Chapter 18, Problem 46Q

To determine

The radius of protosun.

Answer to Problem 46Q

The radius of the protosun in km when it had a luminosity of $1000{\text{ L}}_{\odot }$ and a surface temperature of about $1000\text{ K}$ = $7.330\times {10}^8\text{ km}$

The radius of the protosun in astronomical units when it had a luminosity of $1000{\text{ L}}_{\odot }$ and a surface temperature of about $1000\text{ K}$ = $\text{4}\text{.90 au}$

The radius of the protosun as a multiple of the Sun’s present-day radius when it had a luminosity of $1000{\text{ L}}_{\odot }$ and a surface temperature of about $1000\text{ K}$ = $1.05\times {10}^3{\text{ R}}_{\odot }$

Explanation of Solution

Given data:

Luminosity of the protosun = $1000{\text{ L}}_{\odot }$

Surface temperature of the protosun = $1000\text{ K}$

Formula used:

$\begin{array}{l}L=\sigma A{T}^4\\ \text{For a sphere},A=4\pi r^2\\ 1{\text{ L}}_{\odot }=3.828\times {10}^{26} \text{W}\\ \text{1 au = }1.496\times {10}^8 \text{km}\\ \text{Radius of the sun = }695,510\text{ km}\\ \sigma \text{=}5.670\times {10}^{-8} \text{W}\cdot {\text{m}}^{\text{-2}}\cdot {\text{K}}^{\text{-4}}\end{array}$

Calculation:

$\begin{array}{l}L=\sigma A{T}^4\\ \text{As }A=4\pi r^2\\ L=\sigma \times 4\pi r^2\times T^4\\ \text{When }L=1000{\text{ L}}_{\odot },T\text{=1000 K, and }\sigma \text{=}5.670\times {10}^{-8} \text{W}\cdot {\text{m}}^{\text{-2}}\cdot {\text{K}}^{\text{-4}}\\ 1000{\text{ L}}_{\odot }=5.670\times {10}^{-8}\times 4\pi r^2\times {1000}^4\\ r^2=\frac{1000{\text{ L}}_{\odot }}{5.670\times {10}^{-8}\times 4\pi \times {1000}^4}\\ \text{As }1{\text{ L}}_{\odot }=3.828\times {10}^{26} \text{W}\\ r^2=\frac{1000\times 3.828\times {10}^{26}}{5.670\times {10}^{-8}\times 4\pi \times {1000}^4}\\ r^2=\frac{3.828\times {10}^{29}}{5.670\times {10}^{-8}\times 4\pi \times {1000}^4}\\ r^2=\frac{3.828\times {10}^{29}}{71.251\times {10}^4}\\ r^2=0.05373\times {10}^{25}\\ r^2=5.373\times {10}^{23}\\ r=\sqrt{5.373\times {10}^{23}}\\ r=7.330\times {10}^{11}\text{m = 7}\text{.330}\times {\text{10}}^8\text{km}\\ \text{In astronomical units:}\\ r=\frac{7.330\times {10}^8}{1.496\times {10}^8 }\text{ au}\\ r=\text{4}\text{.90}\text{au}\\ \text{As a measure of the present day sun's radius}\\ r=\frac{7.330\times {10}^8}{695,510}{\text{R}}_{\odot }\\ r=1.0539\times {10}^{-5}\times {10}^8{\text{R}}_{\odot }\\ r=1.0539\times {10}^3{\text{ R}}_{\odot }\\ \text{Rounding off to the second decimal place, }\\ r=1.05\times {10}^3{\text{ R}}_{\odot }\end{array}$

Conclusion The radius of the protosun in km when it had a luminosity of $1000{\text{ L}}_{\odot }$ and a surface temperature of about $1000\text{ K}$ = $7.330\times {10}^8\text{ km}$

The radius of the protosun in astronomical units when it had a luminosity of $1000{\text{ L}}_{\odot }$ and a surface temperature of about $1000\text{ K}$ = $\text{4}\text{.90 au}$

The radius of the protosun as a multiple of the Sun’s present-day radius when it had a luminosity of $1000{\text{ L}}_{\odot }$ and a surface temperature of about $1000\text{ K}$ = $1.05\times {10}^3{\text{ R}}_{\odot }$