Saber Posted February 14, 2023 Posted February 14, 2023 Do planets from Jupiter & beyond receive any heat from the Sun ...? Or are they too far to get any heat from it ? How about Mars ?
Sensei Posted February 14, 2023 Posted February 14, 2023 The Earth's surface receives 1370 W/m^2 per surface area facing the Sun. Some of it is reflected, some is absorbed by the atmosphere, and some is absorbed by the land and sea. Knowing the distance of the Sun from the Earth, the total power generated by the Sun can be calculated. Then you can do similar calculations (check the inverse-square law) for any other space object on your list and find out what Watts/m^2 they receive. https://en.wikipedia.org/wiki/Inverse-square_law https://en.wikipedia.org/wiki/Albedo 1
swansont Posted February 14, 2023 Posted February 14, 2023 Jupiter is ~5.2 astronomical units from the sun, so it gets (1/5.2)^2 as much energy per unit area, or ~1/27, which is about 3.7% 2
Saber Posted February 14, 2023 Author Posted February 14, 2023 (edited) So in that case the heat energy received by each planet in comparison to earth ( we say earth is 1 ) is as followed right ? Edited February 14, 2023 by Saber
exchemist Posted February 14, 2023 Posted February 14, 2023 On 2/14/2023 at 9:30 PM, Saber said: So in that case the heat energy received by each planet in comparison to earth ( we say earth is 1 ) is as followed right ? Expand It's not clear what you have done. Are those diameter measurements?
Sensei Posted February 14, 2023 Posted February 14, 2023 (edited) On 2/14/2023 at 9:30 PM, Saber said: So in that case the heat energy received by each planet in comparison to earth ( we say earth is 1 ) is as followed right ? Expand See my post - you also need to take into account the radius/diameter of the planet (therefor, the area facing the Sun) and albedo. Read both Wikipedia articles carefully. Edited February 14, 2023 by Sensei 1
Saber Posted February 14, 2023 Author Posted February 14, 2023 On 2/14/2023 at 10:33 PM, Sensei said: See my post - you also need to take into account the radius/diameter of the planet (therefor, the area facing the Sun) and albedo. Read both Wikipedia articles carefully. Expand Except from albedo the comparison i meade i meant for the unit area ( same amount of are ) on each planet
Sensei Posted February 15, 2023 Posted February 15, 2023 On 2/14/2023 at 10:57 PM, Saber said: the comparison i meade i meant for the unit area ( same amount of are ) on each planet Expand Different planets have different surface areas facing the Sun. You need to calculate the surface area of the planet and use it in calculating what total Watts per planet they get, not just leave it at Watts/m^2. Then multiply these Watts by the albedo of the planet. Do this in different columns in Excel. Include the raw data too, i.e. distance, radius, albedo, area, for readers and your convenience in calculations (use the column as a data source in the equation). 1
exchemist Posted February 15, 2023 Posted February 15, 2023 (edited) On 2/14/2023 at 10:57 PM, Saber said: Except from albedo the comparison i meade i meant for the unit area ( same amount of are ) on each planet Expand OK, if you are just calculating radiation intensity, your numbers look good to me. (I used these NASA distance numbers.https://www.jpl.nasa.gov/edu/pdfs/scaless_reference.pdf ) Edited February 15, 2023 by exchemist
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