1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Angular diameter

  1. Aug 18, 2008 #1
    1. The problem statement, all variables and given/known data
    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.

    3. 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 = (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
    Last edited: Aug 18, 2008
  2. jcsd
  3. Aug 18, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    Hi yiuscott! :smile:
    I think the book is wrong, and you are right … width = radius * radians. :smile:
    erm … 228K? … I don't think so! :rolleyes:

    Try 288K! :wink:
  4. Aug 18, 2008 #3
    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:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?