Posted by **Luis** on Sunday, July 15, 2012 at 8:23pm.

2. We ultimately want to find the amount of solar energy that hits the observer on the planet. Energy is usually measured in Joules, but there are other appropriate units of energy as well; kilowatt-hours for example, are used to measure the amount of electrical energy that we consume from the power company. For the purposes of this lab, we will invent a convenient unit which we will call the "sunlight-hour". This will be the fraction of direct (perpendicular) sunlight times the number of daylight hours. The amount of perpendicular sunlight is found by considering the angle that the sun strikes the surface. The number of daylight hours we will find using the simulation. Thus,

sunlight-hours = (angle factor)*(hours of daylight)

We will use this formula a lot for this lab (beginning with question 4).

To start off, let us look at the effect of the angle that the sunlight strikes the surface of the planet. We want to be able to quantify the fraction of light that hits perpendicular to the surface.

First, set the tilt of the planet to be 45 degrees and wait until the planet is near the winter solstice (on the far right). Here the sunlight comes in parallel to the surface.

When the light arrives parallel to the surface, what percentage of that light is perpendicular to the surface?

(Points : 2)

0%

30%

50%

100%

- astronomy -
**drwls**, Sunday, July 15, 2012 at 8:26pm
Zero percent

- astronomy -
**ced**, Saturday, July 11, 2015 at 12:42am
30%

- astronomy -
**Dante**, Sunday, November 1, 2015 at 4:06pm
0%

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