Question

The intensity of sunlight reaching the earth is$1360\mathrm{W}/{\mathrm{m}}^2$.

a. What is the power output of the sun$?$

b. What is the intensity of sunlight reaching Mars$?$$?$

Short answer

a. Power output of the sun is $3.85\times {10}^{26}\mathrm{W}$.

b. Intensity of sunlight of mars is$589\mathrm{W}/{\mathrm{m}}^2$.

Step-by-step solution

Given information

Intensity of sunlight reach the earth is${\mathrm{I}}_{\mathrm{avg}}=1360\mathrm{W}/{\mathrm{m}}^2$.

Distance between sun and earth is$\mathrm{d}=150\times {10}^9\mathrm{m}$.

Distance between mars and sun is${\mathrm{d}}^{\text{'}}=228\times {10}^9\mathrm{m}$.

Calculation for power output (part a)

a.

Intensity is,

${\mathrm{I}}_{\mathrm{avg}}=\frac{\mathrm{P}}{\mathrm{A}}$

$\mathrm{P}=\mathrm{I}\times \mathrm{A}$

Substitute all values,

localid="1649960380573" $\mathrm{P}=1360\frac{\mathrm{W}}{{\mathrm{m}}^2}\times 4\mathrm{\pi }{\left(150\times {10}^9\right)}^2{\mathrm{m}}^2$

$\mathrm{P}=3.85\times {10}^{26}\mathrm{W}$

Calculation for intensity of mars (part b)

b.

Intensity is,

$\mathrm{I}=\frac{\mathrm{P}}{\mathrm{A}}$

$\mathrm{I}=\frac{\mathrm{P}}{4{\mathrm{\pi d}}^{\text{'}}}$

localid="1649960411995" $\mathrm{I}=\frac{3.85\times {10}^{26}\mathrm{W}}{4\mathrm{\pi }{\left(228\times {10}^9\right)}^2{\mathrm{m}}^2}$

localid="1649960499973" $\mathrm{I}\approx 589\mathrm{W}/{\mathrm{m}}^2$