Saber Posted February 14, 2023 Share 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 ? Link to comment Share on other sites More sharing options...
Sensei Posted February 14, 2023 Share 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 Link to comment Share on other sites More sharing options...
swansont Posted February 14, 2023 Share 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 Link to comment Share on other sites More sharing options...
Saber Posted February 14, 2023 Author Share 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 Link to comment Share on other sites More sharing options...
exchemist Posted February 14, 2023 Share Posted February 14, 2023 1 minute ago, 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 ? It's not clear what you have done. Are those diameter measurements? Link to comment Share on other sites More sharing options...
Sensei Posted February 14, 2023 Share Posted February 14, 2023 (edited) 1 hour ago, 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 ? 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 Link to comment Share on other sites More sharing options...
Saber Posted February 14, 2023 Author Share Posted February 14, 2023 22 minutes ago, 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. Except from albedo the comparison i meade i meant for the unit area ( same amount of are ) on each planet Link to comment Share on other sites More sharing options...
Sensei Posted February 15, 2023 Share Posted February 15, 2023 10 hours ago, Saber said: the comparison i meade i meant for the unit area ( same amount of are ) on each planet 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 Link to comment Share on other sites More sharing options...
exchemist Posted February 15, 2023 Share Posted February 15, 2023 (edited) 12 hours ago, Saber said: Except from albedo the comparison i meade i meant for the unit area ( same amount of are ) on each planet 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 Link to comment Share on other sites More sharing options...
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