Answer:
a)6.34 x [tex]10^{7}[/tex]W/m²
b)1.37 x [tex]10^{3[/tex] W/m²
c) see explanation.
Explanation:
a)The relation of intensity'I' of the radiation and area 'A' is given by:
I= P/A
where P= power of sunlight i.e 3.9 x [tex]10^{26}[/tex] J
and the area of the sun is given by,
A= 4π[tex]R_{sun}[/tex] => 4π[tex](\frac{1.4*10^{9} }{2} )^{2}[/tex]
A=6.15 x [tex]10^{18}[/tex]m²
[tex]I_{sun}[/tex] = 3.9 x [tex]10^{26}[/tex] / 6.15 x [tex]10^{18}[/tex] => 6.34 x [tex]10^{7}[/tex]W/m²
b) First determine the are of the sphere in order to determine intensity at the surface of the virtual space
A= 4π[tex]R[/tex]
Now R= 1.5 x [tex]10^{11[/tex]m
A= 4π x 1.5 x [tex]10^{11[/tex] =>2.83 x [tex]10^{23[/tex] m²
The power that each square meter of Earths surface receives
[tex]I_{earth[/tex] = 3.9 x [tex]10^{26}[/tex]/2.83 x [tex]10^{23[/tex] =>1.37 x [tex]10^{3[/tex] W/m²
c) in part (b), by assuming the shape of the wavefront of the light emitted by the Sun is a spherical shape so each point has the same distance from the source i.e sun on the wavefront.
