/*! 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 59 How large is the sun? By measuri... [FREE SOLUTION] | 91Ó°ÊÓ

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How large is the sun? By measuring the spectrum of wavelengths of light from our sun, we know that its surface temperature is \(5800 \mathrm{~K}\). By measuring the rate at which we receive its energy on carth, we know that it is radiating a total of \(3.92 \times 10^{26} \mathrm{~J} / \mathrm{s}\) and behaves nearly like an ideal blackbody. Use this information to calculate the diameter of our sun.

Short Answer

Expert verified
The sun's diameter is approximately 1,390,000 km.

Step by step solution

01

Understand the Stefan-Boltzmann Law

To find the sun's diameter, we use the Stefan-Boltzmann Law which relates the power radiated by a blackbody to its temperature and surface area: \[ P = \sigma A T^4 \]where \(P\) is the total power radiated, \(\sigma\) is the Stefan-Boltzmann constant \(5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4\), \(A\) is the surface area, and \(T\) is the temperature.
02

Rearrange the Equation to Solve for Surface Area

We know the total power radiated \(P = 3.92 \times 10^{26}\) J/s and the temperature \(T = 5800\) K, so we solve for the area \(A\): \[ A = \frac{P}{\sigma T^4} \]
03

Calculate the Surface Area

Substitute the values into the equation:\[ A = \frac{3.92 \times 10^{26}}{5.67 \times 10^{-8} \times (5800)^4} \]Compute \(A\) to find the surface area of the sun.
04

Surface Area of a Sphere Relation

For a sphere, the surface area \(A\) is given by \(A = 4\pi R^2\), where \(R\) is the sun's radius.Rearrange this to solve for radius: \[ R = \sqrt{\frac{A}{4\pi}} \]
05

Calculate the Radius of the Sun

Substitute the surface area from Step 3 into the sphere equation:\[ R = \sqrt{\frac{3.83 \times 10^{18}}{4\pi}} \]Compute \(R\) to find the radius of the sun.
06

Convert Radius to Diameter

The diameter \(D\) is twice the radius:\[ D = 2R \]Use the radius found in Step 5 to find the sun's diameter.

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Key Concepts

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

Stefan-Boltzmann Law
The Stefan-Boltzmann Law is foundational in understanding how stars, like our Sun, emit energy. This law states that the total energy radiated per unit surface area of a blackbody is directly proportional to the fourth power of its absolute temperature. Mathematically, it is expressed as:\[ P = \sigma A T^4 \]where:
  • \( P \) is the power output or energy radiated by the blackbody,
  • \( \sigma \) is the Stefan-Boltzmann constant, approximately \(5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4\),
  • \( A \) is the surface area of the blackbody, and
  • \( T \) is the absolute temperature in Kelvin (K).
This law is crucial for determining how much energy a star like the Sun emits, based on its temperature and size. It allows scientists to estimate the Sun's diameter when combined with its known surface temperature and emitted power.
Blackbody Radiation
A blackbody is an idealized physical object that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. It is a perfect emitter, radiating energy at maximum efficiency depending on its temperature. This concept is central in stellar physics, as many stars, including our Sun, approximate blackbody behavior. The spectral distribution of radiation from a blackbody helps determine its temperature. For the Sun, its light spectrum provides a surface temperature of approximately 5800 K. By assuming the Sun behaves as a near-ideal blackbody, we can use the Stefan-Boltzmann Law to calculate characteristics like energy output and surface area, and consequently, its diameter.
Surface Temperature
The surface temperature of a celestial body such as the Sun is critical for understanding its physical properties. It determines how brightly the object shines and directly influences the amount of energy it radiates. In the context of the Sun, its surface temperature is measured to be around 5800 Kelvin. This temperature is derived from analyzing the Sun's emitted spectrum of light and is a key component in applying the Stefan-Boltzmann Law to calculate the Sun's emitted power and size. Knowing the surface temperature allows astronomers to further study and comprehend the processes happening in a star's outer layers and to make broader assumptions about its internal structure and life cycle.
Solar Radius
The solar radius is an essential measurement in astrophysics, representing the Sun's size. Once the surface area is calculated using the Stefan-Boltzmann Law, the solar radius can be found using the formula for the surface area of a sphere:\[ A = 4\pi R^2 \]where \( R \) is the radius. By rearranging this formula, we can solve for the radius:\[ R = \sqrt{\frac{A}{4\pi}} \]Understanding the solar radius is crucial because it helps scientists compare the Sun’s size with other celestial objects. The radius also aids in calculating other significant parameters, such as the Sun's density and gravitational field, enhancing our understanding of stellar structures and dynamics.

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Most popular questions from this chapter

Each pound of fat contains 3500 food calories. When the body metabolizes food, \(80 \%\) of this energy goes to heat. Suppose you decide to run without stopping, an activity that produces \(1290 \mathrm{~W}\) of metabolic power for a typical person. (a) For how many hours must you run to burn up 1 lb of fat? Is this a realistic exercise plan? (b) If you followed your planned exercise program, how much heat would your body produce when you burn up a pound of fat? (c) If you needed to get rid of all of this excess heat by evaporating water (i.e., sweating), how many liters would you need to evaporate? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg}\)

A \(15.0 \mathrm{~g}\) bullet traveling horizontally at \(865 \mathrm{~m} / \mathrm{s}\) passes through a tank containing \(13.5 \mathrm{~kg}\) of water and emerges with a speed of \(534 \mathrm{~m} / \mathrm{s}\). What is the maximum temperature increase that the water could have as a result of this event?

The blood plays an important role in removing heat from the body by bringing this heat directly to the surface where it can radiate away. Nevertheless, this heat must still travel through the skin before it can radiate away. We shall assume that the blood is brought to the bottom layer of skin at a temperature of \(37^{\circ} \mathrm{C}\) and that the outer surface of the skin is at \(30.0^{\circ} \mathrm{C}\). Skin varies in thickness from \(0.50 \mathrm{~mm}\) to a few millimeters on the palms and soles, so we shall assume an average thickness of \(0.75 \mathrm{~mm}\). A \(165 \mathrm{lb}, 6 \mathrm{ft}\) person has a surface area of about \(2.0 \mathrm{~m}^{2}\) and loses heat at a net rate of \(75 \mathrm{~W}\) while resting. On the basis of our assumptions, what is the thermal conductivity of this person's skin?

Camels require very little water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to \(34.0^{\circ} \mathrm{C}\) overnight and rise to \(40.0^{\circ} \mathrm{C}\) during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a \(400 \mathrm{~kg}\) camel would have to drink if it attempted to keep its body temperature at a constant \(34.0^{\circ} \mathrm{C}\) by evaporation of sweat during the day (12 hours) instead of letting it rise to \(40.0^{\circ} \mathrm{C}\). (Note: The specific heat of a camel or other mammal is about the same as that of a typical human, \(3480 \mathrm{~J} /(\mathrm{kg} \cdot \mathrm{K})\). The heat of vaporization of water at \(34^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\)

Much of the energy of falling water in a waterfall is converted into heat. If all the mechanical energy is converted into heat that stays in the water, how much of a rise in temperature occurs in a \(100 \mathrm{~m}\) waterfall?

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