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Chapter 13: Problem 28

Sirius \(\mathrm{B}\) is a white star that has a surface temperature (in kelvins) that is four times that of our sun. Sirius \(\mathrm{B}\) radiates only 0.040 times the power radiated by the sun. Our sun has a radius of \(6.96 \times 10^{8} \mathrm{m} .\) Assuming that Sirius B has the same emissivity as the sun, find the radius of Sirius B.

Short Answer

Expert verified
The radius of Sirius B is approximately \(8.7 \times 10^6 \mathrm{m}\).

Step by step solution

01

Identify Key Information

The exercise provides that Sirius B has a surface temperature that is four times that of the Sun and radiates 0.040 times the power. The radius of the Sun is given as \(6.96 \times 10^8\, \mathrm{m}\). We need to find the radius of Sirius B.
02

Understand the Stefan-Boltzmann Law

The Stefan-Boltzmann Law states that the power radiated by a star is proportional to the fourth power of its temperature and the square of its radius, represented by \( P = \sigma \epsilon A T^4 \), where \( \sigma \) is the Stefan-Boltzmann constant, \( \epsilon \) is the emissivity, \( A \) is the surface area, and \( T \) is the temperature.
03

Establish Relationships for Power and Temperature

For Sirius B, \( P_B = \sigma \epsilon 4\pi R_B^2 (4T_{\odot})^4 \) and for the Sun, \( P_{\odot} = \sigma \epsilon 4\pi R_{\odot}^2 T_{\odot}^4 \). The problem states \( P_B = 0.040 P_{\odot} \). Rewrite the equations in terms of known relationships.
04

Simplify the Equations

Since \( P_B = 0.040 P_{\odot} \), substitute: \( \sigma \epsilon 4\pi R_B^2 (256T_{\odot}^4) = 0.040 \sigma \epsilon 4\pi R_{\odot}^2 T_{\odot}^4 \). The \( \sigma \), \( \epsilon \), \( 4\pi \), and \( T_{\odot}^4 \) cancel out, simplifying to \( 256R_B^2 = 0.040 R_{\odot}^2 \).
05

Solve for Radius of Sirius B

Substituting the known radius of the Sun, \( R_{\odot} = 6.96 \times 10^8 \mathrm{m} \), into the equation gives \( 256R_B^2 = 0.040 (6.96 \times 10^8)^2 \). Solving for \( R_B \), \( R_B = \sqrt{\frac{0.040}{256}} \times 6.96 \times 10^8 \mathrm{m} \).
06

Perform Calculations

Calculate \( R_B \) by first determining \( \frac{0.040}{256} \approx 0.00015625 \) and then \( \sqrt{0.00015625} \approx 0.0125 \). Finally, compute \( R_B = 0.0125 \times 6.96 \times 10^8 \approx 8.7 \times 10^6 \mathrm{m} \).

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Key Concepts

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

Sirius B
Sirius B is a companion star to Sirius A, forming a binary star system. It is often referred to as a white dwarf, which is a remnant of a star that has exhausted most of its nuclear fuel. Despite its small size, Sirius B is incredibly dense. The mass of Sirius B is nearly equal to that of our Sun, but it is compressed into a volume similar to that of the Earth.

Here are some key points about Sirius B:
  • It is much hotter than our Sun, with a surface temperature four times greater.
  • Although hotter, Sirius B radiates significantly less power compared to the Sun, being only 4% of the Sun's power output.
  • Its high temperature and low power output make Sirius B a prime example of dense stellar remnants.
The study of Sirius B helps astronomers understand the late stages of stellar evolution and the physical properties of white dwarfs.
Surface Temperature
The surface temperature of a star is a crucial factor in determining its intrinsic properties. For Sirius B, the temperature is four times greater than that of the Sun, highlighting its intense heat output.

Understanding the impact of surface temperature involves:
  • The Stefan-Boltzmann Law, which states that the radiated power of a star is proportional to the fourth power of its surface temperature.
  • Stars like Sirius B, despite their small size, emit considerable energy due to their high surface temperatures.
  • This higher temperature directly affects the luminosity and color of the star, with Sirius B appearing as a bluish-white star due to its intense heat.
The surface temperature not only affects the star's brightness and color but also its lifespan, with hotter stars typically having shorter life spans.
Power Radiated
The power radiated by a star is vital for understanding its energy output and characteristics. For Sirius B, although its temperature is much hotter than the Sun, it only radiates 0.040 times the power.

This discrepancy can be explained by:
  • The application of the Stefan-Boltzmann Law, which involves the relationship between temperature, radius, and radiated power.
  • Despite its high temperature, Sirius B's small radius results in a significantly lower overall power output compared to the much larger Sun.
  • The computing of radiated power helps astronomers know more about the star's efficiency in energy emission.
Understanding the power radiated by stars like Sirius B provides insight into their evolutionary stage and assists in comparative studies of different celestial bodies.

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Most popular questions from this chapter

How many days does it take for a perfect blackbody cube \((0.0100 \mathrm{m}\) on a side, \(30.0^{\circ} \mathrm{C}\) ) to radiate the same amount of energy that a one-hundred watt light bulb uses in one hour?

A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processes are generating, its temperature will drop. If the drop is severe, it can be life-threatening. Suppose that a person is unclothed and energy is being lost via radiation from a body surface area of \(1.40 \mathrm{m}^{2},\) which has a temperature of \(34^{\circ} \mathrm{C}\) and an emissivity of \(0.700 .\) Also suppose that metabolic processes are producing energy at a rate of \(115 \mathrm{J} / \mathrm{s} .\) What is the temperature of the coldest room in which this person could stand and not experience a drop in body temperature?

In a house the temperature at the surface of a window is \(25^{\circ} \mathrm{C} .\) The temperature outside at the window surface is \(5.0^{\circ} \mathrm{C} .\) Heat is lost through the window via conduction, and the heat lost per second has a certain value. The temperature outside begins to fall, while the conditions inside the house remain the same. As a result, the heat lost per second increases. What is the temperature at the outside window surface when the heat lost per second doubles?

A wall in a house contains a single window. The window consists of a single pane of glass whose area is \(0.16 \mathrm{m}^{2}\) and whose thickness is \(2.0 \mathrm{mm}\). Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are \(18 \mathrm{m}^{2}\) and \(0.10 \mathrm{m},\) respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window?

A baking dish is removed from a hot oven and placed on a cooling rack. As the dish cools down to \(35^{\circ} \mathrm{C}\) from \(175^{\circ} \mathrm{C},\) its net radiant power decreases to \(12.0 \mathrm{W}\). What was the net radiant power of the baking dish when it was first removed from the oven? Assume that the temperature in the kitchen remains at \(22^{\circ} \mathrm{C}\) as the dish cools down.
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