Why does heat lose its energy dramatically as I move back?
Say I have a fire around 0.5 meters in front of me, I can clearly feel the heat, however, as I move even very slightly back, say 1 meter back, I will notice a dramatic fall in the heat I measure from the fire.
Similarly, the Sun radiates very hot radiation that finally reaches us and heats our planet for living things to exist. Now I wonder why heat is lost as it travels slightly.
My first hypothesis is that as light/heat (radiation) travels, the light loses its energy through interactions with particles and air molecules, gives them energy and therefore loses its energy. This may be a valid (or credible) hypothesis locally (on Earth), however, in space, this loses its credibility because there are clearly no masses of particles that normally reduce the energy of the sun's rays.
What actually causes this dramatic fall of heat as I move away, even slightly?
If dispersion happens, what about this heat? Next, if radiation or this heat is carried in photons and photons are particles and waves, how can these waves or particles disperse heat as if photons act as a fluid in large numbers?
Best Answer
It is not lost. It is spread more out.
If you stand so close to the heat source that you are hit by, say 1/10 of it's radiation (1/10 of all photons sent out hit you), then when standing further away you are maybe only hit by 1/100.
The heat radiation sent from the source like the sun spreads out with distance. The same amount of energy every second is spread out at a larger and larger sphere as it travels away from the source.
Surface area of such a sphere is $A=4 \pi r^2$. Doubling your distance to the campfire means:
$$A_2=4 \pi r_2^2=4 \pi (2 r_1)^2=4 *4 \pi r_1^2=4 A_1$$
The area that the radiation is spread over is four times as large for just the double distance.
This of course regards the sun with only space surrounding it. Since the campfire is place on ground, downwards radiation in that case will be smaller.
Solar constant
For your interest, the Suns' intensity on our planet Earth is called the solar constant (or solar coefficient) $S$. In units $\mathrm{W/m^2}$. This tells us how much radiation that reaches a square meter on Earth (or any other planet at the same distance from the Sun) every second. Mercury which is closer will have another solar constant.
To find the Suns' intensity at any distance, you will need to know how much energy is generated per second within the Sun, that is the power of the Sun $P$. This energy will be spread out while travelling:
$$P=S*A=4 S \pi r^2$$
where both intensity $S$ and area $A$ are for a specific sphere at a specific distance. The Suns' own intensity at its surface is then found by insertings the Suns' own radius and isolate S.
Also, our Sun can be viewed as a socalled blackbody. That is, it emits radiation very efficiently. The Stefan–Boltzmann law of blackbody radiation then gives us the emissive power from the Suns' surface:
$$P=\sigma T^4 A= \sigma T^4 4 \pi r^2 $$
with $T$ being the surface temperature of the Sun (around $5800^\mathrm{o C}$ if I remember correctly).
Campfire
Regarding the campfire as pointed out in the comments, convection might be considerable if you are standing very close to the campfire or maybe even reaching over it.
By natural convection, heated air will flow upwards, and this will carry a lot of energy that way. The radiation itself is negligible at that position.
Walking slightly further away might remove the convection effect entirely. This will feel like a huge decrease in heating. Adding a little windy weather, you might not feel any wind while being in "equilibrium" in the convection zone near the campfire. But walking two steps away can have a large cooling effect now that forced convective cooling from the wind acts as well.