Angular Diameter of Sun: Radius, Area, and Power

[yiuscott]

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

 

[tiny-tim]

Hi yiuscott!

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.

… the Earth at a temperature of 228K.

erm … 228K? … I don't think so!

Try 288K!

 

[yiuscott]

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...