Angular Diameter of Sun: Radius, Area, and Power

In summary, the conversation discusses two questions. The first question involves calculating the angular diameter of the sun in radians, its radius in meters, and its surface area assuming it is a sphere. The second question involves calculating the total power and power per m^2 radiated by the Earth at a temperature of 288K, using the Stefan-Boltzmann constant and assuming the Earth is a sphere. The conversation also addresses a possible error in the book's answer for the first question and correcting the temperature for the second question.
  • #1
yiuscott
37
0

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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  • #2
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:
 
  • #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:
 

1. What is the angular diameter of the Sun?

The angular diameter of the Sun, also known as apparent size, is the angle that the Sun subtends at the Earth's center. It is approximately 0.5 degrees, or 30 arcminutes.

2. How is the angular diameter of the Sun calculated?

The angular diameter of the Sun can be calculated using the formula: angular diameter = (2 * radius) / distance. The average distance between the Earth and the Sun is 149.6 million kilometers, and the radius of the Sun is approximately 695,700 kilometers.

3. What is the radius of the Sun?

The radius of the Sun is approximately 695,700 kilometers. This is about 109 times the size of the Earth's radius.

4. How does the angular diameter of the Sun affect its apparent brightness?

The angular diameter of the Sun does not directly affect its apparent brightness. However, a larger angular diameter can make the Sun appear brighter or dimmer depending on atmospheric conditions and the position of the Sun in the sky.

5. How does the area and power of the Sun relate to its angular diameter?

The area and power of the Sun are both directly related to its angular diameter. As the angular diameter increases, the area of the Sun also increases, and therefore, its power or energy output increases as well.

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