Solar irradiation on the Earth as a function of latitude.

In summary, the conversation discusses finding the total power output of the Sun, power received on the Earth, and the power received in Spain, given the solar output and Earth's radius and orbit radius. The method of using the Poynting vector formula is mentioned but deemed insufficient due to the dependence on time and atmospheric conditions. The use of spherical trigonometry is suggested to calculate the power received at a given latitude. The conversation also includes a discussion on how the angle of the sun's light affects the power absorbed by a black square and how it can be calculated using trigonometry. The solution to the first part involves multiplying the power per area by the surface area of the Earth.
  • #1
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Homework Statement


Given the following info:

Solar output(modulus of the Poynting vector): [itex]1.4kW/m^2[/itex]
Earth radius: 6370km
Average Earth orbit radius: [itex]1.49*10^11 m[/itex]

Find the total power output of the Sun, power received on the Earth, and the power received in Spain if its surface area is 5*10^2 km^2 and is at a latitude of 40 N.

Homework Equations


Poynting vector formula?

The Attempt at a Solution


I managed to get the right answer for the power received on the Earth, but I'm not sure how to proceed with the other 2 questions. Wouldn't the power received at a given latitude depend on the time of the year? I found this applet that calculates it for you and apparently at 40 N, the solar output never reaches 1.4kW at any time of the year:
http://pvcdrom.pveducation.org/SUNLIGHT/SHCALC.HTM

... so I'm guessing I'll have to resort to spherical trigonometry some way or another? I am given no other info and this is a course purely on electromagnetic optics.
 
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  • #2
The time of year is important as is the time of day as is the Earth's atmosphere.

Say you have a black square, 1 meter by 1 meter that absorbs all light that falls on it. Say you are at the equator and hold the square perpendicular to the suns light. What is the power absorbed? Now hold the same black square so that the suns light and the unit normal of the black plane make and angle theta. The power now absorbed is rescued by a factor of cos(theta)?
 
  • #3
I think I see it now, drawing a circle (earth) with normal surfaces on the equator and at 60º from the N pole, using trigonometry I get P(at 60 N) = Power(known)*(cos60), which is a halve of the known power output at the given orbital distance (which I got right: 3.9*10^26W).

However, what you're saying assumes the equator is aligned with the sun/the Earth isn't tilted, but since the problem says nothing to that avail so I guess its reasonable? Thanks.

How would you solve the first part? Wouldn't I need the radius or surface area of the Sun?
 
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  • #4
You are given what is needed, the power/m^2 at the distance the Earth is from the sun and the distance to the sun.

power per area times area = power
 
  • #5


I would first clarify the assumptions and limitations of the given information. It seems that the given data is simplified and does not take into account factors such as the Earth's tilt, the varying distance between the Earth and the Sun throughout the year, and atmospheric effects. Therefore, the calculations and results may not be entirely accurate.

To find the total power output of the Sun, we can use the formula for the surface area of a sphere: 4πr^2. Plugging in the Earth's radius (6370km), we get a surface area of approximately 5.1x10^8 km^2. Multiplying this by the given solar output of 1.4kW/m^2, we get a total power output of the Sun of approximately 7.1x10^17 kW.

To find the power received on the Earth, we can use the same formula and plug in the average Earth orbit radius (1.49x10^11 m). This gives us a surface area of approximately 2.8x10^17 m^2. Multiplying this by the solar output, we get a power received on the Earth of approximately 3.9x10^20 kW.

To find the power received in Spain, we need to take into account the angle of incidence of the sunlight at that latitude. As you mentioned, this will vary depending on the time of year. We can use the formula for the projected area of a circle (πr^2cosθ) to find the surface area of Spain that is facing the sunlight at a given time. Plugging in the given latitude of 40 N, we get a projected area of approximately 1.3x10^6 km^2. Multiplying this by the solar output, we get a power received in Spain of approximately 1.8x10^12 kW.

In conclusion, while the calculations above provide a rough estimate of the power output and received on the Earth, the actual values may vary due to the factors mentioned earlier. I would suggest further research and data collection to improve the accuracy of these calculations.
 

What is solar irradiation?

Solar irradiation refers to the amount of solar energy that is received on a specific area of the Earth's surface. It is measured in watts per square meter (W/m²) and is influenced by various factors such as latitude, time of day, atmospheric conditions, and season.

How does solar irradiation vary with latitude?

The amount of solar irradiation decreases as you move away from the equator towards the poles. This is because the sun's rays hit the Earth's surface at a more direct angle near the equator, providing more solar energy, while at the poles, the rays hit at an angle, spreading the same amount of energy over a larger area.

What is the relationship between solar irradiation and the Earth's tilt?

The Earth's tilt (23.5 degrees) is one of the main factors that influence the amount of solar irradiation received at different latitudes. This tilt causes the sun's rays to hit the Earth's surface at different angles throughout the year, resulting in varying levels of solar irradiation at different latitudes.

How does the Earth's atmosphere affect solar irradiation?

The Earth's atmosphere plays a crucial role in determining the amount of solar irradiation that reaches the Earth's surface. It filters out some of the sun's radiation, resulting in a decrease in solar irradiation. The atmosphere also scatters and absorbs solar radiation, which affects the amount of energy that reaches the Earth's surface.

Why is solar irradiation important?

Solar irradiation is essential for sustaining life on Earth. It provides the energy needed for photosynthesis in plants, which is the basis of the food chain. It also plays a significant role in weather patterns, ocean currents, and climate. Additionally, solar irradiation is harnessed as a renewable energy source for electricity and heat production.

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