Shadows, Panel Output and the Distance Misconception
At a school at latitude 46° N, a student drives a 1.00 m vertical rod into the ground and measures its shadow at local noon on two dates: 21 March, 1.04 m; 21 December, 2.67 m. On the same campus a fixed 100 W solar panel logs 6.2 kWh in June and 1.6 kWh in December. Her partner builds a lamp-and-globe model of the seasons: a 40 cm lamp in a darkened room, a tilted globe on a stand moved around the lamp on a circle of constant radius. He concludes that winter is cold because Earth is farther from the Sun then, "because the globe gets less light on the far side of the circle." The teacher asks the pair to use the shadow and panel data to test that conclusion and reminds them that Earth's orbital distance varies by only about 3% and that the Sun's rays arrive essentially parallel.
1. Calculate the Sun's altitude at local noon on both dates and explain the difference.
Read the worked answer
The rod, its shadow and the Sun's ray form a right triangle, so tan(altitude) = rod height ÷ shadow length.
- 21 March: tan θ = 1.00 ÷ 1.04 = 0.962, so θ ≈ 44°.
- 21 December: tan θ = 1.00 ÷ 2.67 = 0.375, so θ ≈ 21°.
The difference is about 23°, which matches Earth's axial tilt of 23.5°.
Check against theory: at the equinox the noon Sun stands at 90° − latitude = 90° − 46° = 44° above the horizon, which matches the March measurement. At the December solstice the northern hemisphere is tilted away from the Sun, so the noon altitude drops by the tilt: 44° − 23.5° ≈ 20.5°, matching the December measurement.
The data therefore support the tilt explanation: the same rod casts a shadow more than twice as long because the sunlight arrives at a much shallower angle, not because the rod moved.
2. Use the shadow data, the panel data and the model itself to test the partner's "greater distance" explanation.
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Four arguments defeat the distance claim:
- Size of the effect. Earth's distance from the Sun changes by only about 3% over the year, which cannot produce the observed change. In the lamp model the globe moves on a small circle around a nearby lamp, so the fractional distance change is enormous - the model exaggerates exactly the variable that does not matter.
- Geometry. A 40 cm lamp is a point source, so its rays spread out and the far side of the globe is dimly lit. Real sunlight arrives essentially parallel; what changes is the angle at which the same beam strikes the ground, not its distance.
- Global seasons. If distance caused seasons, the whole planet would have the same season at once. In December the southern hemisphere is in summer, so the explanation fails immediately.
- Quantitative consistency. Sun altitude falls from 44° to 21°, so energy per square metre of ground falls by sin 44° ÷ sin 21° ≈ 2.0. December days are also shorter (about 8.5 h versus 15.7 h), another factor near 1.8. Together that predicts roughly a 3.6-fold difference, close to the observed 6.2 ÷ 1.6 ≈ 3.9.
The model is fine for showing tilt, but the lamp must be far away with parallel rays for the conclusion to be trustworthy.
3. A classmate proposes tilting the panel to 46° from the horizontal. Use the noon altitudes to judge that choice, and name one limit of your reasoning.
Read the worked answer
A fixed panel collects the most energy when its surface is perpendicular to the incoming rays, which means a tilt from horizontal of 90° − altitude:
- 21 March: 90° − 44° = 46°;
- 21 December: 90° − 21° = 69°.
So 46° is close to ideal at the equinox and during summer, but far too flat in December, when the Sun never rises above about 21° and the rays strike a 46° panel at a glancing angle. Because the June output (6.2 kWh) is nearly four times the December output, optimizing for summer costs little overall. A tilt near 46° is a reasonable year-round compromise for this site; a steeper tilt would help winter but lose summer energy.
Limits of this reasoning:
- it uses noon only, while the Sun sweeps across the sky and most energy arrives before and after noon;
- the monthly figures include clouds, snow cover and panel temperature, not just geometry;
- 100 W is a rated value under standard test conditions, not the real output;
- the optimal angle also depends on whether the goal is total annual energy or winter performance.