Effective temperature
The net emissivity may be low due to surface or atmospheric properties, such as the greenhouse effect.[4][5] The effective temperature and the bolometric luminosity are the two fundamental physical parameters needed to place a star on the Hertzsprung–Russell diagram.The calculation assumes the planet reflects some of the incoming radiation by incorporating a parameter called the albedo (a).This area intercepts some of the power which is spread over the surface of a sphere of radius D. We also allow the planet to reflect some of the incoming radiation by incorporating a parameter a called the albedo.[10] Also note here that this equation does not take into account any effects from internal heating of the planet, which can arise directly from sources such as radioactive decay and also be produced from frictions resulting from tidal forces.The greenhouse effect results from materials in the atmosphere (greenhouse gases and clouds) absorbing thermal radiation and reducing emissions to space, i.e., reducing the planet's emissivity of thermal radiation from its surface into space.Substituting the surface temperature into the equation and solving for ε gives an effective emissivity of about 0.61 for a 288 K Earth.
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