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Chapter 5: Problem 50

The Sun has a radius of \(6.96 \times 10^{5} \mathrm{km}\) and a blackbody temperature of \(5780 \mathrm{K}\). Calculate the Sun's luminosity. (Hint: The area of a sphere is \(4 \pi R^{2}\).)

Short Answer

Expert verified
The Sun's luminosity is \( 3.83 \times 10^{26} \text{ Watts} \).

Step by step solution

01

Identify the Given Data

The radius of the Sun is given as \( 6.96 \times 10^5 \text{ km} \) and its blackbody temperature is \( 5780 \text{ K} \).
02

Convert Radius to Meters

Convert the radius from kilometers to meters: \( R = 6.96 \times 10^5 \text{ km} = 6.96 \times 10^8 \text{ meters} \).
03

Calculate the Surface Area of the Sun

Use the formula for the surface area of a sphere, \( A = 4 \pi R^2 \), to calculate the surface area.
04

Use the Stefan-Boltzmann Law

The Stefan-Boltzmann Law relates the luminosity (L) to the surface area (A) and the blackbody temperature (T) of an object:
05

Insert the Constants

The Stefan-Boltzmann constant (\( \fff4\sigma \)) is \( 5.67 \times 10^{-8} \text{ W} \text{ m}^{-2} \text{ K}^{-4} \).
06

Calculate the Luminosity

Substitute the values into the equation:

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Key Concepts

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

solar radius
The solar radius is a crucial measure for understanding the size of the Sun.
It is the distance from the Sun's center to its surface, which in this exercise is given as \( 6.96 \times 10^5 \text{ km} \).
This value helps us calculate the surface area of the Sun, which is necessary for determining its luminosity.
To make calculations more manageable, we often convert kilometers to meters.
Thus, \( 6.96 \times 10^5 \text{ km} \) becomes \( 6.96 \times 10^8 \text{ m} \).
This conversion is essential because the standard unit for many physical equations, including the Stefan-Boltzmann Law, is meters.
blackbody temperature
The concept of blackbody temperature is vital in astrophysics.
A blackbody is an idealized object that perfectly absorbs and emits all radiation frequencies.
In the case of our Sun, we treat it as a blackbody with a temperature of \( 5780 \text{ K} \).
Kelvin (K) is the base unit of temperature in the International System of Units (SI).
This temperature helps us calculate the Sun's luminosity because it directly affects how much energy is emitted per unit area.
Higher temperatures result in greater energy emissions.
Stefan-Boltzmann law
The Stefan-Boltzmann Law is a cornerstone of calculating stellar luminosity.
This law states that the total energy radiated per unit surface area of a blackbody is proportional to the fourth power of its temperature.
Mathematically, it is expressed as \[L = \fint \textbf \fint} 4 \f}/},4}*A \times T^4 \]
Here, \( L \) is the luminosity, \( \textbf \fint} 4 \) is the Stefan-Boltzmann constant (\( 5.67 \times 10^{-8} \text{ W} \text{ m}^{-2} \text{ K}^{-4} \)), and \( A \) is the surface area.
This law helps us understand how the luminosity and temperature are related.
surface area of a sphere
To find the surface area of the Sun, we use the formula for the surface area of a sphere.
This formula is \( A = 4 \fint} \times R^2 \)
In our exercise, substituting the Sun's radius (\( R \) as \( 6.96 \fint} 10^8 \text{ m} \),)
we have: \[ A = 4 \times 2 \fint} \times \times aint} 10^8 \times } \fint} 10^8 = aint ( 10^18. \]
This surface area is necessary for applying the Stefan-Boltzmann Law accurately.
unit conversion
Unit conversion plays a critical role in scientific calculations.
We often need to convert units to match the required parameters of formulas.
For instance, converting the Sun's radius from kilometers to meters is crucial as the Stefan-Boltzmann constant uses meters as a unit.
General unit conversion steps include:
  • Identify the original unit and the required unit.

  • Use a conversion factor (e.g., \[ 1 \fint} ->1000 \fint}.\fint } \fint2=\]
  • Multiply or divide by the conversion factor as needed.

  • Recalculate with the new units.

    • These steps ensure consistency and accuracy in our calculations.

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