Tanning around the earth

$\vec P \cdot \vec A = I$ $\vec P$ , $\vec A$ , and $I$ stand for power and area vector, and intensity respectively. In summer, the earth is tilted towards the sun, and thus the person lies on the horizontal line connecting the centers of the earth and the sun. In winter, the earth is tilted away from the sun, and thus the person lies $46^\circ$ above the horizontal line connecting the centers of the earth and the sun. Thus, $I_{sum} = \vec P \cdot \vec A = PA cos(0^\circ)$ and $I_{win} = \vec P \cdot \vec A = PA cos(46^\circ)$ . $I_{win} / I_{sum} = PAcos(46^\circ)/PAcos(0^\circ) = 0.695$

In the summer, when the axis is tilted most towards the sun, the angle of the sun relative to the normal at 23 degrees north is $23^\circ-23^\circ = 0 ^\circ$ . Therefore, in the summer, the sun is shining directly above 23 degrees north, which is the maximum possible intensity. We will represent this with $I_{sum}$ .

In the winter, when the axis is tilted most away the sun, the angle of the sun relative to the normal at 23 degrees north is $23^\circ+23^\circ = 46 ^\circ$ .

We wish to solve for the intensity of sunlight vertical to our location in winter. We split the rays of sunlight into horizontal and vertical components (relative to our location). Note that the sum of our components has intensity $I_{sum}$ , since the sun's intensity directly shining at us is constant. Therefore our vertical component has intensity $I_{sum} \cdot \cos 46^\circ = I_{win}$ .

Therefore $I_{win}/I_{sum}= I_{sum} \cdot \cos 46^\circ/I_{sum} = \cos 46^\circ \approx 0.695$ .

In the summer,when the axis of the earth is tilted most towards the sun, the angle of the sun relative to the normal at the $23^{\circ}$ mark is $23^{\circ}-23^{\circ}=0^{\circ}$ Therefore,in the summer,the sun is shining directly over $23^{\circ}N$ which is the maximum possible intensity.Let us represent this as $I_{sum}.$ In the winter when the axis is tilted most away from the sun,the angle relative to the normal at $23^{\circ}N$ is $23^{\circ}+23^{\circ}=46^{\circ}.$ Let us denote the maximum possible intensity in winter as $I_{win}.$ We wish to solve for the intensity of sunlight vertical to our location in winter. We split the rays of sunlight into horizontal and vertical components (relative to our location). Note that the sum of our components has intensity $I_{sum},$ since the sun is shining directly on us at this instant and is constant. Therefore our vertical component has intensity $I_{sum} \times \cos 46^{\circ}=I_{win}.$ Therefore $\frac{I_{win}}{I_{sum}}=\frac{I_{sum} \times \cos 46^{\circ}}{I_{sum}}$ $\Rightarrow \frac{I_{win}}{I_{sum}}=\cos 46^{\circ}$ $\Rightarrow \frac{I_{win}}{I_{sum}}=0.69465$ $\Rightarrow \frac{I_{win}}{I_{sum}}\approx\boxed{0.695}$

The first thing to notice in this problem is the angle at which sun rays fall in each season. At tropics in summer the sun shines vertically overhead at 90 degrees. That is when the intensity is maximum and effective area minimum. At winter the sun rays fall at an angle in tropics. Remember as the earth rotates,the axis stays fixed. And sun rays travel in the plane of the orbit. Using these two hints and by simple geometry we see that at winter in tropics sun rays fall at 47 degrees to normal. Intensity when light shines at an angle theta is Imax*cos(theta). So Isum/Iwin=cos(47)=0.682

Lambert's law tells us that light intensity is proportional to cosine of angle between incidence and normal. In high summer, this angle of incidence is 0 (Earth is tilted towards sun so vertical rays hit the tropic line). Cos(0)=1. In peak of winter, angle of incidence is 2*23=46 degrees. Cos(46)=0.695. Hence ratio is 0.695.

In summer the intensity is Isum so for winter Iwin=Isum cosx Iwin/Isum=cos46=0.69

During the summer, the ray of the sun will strike perfectly normal to said person, because he resides $23$ degrees above the equator, which is in turn $23$ degrees below the line of direct sunlight.

In the winter, the sunlight strikes at an angle $23 + 23 = 46$ to the normal, because the person is $23$ degrees above the equator, which is in turn $23$ degrees above the line of direct sunlight.

Thus, the ratio $I_{win}/I_{sum} = \cos 46 = \boxed{0.695}$