Mars has an albedo of 15\%
15%. Given that it doesn't have much of an atmosphere, estimate the mean surface temperature on Mars. Note: the semimajor axis of the orbit of Mars is 1.52 AU.
The illuminance generated by the Sun on the orbit of the Mars is
"E = \\dfrac{L_{\\odot}}{4\\pi a^2}" .
The energy gathered by the martian surface every second is "(1-A)E\\cdot \\pi r^2" , where r is the radius of the Mars.
Due to the energy balance, the equal amount of energy is released every second. If we assume Mars to behave like a perfect black body, this energy is "4\\pi r^2\\sigma T^4" (Stephen-Boltzmann law).
"\\dfrac{L_{\\odot}}{4\\pi a^2} \\cdot(1- A)\\pi r^2 = 4\\pi r^2\\sigma T^4,\\\\\n\\, \\\\\n\\dfrac{L_{\\odot}}{4\\pi a^2} \\cdot(1- A)= 4\\sigma T^4,\\\\\n\\,\\\\\nT = \\sqrt[4]{\\dfrac{(1-A)L_{\\odot}}{16\\pi \\sigma a^2}} ,\\\\\nT = \\sqrt[4]{\\dfrac{(1-0.15)\\cdot4\\cdot10^{26}}{16\\pi \\cdot5.67\\cdot10^{-8}\\cdot(1.52\\cdot1.5\\cdot10^{11})^2}} = 219\\,\\mathrm{K}."
Comments
Leave a comment