The sun's intentsity is inversely proportional to distance squared. So from aphelion to perihelion it varies by the ratio of those two distances squared, which comes out to about 1.069.
Good thing I used 1.5 x 10^11 m :) I didn't know how big the variation was when making the video, but I just looked it up and the perihelion is 1.47 x 10^11 m, while the aphelion is 1.52 x 10^11 m. When I was making this video, initially I used more precise values (including a more precise value for the mean Earth-sun distance), but the result wasn't quite coming out to what was published in the IB Physics data booklet (which is the course I made this video for). Overall, the level of precision in these estimates is pretty low.
@@danielm9463 I looked up the distances on timeanddate. I am not familiar with the booklet you refer to. So, scientists say the intensity varies by 7%, which I guess comes from that 1.069 value. Seems like that is enough to affect climate and so on. It is just something I find interesting.
Great question! What you've proposed is a perfectly valid way to think about it, but there's something else we still need to add in. Which of the following has more area: the half-sphere of the Earth that is actually illuminated, or the 2D solar radiation disc that hits us? Before I answer that question, consider this: whenever we zoom in close enough on a sphere, the sphere looks flat. For example, if we zoom in far enough on the Earth, then the Earth looks flat and we can't see its curvature. Similarly, if we zoom in far enough on the solar radiation sphere, it looks flat upon reaching the Earth. Hence, the shape of the solar radiation which strikes Earth is a 2D disc, not a 3D sphere or 3D hemisphere. We'll come back to that point in a bit. For now, imagine that the solar radiation sphere is a big yellow balloon in the middle of a room. The Earth is a tiny pea a few feet away. We blow up the balloon bigger and bigger and bigger until it reaches the pea. Then, we cut out a tiny hole in the giant balloon, and we make sure the hole is just the right size so that the pea could just barely pass through the balloon. Because the hole is so tiny compared to the entire balloon, we can neglect any curvature that the hole has and treat it like a 2D shape. When we cut out the 2D hole, we shape it like a disc that matches the **profile** of the pea--that is, we cut a 2D hole whose radius r matches the radius r of the pea. Once the cut is complete, we're holding a yellow 2D rubber disc that we cut from the balloon, and the area of that yellow 2D rubber disc is πr^2. That disc represents the tiny bit of solar radiation that the Earth captures. We take that yellow 2D rubber disc, and we say "only half of the Earth is illuminated at once," so stretch the yellow rubber disc across half of the pea's surface. But notice what we just did: we had to **stretch** that yellow 2D rubber disc across half the pea. The disc's area doesn't match the half-pea's surface area. In fact, the tiny yellow 2D rubber disc that we cut out from the balloon has an area πr^2, and the half-pea has an area 2πr^2. So we divide by 2 to account for this initial stretching. Then we divide by 2 again to account for the second half of the Earth/pea that isn't illuminated.
Sir, So why is the part that is not illuminated is to be considered??? Considering it is equivalent to say that the back(dark) part of the earth's shape (now hemisphere) where there is no light falling upon, effects the intensity on the front(bright) side . And how can the back surface structure effect, the intensity at the front.
When we divide by 4, we're getting an AVERAGE value for every point on Earth. This average is useful, because every point on Earth passes through light and dark times. When looking at climate models, temperature, global warming at a particular point, etc., the average turns out to be very useful. (For those climate models, it would be incorrect to assume that the Sun shines continuously on any single point.)
So, the surface temperature isn't part of this calculation, which means that value wouldn't help us when adjusting the calculation for Venus. (Actually, this calculation is the beginning steps in a larger procedure for predicting Earth's surface temperature.) To adjust the calculation for Venus, we would want to start out with the distance from Venus to the Sun, and the other fact we'd need is the average albedo of Venus. If we know the insolation, then we could short-cut most of the calculation and jump to the later parts of the calculation.
At 1:39 why you took the area of the circle. You should take the area of the 3d sphere because energy is spreading in all directions not only in a 2d plane.
The earth receives 1,360 W per meter squared because every square meter of solar luminosity at one AU receives 1,360 W per meter squared. This is due to the inverse square law. You reduce this to about 1,060 W per meter squared due to the reflectivity and absorption of energy in the atmosphere. If you wanted to calculate the energy total hitting earth at one time, you would use pi times the radius of earth squared. This would give you 2.72x10^16 W or about 27,000 terawatts (not accounting for atmospheric absorption/reflection)
Thanks for the post. A couple clarifications: this is for an IB Physics course, and the values I'm using in this video come from the International Baccalaureate Program (specifically, from the Physics PhDs who write the curriculum). That curriculum is rewritten every 6-7 years, so the values we use do sometimes become outdated. The value you quoted (1,360 W/m^2) isn't the exact same as the IB uses, but you'll notice that the IB wants students to *calculate* the solar intensity at 1 AU, which can involve some estimation (i.e., in the radius of the Sun, the Sun-Earth distance, the Sun's surface temperature, etc.). Within 2 sig figs, the estimated/calculated value of 1397 W/m^2 agrees with the value you cite (1360 W/m^2). They're both ~1400 W/m^2, which is fine for a high school physics course. Secondly, I want to clarify that the purpose of this video isn't to calculate the total energy hitting the Earth at one time. However, I think you're picking up on a subtlety. If we follow your approach, we can multiply the intensity by (pi)(r_Earth)^2 to get the total power striking the Earth at one moment. In this video, I take another step: I calculate the average intensity per square meter on Earth (including the square meters that are in darkness because it's night). For the average intensity on Earth, we take the *total* power and divide by 4(pi)(r_Earth)^2. This gives: (~1400)(pi)(r_Earth)^2 / (4)(pi)(r_Earth)^2. So the method I've used (dividing the intensity of ~1400 by 4) is a mental shortcut for performing the exact same calculation you've begun. The way I show it in this video is how IB has chosen to explain it to students. Last thing--in the IB Physics curriculum, students assume an average global albedo of 0.3. Perhaps that's an outdated value: earthobservatory.nasa.gov/images/84499/measuring-earths-albedo But your numbers suggest an albedo of (1360 - 1060) / (1360) = ~0.22, which seems low to me--where are you getting your values?
Daniel M, Thank you for your reply. I see now that it is a classroom physics problem. I was pointing out that if you were concerned about power output of solar panels, you would not want to use the value for solar irradiance averaged over the entire surface of the Earth. Also, reducing solar intensity due to Earth's albedo in a practical application would include some of the factors that wouldn't be relevant such as snow and other reflectivity. This is because the light striking these surfaces is not applicable to your solar panel. Sorry to comment in on your classwork. Thank you and have a great day.
@@MrShadRiley Ah, yes--I completely see where you were coming from. In the context of solar panels, everything you've said is exactly right, of course. In the IB curriculum, these ideas are a precursor to climate models. (And average intensities are helpful for purposes of predicting Earth's average temperature.) Thanks for the thoughtful comments! Have a great day.
So for the measure of a actual solar radiation reaching the surface as measured by a weather station. Would you take the solar constant, times that by 0.7 and then times by 2 for a maximum value of 820 w/m2? I assume this is only valid for a sunny day where sunlight hits directly overhead. For mid latitudes I assume the calculation is reduced somehow.
I times it by 2 because I am not looking for the average but what is actually striking a point on the earth or in this case where the sun is directly overhead with an albedo of 0.3 and only one half is illuminated
@@brianp2035 The value of the solar constant is how intense the solar radiation is at a given distance from the Sun. If we use that value in a calculation involving the Earth, then we're already assuming that all of the solar radiation is being transmitted. In other words, we're already assuming that it's daytime and the solar rays are striking head-on. So if you do (1360 W/m^2)(0.3), then this is the intensity at noon at the equator when the Sun is directly overhead. Because it only applies at a single time of day, we often divide by 4 to get the average across all points on Earth and across all times of day.
Hi Daniel, thanks for your video! It really helps me to understand this concept. May I have your permission to use the illustration of the Sun and the Earth in my bachelor thesis for visual explanation? I'll be making sure to include your name as the author and the source.
if the earth receives 244w p/m sq, how does that work with a solar panel that is rated at 300w? Can we receive more power than what it receives? Thank you
Great question! There are two important things to consider: (1) The value 244 is the wattage *per each square meter,* while the value 300 is the total wattage *of the entire panel* (which is bigger than just 1 square meter). To make a fair comparison, we wouldn't use 300W--we would use 300W/(area of the solar panel). (2) When comparing to the panel's performance, we wouldn't use a solar intensity of 244W/m^2, because that's not what the company itself is using. The 300W value is a maximum theoretical estimate when the sun is directly overhead, when there are no clouds, etc. By contrast, the 244 W/m^2 value accounts for both night and day, and it accounts for average cloud cover. To make a direct comparison, we'd be better off using 1397 W/m^2 as the solar intensity. Indeed, when we account for both of these factors, I got an efficiency of ~13% for the first 300W solar panel I found online. Here's how I got it: the 300W panel I found has an area of (1.65 m) x (1.02 m) = 1.68 m^2. The solar intensity it coverts into electricity is 300W / 1.68m^2 = 179 W/m^2 at best. The efficiency would be 179/1397 = 13%, which is in range of what we would expect.
hi! please can you help me with this calculations. with your knowledge of solar constant, what is the area in meter square required to generate 676.5watt
Great question! The value 676.5 watts is the power. The solar constant value is in W/m^2, which makes it the intensity. You are looking for area. The relevant equation is I = P/A or intensity = power / area. Which quantities in that equal do you know the value of? Which quantity are you seeking? See if you can use the given values to solve for area.
High can You please do a video how to calculate earth surface and atmosphere temperature on a single layer model please if sigma is 0.9 and atmosphre emmisivity to short wave also n is less than 1 and for long wave also less than 1. Many thanks
You might be looking for one of these videos: ruclips.net/video/JLTvWRl47ng/видео.html&hd=1 ruclips.net/video/2_iZ6mDmKC0/видео.html&hd=1 The second video goes through several different examples.
So, would it be 349.25 W/m^2 for the moon aswell if the moon was right beside the earth in its orbit?
As in the distance from the sun to the moon was the same as the distance from the sun to earth.
Great question! You're exactly right. That value correlates with any spherical object located that particular distance from the Sun.
Got it. Thank you!
Daniel M but the area of moon is comparatively lower than the earth the it should be recieving less than earth’s?
@@zig1792 if you look at the equation the pir^2 cancels out. So it only depends on the distance from the sun. It is average per metre squared
Unlike Other RUclipsr Without Any Annoying Intros Directly Into The Content..Appreciatable One...BEST OF LUCK😘
Thanks, Buddy!!!! really helped me out...
You're welcome!
The sun's intentsity is inversely proportional to distance squared. So from aphelion to perihelion it varies by the ratio of those two distances squared, which comes out to about 1.069.
Good thing I used 1.5 x 10^11 m :) I didn't know how big the variation was when making the video, but I just looked it up and the perihelion is 1.47 x 10^11 m, while the aphelion is 1.52 x 10^11 m. When I was making this video, initially I used more precise values (including a more precise value for the mean Earth-sun distance), but the result wasn't quite coming out to what was published in the IB Physics data booklet (which is the course I made this video for). Overall, the level of precision in these estimates is pretty low.
@@danielm9463 I looked up the distances on timeanddate. I am not familiar with the booklet you refer to. So, scientists say the intensity varies by 7%, which I guess comes from that 1.069 value. Seems like that is enough to affect climate and so on. It is just something I find interesting.
@@comic4relief thanks for sharing!
Thank you for sharing your ideas
thank you very much
Thanks a lot , by this small animation you cleared all my doubts 🙏🙏
Helping hand ✌
Only half the surface area of the earth is illuminated at any one time so shouldn't you divide I by 2 instead of 4 at 4:15?
Great question! What you've proposed is a perfectly valid way to think about it, but there's something else we still need to add in. Which of the following has more area: the half-sphere of the Earth that is actually illuminated, or the 2D solar radiation disc that hits us? Before I answer that question, consider this: whenever we zoom in close enough on a sphere, the sphere looks flat. For example, if we zoom in far enough on the Earth, then the Earth looks flat and we can't see its curvature. Similarly, if we zoom in far enough on the solar radiation sphere, it looks flat upon reaching the Earth. Hence, the shape of the solar radiation which strikes Earth is a 2D disc, not a 3D sphere or 3D hemisphere. We'll come back to that point in a bit. For now, imagine that the solar radiation sphere is a big yellow balloon in the middle of a room. The Earth is a tiny pea a few feet away. We blow up the balloon bigger and bigger and bigger until it reaches the pea. Then, we cut out a tiny hole in the giant balloon, and we make sure the hole is just the right size so that the pea could just barely pass through the balloon. Because the hole is so tiny compared to the entire balloon, we can neglect any curvature that the hole has and treat it like a 2D shape. When we cut out the 2D hole, we shape it like a disc that matches the **profile** of the pea--that is, we cut a 2D hole whose radius r matches the radius r of the pea. Once the cut is complete, we're holding a yellow 2D rubber disc that we cut from the balloon, and the area of that yellow 2D rubber disc is πr^2. That disc represents the tiny bit of solar radiation that the Earth captures. We take that yellow 2D rubber disc, and we say "only half of the Earth is illuminated at once," so stretch the yellow rubber disc across half of the pea's surface. But notice what we just did: we had to **stretch** that yellow 2D rubber disc across half the pea. The disc's area doesn't match the half-pea's surface area. In fact, the tiny yellow 2D rubber disc that we cut out from the balloon has an area πr^2, and the half-pea has an area 2πr^2. So we divide by 2 to account for this initial stretching. Then we divide by 2 again to account for the second half of the Earth/pea that isn't illuminated.
Sir, So why is the part that is not illuminated is to be considered??? Considering it is equivalent to say that the back(dark) part of the earth's shape (now hemisphere) where there is no light falling upon, effects the intensity on the front(bright) side . And how can the back surface structure effect, the intensity at the front.
When we divide by 4, we're getting an AVERAGE value for every point on Earth. This average is useful, because every point on Earth passes through light and dark times. When looking at climate models, temperature, global warming at a particular point, etc., the average turns out to be very useful. (For those climate models, it would be incorrect to assume that the Sun shines continuously on any single point.)
Thank you so much :) This video helped me so much
Feel free to post any questions you have as you're studying!
Thank you so much .well explained 🔥
I'm glad it helped!! This is a concept I struggled with until I thought about it visually.
So can you calculate that same for Venus knowing Venus is receiving 1.97 times more insolation but surface temperature is 460 degeree?
So, the surface temperature isn't part of this calculation, which means that value wouldn't help us when adjusting the calculation for Venus. (Actually, this calculation is the beginning steps in a larger procedure for predicting Earth's surface temperature.) To adjust the calculation for Venus, we would want to start out with the distance from Venus to the Sun, and the other fact we'd need is the average albedo of Venus. If we know the insolation, then we could short-cut most of the calculation and jump to the later parts of the calculation.
At 1:39 why you took the area of the circle. You should take the area of the 3d sphere because energy is spreading in all directions not only in a 2d plane.
The earth receives 1,360 W per meter squared because every square meter of solar luminosity at one AU receives 1,360 W per meter squared. This is due to the inverse square law.
You reduce this to about 1,060 W per meter squared due to the reflectivity and absorption of energy in the atmosphere.
If you wanted to calculate the energy total hitting earth at one time, you would use pi times the radius of earth squared. This would give you 2.72x10^16 W or about 27,000 terawatts (not accounting for atmospheric absorption/reflection)
Thanks for the post. A couple clarifications: this is for an IB Physics course, and the values I'm using in this video come from the International Baccalaureate Program (specifically, from the Physics PhDs who write the curriculum). That curriculum is rewritten every 6-7 years, so the values we use do sometimes become outdated. The value you quoted (1,360 W/m^2) isn't the exact same as the IB uses, but you'll notice that the IB wants students to *calculate* the solar intensity at 1 AU, which can involve some estimation (i.e., in the radius of the Sun, the Sun-Earth distance, the Sun's surface temperature, etc.). Within 2 sig figs, the estimated/calculated value of 1397 W/m^2 agrees with the value you cite (1360 W/m^2). They're both ~1400 W/m^2, which is fine for a high school physics course.
Secondly, I want to clarify that the purpose of this video isn't to calculate the total energy hitting the Earth at one time. However, I think you're picking up on a subtlety. If we follow your approach, we can multiply the intensity by (pi)(r_Earth)^2 to get the total power striking the Earth at one moment. In this video, I take another step: I calculate the average intensity per square meter on Earth (including the square meters that are in darkness because it's night). For the average intensity on Earth, we take the *total* power and divide by 4(pi)(r_Earth)^2. This gives: (~1400)(pi)(r_Earth)^2 / (4)(pi)(r_Earth)^2. So the method I've used (dividing the intensity of ~1400 by 4) is a mental shortcut for performing the exact same calculation you've begun. The way I show it in this video is how IB has chosen to explain it to students.
Last thing--in the IB Physics curriculum, students assume an average global albedo of 0.3. Perhaps that's an outdated value: earthobservatory.nasa.gov/images/84499/measuring-earths-albedo But your numbers suggest an albedo of (1360 - 1060) / (1360) = ~0.22, which seems low to me--where are you getting your values?
Daniel M,
Thank you for your reply. I see now that it is a classroom physics problem. I was pointing out that if you were concerned about power output of solar panels, you would not want to use the value for solar irradiance averaged over the entire surface of the Earth. Also, reducing solar intensity due to Earth's albedo in a practical application would include some of the factors that wouldn't be relevant such as snow and other reflectivity. This is because the light striking these surfaces is not applicable to your solar panel.
Sorry to comment in on your classwork. Thank you and have a great day.
@@MrShadRiley Ah, yes--I completely see where you were coming from. In the context of solar panels, everything you've said is exactly right, of course. In the IB curriculum, these ideas are a precursor to climate models. (And average intensities are helpful for purposes of predicting Earth's average temperature.) Thanks for the thoughtful comments! Have a great day.
How to find the solar energy received per unit area per unit time at different latitudes of earth?
very helpful!!
So glad it helped!
So for the measure of a actual solar radiation reaching the surface as measured by a weather station. Would you take the solar constant, times that by 0.7 and then times by 2 for a maximum value of 820 w/m2? I assume this is only valid for a sunny day where sunlight hits directly overhead. For mid latitudes I assume the calculation is reduced somehow.
I times it by 2 because I am not looking for the average but what is actually striking a point on the earth or in this case where the sun is directly overhead with an albedo of 0.3 and only one half is illuminated
@@brianp2035 The value of the solar constant is how intense the solar radiation is at a given distance from the Sun. If we use that value in a calculation involving the Earth, then we're already assuming that all of the solar radiation is being transmitted. In other words, we're already assuming that it's daytime and the solar rays are striking head-on. So if you do (1360 W/m^2)(0.3), then this is the intensity at noon at the equator when the Sun is directly overhead. Because it only applies at a single time of day, we often divide by 4 to get the average across all points on Earth and across all times of day.
Hi Daniel, thanks for your video! It really helps me to understand this concept. May I have your permission to use the illustration of the Sun and the Earth in my bachelor thesis for visual explanation? I'll be making sure to include your name as the author and the source.
You have my permission :) Good luck in your thesis! I made this originally in powerpoint if you want to try re-creating the original diagram.
PS -- what's the topic of your thesis/
Thank you Daniel! I’m doing “Solar Power in Space” :) and I’m using your illustration to explain about solar constant.
if the earth receives 244w p/m sq, how does that work with a solar panel that is rated at 300w? Can we receive more power than what it receives? Thank you
Great question! There are two important things to consider: (1) The value 244 is the wattage *per each square meter,* while the value 300 is the total wattage *of the entire panel* (which is bigger than just 1 square meter). To make a fair comparison, we wouldn't use 300W--we would use 300W/(area of the solar panel). (2) When comparing to the panel's performance, we wouldn't use a solar intensity of 244W/m^2, because that's not what the company itself is using. The 300W value is a maximum theoretical estimate when the sun is directly overhead, when there are no clouds, etc. By contrast, the 244 W/m^2 value accounts for both night and day, and it accounts for average cloud cover. To make a direct comparison, we'd be better off using 1397 W/m^2 as the solar intensity.
Indeed, when we account for both of these factors, I got an efficiency of ~13% for the first 300W solar panel I found online. Here's how I got it: the 300W panel I found has an area of (1.65 m) x (1.02 m) = 1.68 m^2. The solar intensity it coverts into electricity is 300W / 1.68m^2 = 179 W/m^2 at best. The efficiency would be 179/1397 = 13%, which is in range of what we would expect.
@@danielm9463 thank you
Thank q sir
You're welcome!
hi! please can you help me with this calculations.
with your knowledge of solar constant, what is the area in meter square required to generate 676.5watt
Great question! The value 676.5 watts is the power. The solar constant value is in W/m^2, which makes it the intensity. You are looking for area. The relevant equation is I = P/A or intensity = power / area. Which quantities in that equal do you know the value of? Which quantity are you seeking? See if you can use the given values to solve for area.
High can You please do a video how to calculate earth surface and atmosphere temperature on a single layer model please if sigma is 0.9 and atmosphre emmisivity to short wave also n is less than 1 and for long wave also less than 1. Many thanks
You might be looking for one of these videos:
ruclips.net/video/JLTvWRl47ng/видео.html&hd=1
ruclips.net/video/2_iZ6mDmKC0/видео.html&hd=1
The second video goes through several different examples.
Thank you very much
You're welcome--glad to help!
HIGHLY RECOMMENDED FOR THE CONCEPT
the best video with in-depth visual learning yet with simplest approach possible
"highly appreciated"
thanks Daniel
Thanks @dragon m!