Based on the simple blackbody radiation model described in class, answer the following question. The planets Mars and Venus have albedo values of 0.15 and 0.75, and observed surface temperatures of approximately 220 K and 700 K, respectively. The average distance of Mars from the sun is 2.28 x 108 km, and the average distance of Venus is 1.08 x 108 km. Given that the radius of the sun is 7 x 108 m, and the energy flux at the surface of the sun is 6.28 x 107 W/m2 , what is the extent of the greenhouse effect for Mars

Respuesta :

Answer:

The extent of greenhouse effect on mars is [tex]G_m = 87 K[/tex]  

Explanation:

From the question we are told that

     The albedo value of Mars is  [tex]A_1 = 0.15[/tex]

      The albedo value of Mars is  [tex]A_2 = 0.15[/tex]

       The surface temperature of Mars is  [tex]T_1 = 220 K[/tex]

        The surface temperature of Venus is  [tex]T_2 = 700 K[/tex]

          The distance of Mars from the sun is [tex]d_m = 2.28*10^8 \ km = 2.28*10^8* 1000 = 2.28*10^{11} \ m[/tex]

          The distance of Venus from sun is  [tex]d_v = 1.08 *10^{8} \ km = 1.08 *10^{8} * 1000 = 1.08 *10^{11} \ m[/tex]

       The radius of the sun is [tex]R = 7*10^{8} \ m[/tex]

        The energy flux is   [tex]E = 6.28 * 10^{7} W/m^2[/tex]

The solar constant for Mars is mathematically represented as

 

          [tex]T = [\frac{E R^2 (1- A_1)}{\sigma d_m} ][/tex]

Where [tex]\sigma[/tex] is the Stefan's constant with a value  [tex]\sigma = 5.6*10^{-8} \ Wm^{-2} K^{-4}[/tex]

So substituting values

         [tex]T = \frac{6.28 *10^{7} * (7*10^8)^2 * (1-0.15)}{(5.67 *10^{-8}) * (2.28 *10^{11})^2)}[/tex]

          [tex]T = 307K[/tex]

So the greenhouse effect on Mars is  

           [tex]G_m = T - T_1[/tex]

           [tex]G_m = 307 - 220[/tex]

          [tex]G_m = 87 K[/tex]