Who much solar energy is absorbed and how much is reradiated by the Earth in a typical 24 hour period?
446503
if you had a solar panel you would only achieve 14% absorbs-ion max.
273539
themal energy is the vibrational energy of the disordered motion of microscopic particles such as molecules and atoms.\nThe sum of all the kinetic energies.\nthere are many forms one is:\nocean thermal energy;\nThe oceans cover a little more than 70 percent of the Earth's surface. This makes them the world's largest solar energy collector and energy storage system. On an average day, 60 million square kilometers (23 million square miles) of tropical seas absorb an amount of solar radiation equal in heat content to about 250 billion barrels of oil. If less than one-tenth of one percent of this stored solar energy could be converted into electric power, it would supply more than 20 times the total amount of electricity consumed in the United States on any given day.\nthermal energy is in every matter. it has been proven that everything that is matter is always vibrating.
1142280
in North America the average power of the solar radiation lies somewhere between 125 and 375 W/m², between 3 and 9 kWh/m²/day. [3]\n\nIt should be noted that this is the maximum available power, and not the power delivered by solar power technology. For example photovoltaic panels currently have an efficiency of ca. 15% and, hence, a solar panel delivers 19 to 56 W/m² or 0.45-1.35 kWh/m²/day (annual day and night average).\n\n\nGaAs multijunction devices are the most efficient solar cells to date, reaching as high as 39% efficiency[9]. They are also some of the most expensive cells per unit area (up to US$40/cm2).
235735
Depending on its position in the oval orbit it has around the sun, the earth gets sunlight 8.317 minutes after it is emitted from the sun.
250338
I would guess the sun would be the energy in SOLAR powered anything. Solar = sun
971361
Today is November 16, not April 1. Everything expending energy eventually uses up the energy. Even the sun will eventually die. You and I and the rest of the current human race will beat it to the punch, however.
1871
To put it plainly and simply quoted:  \n\nF = Gm1m2/r2 = ma\n\nGm/r2 = g\n\nm = gr2/G\n\nm = (9.8 m/s2)(6.37 x 106 m)2/(6.673 x 10-11 Nm2/kg2) \n\nm = 5.96 x 1024 kg\n\nThe Earth gains mass each day, as a result of incoming debris from space. This occurs in the forms of "falling stars", or meteors, on a dark night. The actual amount of added material depends on each study, though it is estimated that 10 to the 8th power kilograms of in-falling matter accumulates every day. The seemingly large amount, however, is insignificant to the Earth's total mass. The Earth adds an estimated one quadrillionth of one percent to its weight each day.
824682
The energy absorbed is that which is not outputted.  In this case, 75% of the energy is absorbed, so the energy absorbed per cycle is .75*8500 J.  One watt = 1J/sec.  So seconds = J/watt.  The power is constant, so the power in each cycle is 6,00 kW; this times the duration of the cycle will be the energy in each cycle.
984374
Well, that depends.  There is evidence of algae at 3.5 billion, probably photosythetic cells by 2.0, and clear plants by 400 million.  With a 4.6 billion year old earth, that's 76%, 43%, or 8.7% of the time.
475577
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.
969774
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.
637131
About  93 million miles depending on where we are in the orbit as the orbit is not circular. The sun is made of gas, and actually more of a plasma. The earth has a solid iron, nickel core with a molten iron nickel outer core and then then mantel then the crust with rock on it.
916850
the earth rotational energy is calculated to be  2.58×10^29 J , thus 0.01 of this is 2.58*10^27J , and this enough for human needs for about  2580000000 years .\n\nand if we asume that the energy decrease the rotational energy by the same portion (0.01%) then the day length will decrease by 14.4 mins.
829976
You'll need this number:\n\n1.740×10^17 W.\n\nThis is the solar constant, the amount of energy received at the earth per second per area.\n\nAnd this number:\n\nsigma=5.67x10^-7 J/s/m^2/K^4\n\nThis is the Stefan Boltzmann constant which goes into the (surprise) Stefan Boltzmann law:\n\nJ=e*sigma*T^4\n\nSo, J is the energy received by the body per second per area, and T is it is temperature. The e stands for the efficiency of the body which is the ratio between the power it absorbs and then reemits.\n\nSo, you have an albedo which is an indication of how much light an object reflects from its surface versus how much it absorbs. With these numbers, you should be able to piece together your information with the equation above to get an answer. Let me know if you have further questions.
736686
Material from the sun's surface, about 75% hydrogen, 24% helium, and 1% everything else
478214
Insolation is the incoming solar radiation that reaches a planet and its atmosphere or, by extension, any object exposed to solar rays, such as watts per square meter of Sun-facing cross section, across the entire electromagnetic spectrum; most of that power is in the visible light portion of the spectrum. On Earth's surface, the Sun's rays are attenuated by the atmosphere. The midday insolation on clear days in temperate latitudes may be estimated as 1000 watts per square meter directly facing the Sun. The actual figure varies with the Sun angle at different times of year (because the sunlight must travel through varying distances of air), and with atmospheric haze and cloud cover. Long-term time-averaged insolation in sunny locations is closer to 250 watts per square meter, taking into account the lower insolation in early morning and evening, and the presence of night.\n\nThe insolation into a fixed surface (such as a window or roof) depends on the projected area of the surface perpendicular to the incoming sunlight, not directly on the area of the surface. This is the reason that the polar regions are much colder than equatorial regions on Earth: on an annual average the poles receive less insolation than does the equator, because the Earth's surface at the poles never faces the Sun directly. The projection effect can also be used in architecture to design buildings that are cool in summer and warm in winter, by providing large vertical windows on the equator-facing side of the building (the south face, in the northern hemisphere): this maximizes insolation in the winter months when the Sun is low in the sky, and minimizes it in the summer when the noonday Sun is high in the sky. (The Sun's north/south path through the sky spans over 45 degrees through the year).
64722
A radiometer does what you describe. \n\nhttp://store.sundancesolar.com/solrad.html\n\n\nDirections to build a simple radiometer:\n\nMake a radiometer to measure solar energy. Colour the paper side of a gum wrapper black and cut in four pieces. Stick the pieces to each side of a match stick with the shiny sides all facing the same way. Wrap a thread around a pencil and attach to the match stick. Suspend in the jar and place in a sunny location. The radiometer will turn as it absorbs energy.