Want to know about distance from Sun and temperature of Planet

[robert Ihnot]

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?

 

[Clausius2]

You should employ:

T=Ts*(Rs/(D*2))^0.5

where Ts=average sun temperature. I don't remember it, but it is calculated of Steffan law, aprox 6000K using Black Body hypothesis and the Abbot Constant. You can search it in books as equivalent sun temperature. This means at this temperature a black body emites the same spectrum of the sun.

Rs=sun radius
D=average distance sun-planet
T=average planet temperature.

If you want to know of where surges this formula, I will explain you, although you can consider an energy balance of radiation between sun and planet. But first please check it's a good aproximation.

 

[LURCH]

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?

Ideally, yes. However, the key to your question is that we disregard atmosphere. In reality, atmosphere is what makes the huge difference between mercury, Venus, Earth and Mars. Other factors that could affect surface temperature on the planet might include internal heating.

 

[Nereid]

Albedo, don't forget the amount of radiation reflected back into space by the surface!

If the surface of the planet - with no atmosphere - were perfectly reflective (all incident photons reflected back the way they came), how hot would you be if you were in a room (with a proper atmosphere, heating and aircon and so on) that was 1m below the surface? What about 1cm, 1mm??

If the surface reflected merely 99.99% of the incident energy, how hot would it get? (enter: thermal intertia!) What assumptions do you need to make about the rotation rate to get any decent answers?

 

[robert Ihnot]

Checking this out with regards to the Earth temperature

You should employ:

T=Ts*(Rs/(D*2))^0.5

where Ts=average sun temperature. I don't remember it, but it is calculated of Steffan law, aprox 6000K using Black Body hypothesis and the Abbot Constant. You can search it in books as equivalent sun temperature. This means at this temperature a black body emites the same spectrum of the sun.

Rs=sun radius
D=average distance sun-planet
T=average planet temperature.

If you want to know of where surges this formula, I will explain you, although you can consider an energy balance of radiation between sun and planet. But first please check it's a good aproximation.

Well, we have Ts = 6000K, Rs = 695,000 km, Distance of Earth = 149,600,000 km, so when I work this out we get Te =289.18K = 16.18C, which is very good approximation!

 

[Clausius2]

Allright: (s=steffan constant)

Watts per surface unit that reach an imaginary sphere at distance D:

q''=s*Ts^4*(Rs/D)^2

Watts absorbed by the Earth:

q=s*Ts^4*pi*(Rt*Rs/D)^2

note that I suppose an Earth exposed area of pi*Rt^2 (1/4 of total area).

Supposing that the Earth emits like a Black Body:

watts emited by the Earth qe=4*pi*Rt^2*s*T^4

in the thermal equilibrium we have q=qe

T=Ts*(Rs/(D*2))^2

You can realize we made large amount of aproximations. I think we had to take into account geometrical factors to obtain better aproximation. It would be a very interesting problem playing with shape factors and emisivities different of 1.

 

[ArmoSkater87]

Does that model have such good approximations for all the planets? It seems like it wont, considering that the atmospheres of all the planets are so different. Especailly how Venus has a much higher temp. than Mercury even though Mercury is closer. Since Venus has the strongest greenhouse effect in the solar system, that makes it much hotter than Mercury. In other words, i think that model would work very well if all the planets had an atmosphere similar to earth, in which case the only factor in the temp. variations would be the amount of light gets there. Thats my thought. :D

 

[Clausius2]

I am agree with you. It likely does not work for all planets. But I was not pretending it. If you watch the formulae derivation I have not had into account scattering effects into a particular atmosphere, its composition, emisivities and reflectivities. Although Mercury is nearer sun than Venus, Perhaps this one has an atmosphere very absorbantly, and average temperature could be greater.
It is embarrasing to try to explain temperature distributions in the Solar System with this simple model. Anyway, it works if you want a first aproximated number.

 

[ArmoSkater87]

robert Ihnot: I don't know if this helps but, basically the amount of light that gets to a planet is what makes it heat up in the first place. Depending on the planet and atmosphere, more or less is reflected. But if you had a uniform atmosphere that stretches from near the sun, to the end of the solar sytem, the amount of light that hit the air molecules would have the influence of the variation of temperatures in diffrent parts of the atmosphere. Since light from the sun is emitted in all directions, less light will get to farther places.

 

[Clausius2]

What the f .. are you saying? :surprise:

That is just what I had into account in order to derive the equation!.

Please, decide yourself and your answers before posting, unless you want me to think you only write to criticize people.

 

[ArmoSkater87]

no no no, i was just kind of summing up everything for him. I didnt mean to take your work, lol. :D