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Chapter 16: Problem 66

A star whose surface temperature is \(50 \mathrm{kK}\) radiates \(4.0 \times 10^{27} \mathrm{W}\) If the star behaves like a blackbody, what's its radius?

Short Answer

Expert verified
The radius of the star is \(6.18 x 10^{8} m\).

Step by step solution

01

Understanding the Problem

We know that a star radiates like a blackbody, and hence its radiated power \(P\) is given by the Stefan-Boltzmann law: \(P = 4πr^2σT^4\), where \(r\) is the radius of the star, \(T\) is the temperature, and \(σ\) is Stefan-Boltzmann constant, \(σ = 5.67 x 10^{-8} W/m^2K^4\). We‘re given a surface temperature \(T\) of 50kK or 50000K, and a power output of \(4.0 x 10^{27}W\). The goal is to find the radius \(r\) of the star.
02

Rearranging the Stefan-Boltzmann Equation

Given the equation \(P = 4πr^2σT^4\) , we rearrange to find \(r = \sqrt{\frac{P} { 4πσT^4}}\).
03

Substituting Values into the Equation

Now substitute the given values of \(P\), \(σ\), and \(T\) into the equation to solve for \(r\).
04

Calculation

We plug in the numbers and get \(r = \sqrt{\frac{4.0 x 10^{27}} { 4π x 5.67 x 10^{-8} x (50000)^4}}\). The calculation gives \(r = 6.18 x 10^{8} m\)

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