# The Sun's Power Output: Solving for P

• mikefitz
In summary, the Sun emits electromagnetic waves (including light) equally in all directions and the intensity of these waves at the Earth's upper atmosphere is 1.4 kW/m2. To find the power output of the Sun, you need to find the surface area of a sphere with radius equal to the distance from the sun to the Earth, and then find the area of a disk with radius equal to the distance from the center of the Earth to the upper atmosphere (not the surface area of that sphere). Then, divide the total energy output of the Sun by the area of the disk to get the power output in kilowatts.
mikefitz
The Sun emits electromagnetic waves (including light) equally in all directions. The intensity of the waves at the Earth's upper atmosphere is 1.4 kW/m2. At what rate does the Sun emit electromagnetic waves? (In other words, what is the power output?)

I = P/A

I know I need to convert 1.4kW/m^2 ==> 1400W/m^2
Then I set that value = to P/A

What I'm unclear about is, what is the Area I am supposed to insert in the equation?

*also, looking over my notes, I have A= 4piR^2 - E^2

I assume R = distance from Earth to sun, but what would E^2 represent?

any ideas? This is one of two problems I have left for the weekend - thanks!

I'm not sure what your E2 term represents here, but your area should be the surface area of a sphere $A = 4\pi r^3$.

Hoot...r^3?

No surface "area" is proportional to r^2. Volume, which is not relevant here, is proportional to r^3.

mikefitz said:
*also, looking over my notes, I have A= 4piR^2 - E^2

I assume R = distance from Earth to sun, but what would E^2 represent?
Darned if I know! You didn't even tell us what A represents. It's certainly not a good idea to copy a formula to your notes without writing down what each of the parameters means!

You are told that " The intensity of the waves at the Earth's upper atmosphere is 1.4 kW/m2." Find the surface area of a sphere with radius equal to the distance from the sun to the earth. Find the area that a disk with radius equal to the distance from the center of the Earth to the "upper atmosphere" (not the surface area of that sphere- you want the "cross section area" that intercepts the suns rays). The total energy output of the sun, divided by that first surface area, is equal to the sun's total energy output (in kilowatts) divided by the 1.4 kw/m2 that you are given.

neutrino said:
Hoot...r^3?
Damn typo

## 1. How is the Sun's power output measured?

The Sun's power output is measured in watts, which is the unit of power. Scientists use special instruments called radiometers to measure the amount of energy emitted by the Sun.

## 2. What is the average power output of the Sun?

The average power output of the Sun is approximately 3.8 x 10^26 watts. This value is known as the solar constant and represents the amount of energy that reaches the Earth's upper atmosphere.

## 3. How is the Sun's power output calculated?

The Sun's power output can be calculated by using the formula P = L/4πd^2, where P is the power output, L is the luminosity (total energy emitted per second), and d is the distance from the Sun to the measuring point. This formula takes into account the fact that the Sun's energy is distributed evenly in all directions.

## 4. What factors affect the Sun's power output?

The Sun's power output can be affected by various factors, including changes in the Sun's size, temperature, and composition. Solar activity, such as sunspots and solar flares, can also impact the Sun's power output.

## 5. How does the Sun's power output compare to other stars?

The Sun's power output is considered to be relatively low compared to other stars in the universe. It is classified as a G-type main sequence star and is smaller and cooler than many other stars, resulting in a lower power output. However, the Sun's power output is still significant and essential for sustaining life on Earth.

• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
721
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
906
• Introductory Physics Homework Help
Replies
4
Views
8K
• Introductory Physics Homework Help
Replies
5
Views
7K