Yea I'm, asking you to give me a couple examples of any reasonably respected mathematical resource that use the notation you are using.

that does not address the fact you used a relative scale to make a proportion argument. Also as a footnote, the heat you get from the sun, is in the form of photons as well.

It's called a google search - it's a very scientific tool - do a search for the phrase you think is incorrect and see if google shows circumference or area results.
Then explain to me why you think you're smarter than google.



And how does a relative scale not work on a proportion augment in this case, Does Mars not receive a portion of the Sun's energy relative to Earth? Is heat not a form of energy?
I thought you were just making wild assumptions, how do you account for the lack of what I was inferring or you're not quite answering the op's question?

This post was edited by card_sultan on Mar 12 2016 04:49pm

Because 10 degrees F is not twice as much heat as 5 degrees F.

and maybe I should add: 75F is not 75% as hot as 100F.

According to the kelvin scale or the Fahrenheit scale? Are you saying there can only be a mathematical solution based on the surface area of the sun divided by the distance away from it?
What is wrong with a relative solution? What is the actual number Nasa is using to predict the efficiency of its Mars rover solar panels? Do you even care about answering questions or is it simply some ego thing for you? Why haven't you even made a prediction of the intensity of light on Mars, but you're questioning my relative one?

Fahrenheit, that is why there is an F next to the number...

Get this through your head: I NEVER CALCULATED THE SURFACE AREA OF THE SUN, NO PART OF MY MODEL RELIES ON IT IN ANY WAY.

Get this through your head: 75F isn't 75% 'as hot' as 100F.

I don't work for NASA, ask them; but I bet you they did it in a very similar way to how I did it. Although funny story: they still owe me $100.

I laid the ground work for one, but if you want I'll finish it:
As I established a unit area in orbit around the Earth receives about 2.3 times as much sunlight as a unit area orbiting Mars. However, on the surface of the Earth loses an additional 25% do to the atmosphere while Mars loses essentially 0% from its atmosphere. Therefore: for each units that arrives at the surface of the Earth, 1/.75 ≈ 1.33 arrives in orbit around the Earth, and 1.33/2.3 ≈ .58 ≈ 58% reaches the surface of Mars.

You'll learn in any introductory science class you can't do calculations that rely on temperature with anything but Kelvin, unless you're equation uses temperature change instead of just temperature in which case you can also use celsius.

I told the exact same thing to the weather channel, but they keep telling me Fahrenheit, idiot Meteorologists.