- #1
robert Ihnot
- 1,059
- 1
Seeing "The Chronicles of Riddick" there is described a planet subject to melting distruction should one be caught out in the sunlight during its sun-star passover. Apparently things are O.K. if one stays in the shade. This has been thought not too scientific even by Ebert, a movie critic, since the planet has a breathable atmospher.
But, it does bring up the question of just how the distance from the Sun effects a planet's temperature. I found these figures: Mercury, mean temp=179 C, .3871 distance of Earth from Sun. Venus: mean temp =482C, distance .7273 of Earth. Earth: distance 1, mean temp 15C. Mars, mean temp -63C, distance 1.5273 of Earth. Mars -63C, distance 1.5273 of Earth.
Disregarding axes tilt and atmospher, would we expect as a general rule that the planet temperature would fall by the square of the distance from the Sun?
Certainly it does not work very well with Venus and Mercury. Now for Mars I calculated that we should take the absolute temperature of 273C and add it to the 15C Earth temperature and then divide by the square of the distance from the Sun getting: 288/(1.5273)^2 = 123.46 above absolute 0, or -149.5C. Which is more than twice as cold as the figure given: -63C.
Anyway, I wondered if there was any kind of sense in this?
But, it does bring up the question of just how the distance from the Sun effects a planet's temperature. I found these figures: Mercury, mean temp=179 C, .3871 distance of Earth from Sun. Venus: mean temp =482C, distance .7273 of Earth. Earth: distance 1, mean temp 15C. Mars, mean temp -63C, distance 1.5273 of Earth. Mars -63C, distance 1.5273 of Earth.
Disregarding axes tilt and atmospher, would we expect as a general rule that the planet temperature would fall by the square of the distance from the Sun?
Certainly it does not work very well with Venus and Mercury. Now for Mars I calculated that we should take the absolute temperature of 273C and add it to the 15C Earth temperature and then divide by the square of the distance from the Sun getting: 288/(1.5273)^2 = 123.46 above absolute 0, or -149.5C. Which is more than twice as cold as the figure given: -63C.
Anyway, I wondered if there was any kind of sense in this?