Question

Problem 2

a) What is the radiant power of the sun?

b) What is the radiant power at the surface of the earth?

c) Calculation of the solar constant

d) How much energy does the earth absorb?

e) Using the equilibrium condition to calculate the earth’s temperature (W/O atmosphere) ? (Give a brief comment regarding your findings)

f) Using the equilibrium condition to calculate the earth’s real temperature (W/ atmosphere)? (Give a brief comment regarding your findings)

Problem 3 (Very similar to problem 2)

Now you know the solar constant of the Earth from the second problem the value of the total radiant energy flux density at the earth from the Sun normal to the incident rays. The observed value usually integrated over all emission wavelengths and refereed to the mean Earth-Sun distance is, solar constant, 0.136 J s-1 cm-2 . (a)Show that the total rate of energy generation of the Sun is 4x1026Js-1 . (b) Show the effective temperature of the surface of the Sun treated as a black body is T=6000K. (Take the distance of the Earth from the Sun as 1.5x1013 cm and the radius of the Sun as 7x1010cm.) (c) Calculate the temperature of the surface of the Earth on the assumption that as a black body in thermal equilibrium it reradiates as much thermal radiation as it receives from the Sun. Assume also that the surface of the Earth is at a constant temperature over the day-night cycle. (e) Assume that Sun’s surface temperature is about 12000K, rather than 6000K. How much more thermal radiation would the Sun emit? Where would be the peak emission wavelength? What would be earth temperature? Does Earth will be still habitable?

Answer 1

a] Radiant power of Sun is approximately, P = 3.85 x 10²⁶ W

b] The power mentioned above is radially distributed. So, at a distance of L = distance between Sun and Earth, the intensity will be:

now, at any given time, this intensity is scattered over half of the Earth. Therefore, area illuminated =

where R is the radius of Earth.

Therefore, power at the surface of the Earth is:

L = 1.496 x 10¹¹ m and R = 6378100 m

so, P_E = 1.75 x 10¹⁷ W

c] Solar constant = Intensity at the surface =

d] Earth absorbs about 70% of this energy received.

So, Energy absorbed per unit time per unit area is: E = 1369 x 0.7 = 958.52 J.