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Angular diameter of the Sun from the Earth orbit is approximately 32′ (arc minutes), and the solar constant is 1.36kWt/m2 . (a) What is the Sun’s radius? (b) How much total power does the Sun emit? (c) What is the power flux on the surface of the Sun?

Answer :

andreivillamil

Answer:

a) 1392.54 *10^6 m

b) 3.8248284*10^26 W

c) 6.2783674*10^7 W/m^2

Explanation:

from the exercise we have the following values:

angular diameter of sun from earth, phi = 32'

solar constant, S = 1.36 kW/m^2

a. distance of sun from earth, d = 149.6*10^9 m

therefore

phi = D/d

clear D

D = d*phi

so we have

a) diameter of sun = D

D = 1392.54 *10^6 m

b. How much total power does the Sun emit? =P

P = S*4*pi*d^2

P= = 3.8248284*10^26 W

c. power flux on surface of sun = P/pi*D^2

c. power flux on surface of sun  = 6.2783674*10^7 W/m^2

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