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If a piece of iron is right next to the Sun, it will melt. But far away, it'll just stay iron. My question is, in terms of big $\Theta$ notation, how does the temperature a piece of iron feels from the Sun vary with distance from the sun? E.g. $T\in\Theta(1/r^3)$ means that, roughly, doubling the distance means experiencing 1/8th the temperature.

To be clear, I don't want crazy specifics in terms of the radius of the Sun, etc. Assume free space, and a point heat source. E.g. gravity's effect goes as $\Theta(1/r^2)$.

Edit: For those not familiar, Big $\Theta$ notation is used to say that something is roughly proportional in the limit.

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  • $\begingroup$ I think this is related to the inverse square law. So the heat energy reaching the object would be inversely proportional to the square of the distance. $\endgroup$ Commented Feb 21, 2022 at 12:21
  • $\begingroup$ Minor comment: for order-of-magnitude statements, it’s more common to use a cursive $\mathcal O$ (produced via $\mathcal O$) than an upper-case theta $\Theta$. $\endgroup$
    – rob
    Commented Feb 24, 2022 at 14:21
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    $\begingroup$ @rob it is, in fact, a common misuse of notation. Big O is only for giving an upper bound. Big Theta is for bounding from above and below. That's the convention in Math and Computer Science. en.wikipedia.org/wiki/Big_O_notation $\endgroup$
    – chausies
    Commented Feb 25, 2022 at 3:09
  • $\begingroup$ Oh, very interesting, thank you. $\endgroup$
    – rob
    Commented Feb 25, 2022 at 5:20

3 Answers 3

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The absolute temperature, $T$ of the iron varies with its distance, $r$, from the Sun according to $$T\propto \frac 1{\sqrt r}.$$

My reasoning is as follows.

If the radiant power of the sun is $L$, the solar energy passing through an imaginary spherical surface of radius $r$, centred on the Sun, per unit time, per unit area will be $$I=\frac L{4\pi r^2}.$$ So we have an inverse square law of intensity. If we place an iron sphere of radius $a$ at distance $r$ from the Sun, the solar power intercepted will be $$P=\pi a^2 I=\frac{\pi a^2 L}{4\pi r^2}.$$ The sphere will reach a temperature at which the rate of solar gain is equal to the rate of loss of energy by radiation. We will assume that the sphere absorbs all the intercepted radiation and that it emits as a 'black body' obeying Stefan's law, that is at a power of $$P_{em}=\sigma A T^4= \sigma 4\pi a^2 T^4$$ in which $\sigma$ is the Stefan constant and $A$ is the surface area. So putting $P=P_{em}$ we get $$T^4=\frac L {16\pi \sigma r^2},$$ from which follows the relationship given at the top.

The above result will not be accurate if the sphere doesn't absorb and emit as a black body, unless the sphere happened to absorb the same fraction $f$ of the solar power at all wavelengths and angles of incidence (the grey body idealisation). In this case it would also emit only a fraction $f$ of the power predicted by Stefan's law, so the factor of $f$ would cancel. [My misleading original remarks on non-black bodies were pointed out by Bert Barrois.]

Numerical example: We know that $\sigma =5.67 \times 10^{-8}\ \text{W m}^{-2}\text K^{-4}$ and $L=3.63 \times 10^{26}\ \text W$, so if $r=1.50 \times 10^{11}\ \text{m}$ (that is the distance to the Earth from the Sun), we find that $T=274$ K. Plausible?

Note that this answer applies to a macroscopic body, not to the particles making up the interplanetary plasma. Nor should it be applied to a body close to the Sun's surface. These points were raised by the contributor Math Keeps Me Busy.

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  • $\begingroup$ @Chausies I've not met your 'Big theta' notation before, but, as I'm sure you've seen, my result translates as $$T\in \Theta \left(\frac 1{\sqrt r}\right)$$ $\endgroup$ Commented Feb 21, 2022 at 13:22
  • $\begingroup$ @Math Keeps Me Busy Edit made (see end of answer). Hope it does the job. $\endgroup$ Commented Feb 22, 2022 at 12:48
  • $\begingroup$ Major pitfall: Absorptivity & emissivity may vary with wavelength and angle of incidence. In that case, you cannot assume that f/f'=1. $\endgroup$ Commented Feb 22, 2022 at 12:53
  • $\begingroup$ @Bert Barrois. Many thanks for this comment. What I wrote was misleading. Hope I've taken care of it in my edit. $\endgroup$ Commented Feb 22, 2022 at 13:36
  • $\begingroup$ Shouldn't we only consider the surface area of the iron sphere that is in the direct line of sight of the sun (i.e. the cross section)? The sun is only shining from one direction and not from all directions which would cover the entire surface area. $\endgroup$
    – Tachyon
    Commented Apr 3 at 22:46
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Answer by Philip Wood is good, so I won't repeat it, but I will add an extension to deal with the comment made in the question concerning gravity.

The gravitational force from the Sun varies as $1/r^2$ and that is the end of the story (within Newtonian physics). Nothing can change that fact.

But the temperature of a body at some distance from the Sun, and in thermal equilibrium, is not dictated by the Sun. It depends a lot on the nature of the body. The answer given by Philip Wood is correct for a black body or a grey body. But you can also have a body whose absorptivity (and therefore also emissivity) is larger in the infra-red than in the visible region, and such a body will be colder. Or it can be the other way around. The reason why this is important for this question is that such properties can cause the equilibrium temperature at some distance $r$ from the Sun to vary as some function of $r$ different from $r^{-1/2}$. The $r^{-1/2}$ result is a good ball-park figure to give a sense of what one might expect, but it is nothing like as precise or as universal as the $r^{-2}$ result for gravitational force.

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  • $\begingroup$ A welcome extension. Would that we could quickly and painlessly increase the Earth-system's emissivity in the infrared. $\endgroup$ Commented Feb 24, 2022 at 17:47
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Interpreting the question as "What would be the temperature of a blackbody object at some distance from the Sun?":

The all-band radiance [W/m$^2$/sr] of the Sun is $$ L_{Sun} = \frac{\sigma T_{Sun}^4}{4 \pi} $$ where $\sigma$ is the Stefan-Boltzmann constant [W/m$^2$/K$^4$] and $4 \pi$ is the number of steradians in a sphere. The solid angle of a cone with its apex at the object would be $$ \omega=\frac{\pi r_{Sun}^2}{d^2} $$

where $d$ is measured from the center of the Sun. By definition, the blackbody exitance [W/m$^2$, outgoing] will equal the blackbody irradiance [W/m$^2$, incoming], $$ \sigma T_{obj}^4 = \omega L_{Sun} $$ Solving for object temperature, $$ T_{obj} = T_{Sun} \sqrt{\frac{r_{Sun}}{2d}} $$ where $T_{Sun}$ = 5778 K and $r_{Sun}$ = 6.963E8 m.

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