Solar Radiation Change Calculation

The intensity of solar radiation ($I$) reaching Earth follows the inverse square law with respect to the Earth-Sun distance ($r$). This means:

$ I \propto \frac{1}{r^2} $

To find the percentage change in radiation intensity when the distance changes, we compare the intensity at the closest point (perihelion, $r_p$) to the intensity at the mean distance ($r_{mean}$).

Given Distances:

• Perihelion distance ($r_p$): $147 \times 10^6 \text{ km}$
• Mean distance ($r_{mean}$): $149.6 \times 10^6 \text{ km}$

The ratio of intensities ($I_p$ at perihelion and $I_{mean}$ at mean distance) is given by:

$ \frac{I_p}{I_{mean}} = \left(\frac{r_{mean}}{r_p}\right)^2 $

Substituting the given values:

$ \frac{I_p}{I_{mean}} = \left(\frac{149.6 \times 10^6 \text{ km}}{147 \times 10^6 \text{ km}}\right)^2 $

$ \frac{I_p}{I_{mean}} = \left(\frac{149.6}{147}\right)^2 $

Calculate the numerical value:

$ \frac{I_p}{I_{mean}} \approx (1.017687)^2 \approx 1.035689 $

The percentage change is calculated as:

$ \text{Percentage Change} = \left( \frac{I_p}{I_{mean}} - 1 \right) \times 100\% $

$ \text{Percentage Change} \approx (1.035689 - 1) \times 100\% $

$ \text{Percentage Change} \approx 0.035689 \times 100\% \approx 3.57\% $

This shows an approximate 3.6% increase in solar radiation when Earth is closest to the Sun.