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Magazine Chapter 20: Problem 3
What would be the luminosity of the Sun if its surface temperature were \(3000 \mathrm{K}\) and its radius were \((\mathrm{a}) 1 \mathrm{AU},(\mathrm{b}) 5 \mathrm{AU} ?\)
Short Answer
Expert verified
The luminosity for 1 AU is about \(3.98 \times 10^{33}\) W and for 5 AU is about \(9.95 \times 10^{34}\) W.
Step by step solution
01
Understanding the Problem
We need to find the luminosity of the Sun using the Stefan-Boltzmann law. The luminosity can be affected by changes in the surface temperature and radius of the Sun.
02
Applying the Stefan-Boltzmann Law
The Stefan-Boltzmann law states that the luminosity (L) of a star or any radiating object is given by \( L = 4\pi R^2 \sigma T^4 \), where \( R \) is the radius, \( \sigma \) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \; \text{W m}^{-2} \text{K}^{-4})\), and \( T \) is the surface temperature.
03
Calculate Luminosity for Radius 1 AU
Convert 1 AU to meters: \( 1 \text{ AU} = 1.496 \times 10^{11} \text{ m} \). Substitute \( R = 1.496 \times 10^{11} \text{ m} \) and \( T = 3000 \text{ K} \) into the Stefan-Boltzmann formula: \[ L = 4\pi (1.496 \times 10^{11})^2 (5.67 \times 10^{-8}) (3000)^4 \].Calculate the result for luminosity.
04
Substitute the Values and Compute for 1 AU
Replacing the values, \[ L = 4\pi \times (1.496 \times 10^{11})^2 \times 5.67 \times 10^{-8} \times (3000)^4 \]Calculate the result using a calculator for precision.
05
Calculate Luminosity for Radius 5 AU
Convert 5 AU to meters: \( 5 \text{ AU} = 5 \times 1.496 \times 10^{11} \text{ m} = 7.48 \times 10^{11} \text{ m} \). Substitute \( R = 7.48 \times 10^{11} \text{ m} \) and \( T = 3000 \text{ K} \) into the equation.
06
Substitute the Values and Compute for 5 AU
Using similar steps as before, \[ L = 4\pi \times (7.48 \times 10^{11})^2 \times 5.67 \times 10^{-8} \times (3000)^4 \]Calculate the luminosity for this radius.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Luminosity Calculation
Luminosity is a measure of the total energy emitted by an astronomical object like the Sun every second. To find this, we use the Stefan-Boltzmann Law. This law helps us understand how changes in surface temperature and radius affect the luminosity. The formula:
- Given by \[ L = 4\pi R^2 \sigma T^4 \]
- Where \( L \) is luminosity in watts, \( R \) is the radius in meters, \( \sigma \) is the Stefan-Boltzmann constant \( (5.67 \times 10^{-8} \, \text{W m}^{-2} \text{K}^{-4}) \), and \( T \) is the surface temperature in Kelvin.
Surface Temperature
Surface temperature plays a crucial role in determining the luminosity of the Sun or any star. It is one of the determining factors in the Stefan-Boltzmann law equation. A small increase in surface temperature means a huge increase in luminosity since it is raised to the fourth power in the formula \( T^4 \). Let's say, if the temperature of the Sun were changed to \( 3000 \text{ K} \), different than its actual temperature, it would drastically affect its luminosity.Understanding temperature's exponential influence helps us predict how a star's brightness might change due to temperature fluctuations over its lifespan.
Solar Radius
The solar radius is a key variable in the luminosity calculation. It's part of the surface area term in the Stefan-Boltzmann law, \( 4\pi R^2 \). This tells us how much surface area an object like the Sun has to emit energy.The solar radius changes for our exercise based on the given astronomical units (AU).
- 1 AU (astronomical unit) is approximately \( 1.496 \times 10^{11} \text{ m} \).
- 5 AU is calculated by multiplying the distance by 5, resulting in \( 7.48 \times 10^{11} \text{ m} \).
Astronomical Unit
An astronomical unit (AU) is a standard measure of distance in astronomy. It is the average distance from the Earth to the Sun, approximately \( 1.496 \times 10^{11} \text{ meters} \).In the context of luminosity calculations, AU helps in scaling the size of the object we are examining. For instance:
- A radius of 1 AU means using the actual average radius equivalent of the Earth-Sun distance.
- A radius of 5 AU expands this to five times, which showcases how increasing the size of a radiating body impacts energy emission.
