Angular Diameter of Sun: Radius, Area, and Power

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The discussion revolves around calculating the angular diameter, radius, and surface area of the Sun, as well as the luminosity of the Earth based on its temperature. The angular diameter of the Sun is given as 0.52 degrees, which converts to 0.00908 radians. The radius of the Sun is debated, with one participant calculating it as 7X10^8m, while the textbook states it should be 1.4X10^9m, leading to suspicions about the book's accuracy. For the luminosity calculation, the participant initially used 228K for Earth's temperature, which was corrected to 288K, indicating a potential error in the textbook. The conversation highlights the importance of verifying textbook answers and understanding the underlying calculations.
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Homework Statement


This is the questions:

1) The angular diameter of the sun measured from the Earth is 0.52 degrees. Calculate:
i) the angular diameter in radians
ii) the sun's radius in meters
iii) The surface area of the sun, assuming that it is a sphere

2) Using L=aAT4 where L is the luminosity, a is the stefan-Boltzmann constant: 5.67X10^-8, A is the surface area, T is the temperature.

Calculate the total power and the power per m^2 radiated by the Earth at a temperature of 228K. You can assume that the Earth is a sphere of radius 6400km.

The Attempt at a Solution



Question 1:
Question i) is quite easy and found out that it is 0.00908
Question ii) is the major problem. I tried to assume that it is a triangle and used the sine rule (since i know the distance of the sun to the Earth is 149X10^11m while the other angles in the triangle should be (180-0.52)/2. However, this gave me an answer of 1.4X10^9. If i divide by 2 (to find the radius), i ger 7X10^8m.
The answer at the back of the book is 1.4X10^9m. which is double my answer. I suspect that the book may be wrong but just wish to double check
Question iii) if i knew the answer to question ii), i would be able to do this.

Question 2: I thought that this question should be easy where i just put in the numbers into the equation... but i couldn't get the right answer...
L=aAT^4
L = (5.67X10^-8)*(4*pi*6400000^2)*(228^4)
L = 7.89X10^16
The book's answer is 201X10^15 W

Thank you very much
 
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Hi yiuscott! :smile:
yiuscott said:
The answer at the back of the book is 1.4X10^9m. which is double my answer. I suspect that the book may be wrong but just wish to double check

I think the book is wrong, and you are right … width = radius * radians. :smile:
… the Earth at a temperature of 228K.

erm … 228K? … I don't think so! :rolleyes:

Try 288K! :wink:
 
Thanks for the reply.

lol thanks. I can't imagine a textbook getting 2 questions wrong in a row. I should have spotted the problem with a 228K Earth though... :rolleyes:
 
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