Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why does the sun's heat & light intensity decrease over AU distances?

  1. Nov 14, 2013 #1
    I'm a bit confused because the sun's radiation & heat are passing through space (with some passing through planets' atmospheres), and there is almost nothing in space, therefore, how does that immense heat & light just diminish over AU distances when its going through almost nothingness? What absorbs or transforms both of these energies (especially heat, since we can still see the sun from much further away)? Do the planets, asteroids, gasses/dust, and other small objects in space absorb the sun's total energy? So basically, where does all that heat go?
    I guess a more broad/general question would be why is outer space so cold when there's billions of stars constantly warming it? An even broader question would be why do some scientists believe that the universe is going to end in a "big freeze"? (you don't have to answer that last one lol)
  2. jcsd
  3. Nov 14, 2013 #2


    User Avatar

    Staff: Mentor

    The light doesn't decrease, it spreads out. Consider a spherical shell of a certain radius. All of the sun's energy hits it. If you double the radius, then the same energy hits a surface 4x the size and is thus 4x less intense.
  4. Nov 14, 2013 #3


    User Avatar
    2017 Award

    Staff: Mentor

    Space is huge. The light from some tiny stars is not sufficient to heat all of the universe significantly.

    At some point, there won't be enough material left to form new stars, and stars don't live forever. Together with the expansion of space, the universe will become a very dark, cold place.
  5. Nov 14, 2013 #4
    well... that's inspiring lol. But seriously though, you guys are saying that the main reason that stars' heat diminish is because the energy waves simply just spread out everywhere? So instead of heating a small area significantly (such as a planet close to it), the stars heat large areas insignificantly (such as...well... space)?
  6. Nov 14, 2013 #5


    User Avatar

    Staff: Mentor


    Surface that is heated grows as a square of the distance, but the power doesn't change. That means when you move twice further, amount of energy falling on a unit surface gets 4 times lower.
  7. Nov 14, 2013 #6


    User Avatar

    Staff: Mentor

    A similar effect happens with water waves. Throw a rock into a still pond, lake, or other body of water. The amplitude of the water waves will be highest near the center and will gradually lessen as the wavefront propagates outwards.
  8. Nov 14, 2013 #7
    Ok, well thank you guys... such as simple answer for something I've been wondering about for a while
    ahh...life never ceases to amaze lol
  9. Nov 14, 2013 #8


    User Avatar
    Science Advisor

    When I was in highschool, my physics teacher had what he called a "butter gun". It was a squirt gun with four sticks spreading out from the muzzle with places to put "toasts" inside them. The point was that there was a point where you could fit one piece of toast and "spray" butter on it. Twice that distance from the muzzle was were wires where you could put four pieces of toast- at twice the distance, by "similar triangles" (which we had learned the previous year in geometry!) the area between the sticks was four times as great. So the same amount of butter coming out of the muzzle had to be spread over fourth the distance. Twice the distance, 1/4 the density of butter, an inverse square relation!

    In fact, spreading out a constant amount of anything in space results in an inverse square relation.
  10. Nov 14, 2013 #9


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    There's nothing in space to keep warm. That's why it's called 'space'. All of the energy emitted by stars radiates unimpeded into the void.
  11. Nov 15, 2013 #10
    To expand on SteamKing's answer, what we call temperature is a measure of how fast the particles in a substance are moving or vibrating and is related to the amount of heat energy stored within that substance or gas. We feel things are hot when these particles are travelling at high speeds and collide with our hands or thermometers. These collisions transfer energy. That's what we feel.

    Because there are very few particles in space, there is very little capacity to store heat energy and if we put a temperature there, we would register very few of these collisions and would therefore measure almost no temperature.

    That is why the earth is hot (dense atmosphere with lots of particles) while empty space at the same distance from the Sun is very cold.
  12. Nov 15, 2013 #11


    User Avatar
    Gold Member

    We would still get the energy by radiation,increasing the temperature of the liquid in the thermometer.No,almost no temperature means it's just above absolute zero,It can't be unless sun's or other star's radiation doesn't reach us
  13. Nov 15, 2013 #12
    For the right triangle where "A" is the radius of the Earth and "B" is 1AU, then the similar triangle is where "a" is the radius of the radiating spot on the Sun's surface, and "b" is the radius of the Sun (approximately).

    This makes the diameter of this central radiating spot on the Sun 592m.

    So a spot on the Sun about 27,500 square meters is projecting to half the surface of the Earth, about 510 million square kilometers.

    That is a proportion of reduction of 0.00000005 at 1AU
  14. Nov 15, 2013 #13


    User Avatar
    2017 Award

    Staff: Mentor

    The gas in that "nothing" has 20 times more mass than all stars combined.

    That is just wrong.
    Few collisions are also few opportunities to lose energy, so the equilibrium does not depend on the particle density. In addition, there is always thermal radiation, incoming and outgoing. Far away from stars, that equilibrium temperature is currently 3K, and mainly coming from the cosmic microwave background.

    Source: The cosmic energy inventory

    @bahamagreen: I don't know how you got that number, the reduction is (radius of sun / distance to sun)^2 = 2*10-5. Did you calculate the reverse quantity (or the fraction of the solar radiation that reaches earth)?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook