 10^{26} \mathrm{~W}\). Does your answer depend on the distance of the particle from the sun? Why or why not? (d) Explain why dust particles with a radius less than that found in part (c) are unlikely to be found in the solar system. [Hint: Construct the ratio of the two force expressions found in parts (a) and (b).]

Two square reflectors, each \(1.50 \mathrm{~cm}\) on a side and of mass \(4.00 \mathrm{~g},\) are located at opposite ends of a thin, extremely light, \(1.00 \mathrm{~m}\) rod that can rotate without friction and in vacuum about an axle perpendicular to it through its center (Fig. \(\mathbf{P 3 2 . 3 7}\) ). These reflectors are small enough to be treated as point masses in moment-of-inertia calculations. Both reflectors are illuminated on one face by a sinusoidal light wave having an electric field of amplitude \(1.25 \mathrm{~N} / \mathrm{C}\) that falls uniformly on both surfaces and always strikes them perpendicular to the plane of their surfaces. One reflector is covered with a perfectly absorbing coating, and the other is covered with a perfectly reflecting coating. What is the angular acceleration of this device?

Electromagnetic waves propagate much differently in conductors than they do in dielectrics or in vacuum. If the resistivity of the conductor is sufficiently low (that is, if it is a sufficiently good conductor), the oscillating electric field of the wave gives rise to an oscillating conduction current that is much larger than the displacement current. In this case, the wave equation for an electric field \(\vec{E}(x, t)=E_{y}(x, t) \hat{\jmath}\) propagating in the \(+x\) -direction within a conductor is $$ \frac{\partial^{2} E_{y}(x, t)}{\partial x^{2}}=\frac{\mu}{\rho} \frac{\partial E_{y}(x, t)}{\partial t} $$ where \(\mu\) is the permeability of the conductor and \(\rho\) is its resistivity. (a) \(\mathrm{A}\) solution to this wave equation is \(E_{y}(x, t)=E_{\max } e^{-k_{\mathrm{C}} x} \cos \left(k_{\mathrm{C}} x-\omega t\right)\) where \(k_{\mathrm{C}}=\sqrt{\omega \mu / 2 \rho}\). Verify this by substituting \(E_{y}(x, t)\) into the above wave equation. (b) The exponential term shows that the electric field decreases in amplitude as it propagates. Explain why this happens. (Hint: The field does work to move charges within the conductor. The current of these moving charges causes \(i^{2} R\) heating within the conductor, raising its temperature. Where does the energy to do this come from?) (c) Show that the electric-field amplitude decreases by a factor of \(1 / e\) in a distance \(1 / k_{\mathrm{C}}=\sqrt{2 \rho / \omega \mu},\) and calculate this distance for a radio wave with frequency \(f=1.0 \mathrm{MHz}\) in copper (resistivity \(1.72 \times 10^{-8} \Omega \cdot \mathrm{m} ;\) permeability \(\mu=\mu_{0}\) ). Since this distance is so short, electromagnetic waves of this frequency can hardly propagate at all into copper. Instead, they are reflected at the surface of the metal. This is why radio waves cannot penetrate through copper or other metals, and why radio reception is poor inside a metal structure.

Consider each of the following electric- and magnetic-field orientations. In each case, what is the direction of propagation of the wave? (a) \(\overrightarrow{\boldsymbol{E}}=E \hat{\imath}, \overrightarrow{\boldsymbol{B}}=-B \hat{\jmath} ;\) (b) \(\overrightarrow{\boldsymbol{E}}=E \hat{\jmath}, \overrightarrow{\boldsymbol{B}}=B \hat{\imath} ;(\mathrm{c}) \overrightarrow{\boldsymbol{E}}=-E \hat{\boldsymbol{k}}, \overrightarrow{\boldsymbol{B}}=-B \hat{\imath}\) (d) \(\overrightarrow{\boldsymbol{E}}=E \hat{\imath}, \overrightarrow{\boldsymbol{B}}=-B \hat{\boldsymbol{k}}\)

Consider each of the electric- and magnetic-field orientations given next. In each case, what is the direction of propagation of the wave? (a) \(\vec{E}\) in the \(+x\) -direction, \(\overrightarrow{\boldsymbol{B}}\) in the \(+y\) -direction; (b) \(\overrightarrow{\boldsymbol{E}}\) in the \(-y\) -direction, \(\vec{B}\) in the \(+x\) -direction; (c) \(\vec{E}\) in the \(+z\) -direction, \(\vec{B}\) in the \(-x\) -direction; (d) \(\overrightarrow{\boldsymbol{E}}\) in the \(+y\) -direction, \(\overrightarrow{\boldsymbol{B}}\) in the \(-z\) -direction.
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Electricity and Magnetism

Read Explanation

Particle Model of Matter

Read Explanation

Nuclear Physics

Read Explanation

Radiation

Read Explanation

Thermodynamics

Read Explanation
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.