What absorbs the energy that is radiated from earth's surface?
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Heat energy radiated by the Earth is in the infra-red wavelength of the electromagnetic spectrum. \n\nThe energy passes through bulk of the gases in the atmosphere that are transparent to infra-red radiation and is radiated out into space. \n\nAn increasing condition/problem is that there is a slow increase in gases such as Carbon dioxide and Methane that are opaque at the infra-red spectrum and so they absorb the energy and re-radiate it back down to the Troposphere and out into space.\n\nThe growing ability of the atmosphere to trap heat energy and re-radiate to earth is generally compared  to the principle of a green house and the hence the term "greenhouse effect"
777007
It absorbs the main energy from waves and storm surges.
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The radiations emitted from the sun and also the earth are usually blocked by layers of atmospheric gases and prevent them from escaping into space.This causes 'GLOBAL WARMING'-due to the reflected radiations.This is also known as 'GREEN HOUSE EFFECT'.
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Yes. Radiant energy strikes the matter surface it repel electrons from the matter and it release heat energy.
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Almost all of Earth's internal temperature is radiated outward from its molten core. There is also some heating, particularly nearer the Earth's surface, resulting from physical stresses brought on by Earth's rotation, motion around the Sun, and by gravitational interaction with the Moon.
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All living things emit energy - mostly in the form of heat.\nAll living things produce energy and even the ones most efficient at using the energy the produce lose some to the surronding environment.
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The Sun.\n\nEDIT: Yes, what other people have said about the Earth's tilt and whatnot are true in that they affect the atmosphere, but you asked for the source of energy. That is solely the Sun.
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The sun converts millions of tons of matter into energy every second via fussion processes; this energy is radiated out in every direction in the form of photons and a slew of other particles. \n\nWhen these photons reach earth, we call it "incident light". Light is easily reflected, refracted, and absorbed by many materials on earth, and at every level of interaction, energy is lost to heat.\n\nEdit - what's with all this "pan heating an egg" stuff? What does that even mean?
1074251
Simple answer: Carbon dioxide and other "greenhouse gases" get caught up in the upper levels of earth's atmosphere. When rays from the sun enter earth's atmosphere they pass through to the earth's surface and are supposed to reflect off the earth back into space but do not because the greenhouse gases trap them in the atmosphere.  This causes the earth to heat up due to the cosmic radiation in these rays.
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very simple, land had widely different albedo's.\nthis is the amount of energy which is absorbed and re-emitted as heat. The ocean has a fairly constant albedo, where as on land you have different coloured rocks, which have different albedo's.\nTrees, grass, black soil, all have different albedo's. This varying amount of energy gives rise to different amounts of convective heating of the air above it.\n\nEvery try walking from grass onto a sandy beach on a sunny day? well then you have first hand experience of the difference in absorbed energy and albedo.\n\ndo a web search for albedo, here is the wiki entry for it\nhttp://en.wikipedia.org/wiki/Albedo
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Think of it this way.  The earth has 4 sides, just because ONE OF THEM is looking at the moon and the Sun is facing another side, means nothing.  There is over 90,000,000 miles between the earth and the sun, so we only get a fraction of a percent of its total output of energy.  The Moon while it is in space going around us, is no longer "blocked" by the earth and can now receive its share of light from the sun.  HOWEVER, you may be surprised to learn that the albedo (reflection of light) from the moon is mostly caused by the Earth as much as the Sun.  That is, where it is now daylight on Earth is relecting off of the moon also.
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many materials absorb better at the infrared wavelengths than the shorter wavelengths so the absorbed energy is conserved as heat.  There are materials that absorb shorter wavelengths and convert the energy to heat, such as a microwave oven???  Water absorbs at that frequency (1-2 GHz) and the water molecule resonates.  The vibrational energy is re-emitted as heat due to a Stokes shift.\n\nThe shorter wavelengths are more energetic and more capable of depositing energy that might show up as heat but it depends on the absorption of the material itself.
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Yes, heat can dissipate through radiation, just like a radiator in your house. Each body (Earth, moon, satellite, etc.) has a number called the albedo, the ratio of how much heat it absorbs to how much heat it reflects. And when somethnig passes into the shadow, it radiates out its heat and gets cold.
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The sun emits energy at a rate of P = 4x10^26 W. The amount of power density that reaches the icy planey is then this\n\nPd = P/(4*pi*R^2)\n\nwhere R is the distance from the sun to the planet. This is the total flux of energy emitted by the sun spread over the surface area of a sphere of radius R. The albedo (A) of a planet indicates how much of the incident energy is reflected back into space. The absorbed power density is then\n\nPa = (1 - A)*Pd = (1 - A)*P/(4*pi*R^2).\n\nThe absorbed energy heats the planet and also leads to radiation. The radiated power follows the Stefan-Boltzmann law\n\nPr = e*s*T^4\n\nwhere e is the emissivity, s is Stefan's constant, and T is the temperature of the radiating body. Stefan's constant is s = 5.67x10^-8 W/(m^2*K^4), and for a blackbody, the emissivity is one.\n\nAt equilibrium, the absorbed power and the radiated power are equal\n\n(1 - A)*P/(4*pi*R^2) = s*T^4\n\nwhich can be solved for T, the temperature of the planet.\n\nThe temperature of the planet will impart kinetic energy to gas molecules in the air. The speed of the molecules will be distributed with an rms spedd os\n\nVrms = sqrt(3*k*T/m) = sqrt(3*R*T/M)\n\nwhere k is Boltzmann's constant, R is the universal gas constant, m is the mass of the molecule, and M is the molar mass.\n\nFor a planet the escape velocity for an object is given by\n\nVesc = sqrt(2*G*M/r)\n\nwhere G is the universal gravitation constant, M is the mass of the planet, and r is the planet's radius.\n\nIf the speed of the gas molecules is greater than the escape velocity of the planet, then that planet will not be able to retain that gas at the given temperature.
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There are two main sources of heat on the Earth.  The first obvious one is the Sun.  The second is the Earth itself.  The ground holds the heat from the Sun, but also, radioactive decay within the Earth produces massive amounts of heat.  Most of the heat on the Earth comes from within it.  Because most of the heat comes from the ground itself, as you get farther away from it, it gets colder.
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You would surmise, that since the sun is the source of the heat for the\nearth and atmosphere, that the atmosphere should be warmer at points closer\nto the sun, like at the TOP of the atmosphere, instead of the reverse.\n\n   But the atmosphere is not warmed directly by the sun...it is warmed by the\nEARTH...!  Most of the energy from the sun arrives at the earth in the form\nof light energy, or short-wave energy. The atmosphere is transparent to\nshort-wave energy, and most of the solar energy reaches the surface of the\nearth, where it is absorbed. This energy warms the surface of the earth,\nwhich, in turn, warms the atmosphere with long-wave energy. The type of\nenergy radiated by a star primarily depends on the temperature. So the\nenergy from the sun arrives at the earth as short-wave (in the visible part\nof the spectrum) energy, and leaves the earth as long-wave (in the infra-red\npart of the spectrum) energy.\n\n   The atmosphere is not transparent to long-wave radiation, and absorbs the\nheat energy from the earth. That part of the atmosphere closest to the earth\ngets the most energy and is the warmest.\n\n   Because the sun is nearly overhead in the latitudes nearest the equator,\nthose locations receive more solar energy per unit area, and become warmer\nthan those locations near the poles. This difference in the amount of energy\nreceived and subsequent heating of the atmosphere and evaporation of water\nfrom the lakes and oceans, is what drives the "weather machine" on the\nearth.
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In the morning/evening the sun rays are falling obliquely on to the earth, \nthereby passing through a thicker layer of the atmospheric air because of which (oblique incidence) a lot of its blue component is scattered off \n\nIts 1/lambda^4 scattering - smaller the lambda, larger is the energy scattered off/lost as compared to shorter path transversed during the day time \n\n[This is a standard class XII Physics question]