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%
Zero percent
30%
To find the percentage of light that is perpendicular to the surface when the light arrives parallel to the surface, we need to consider the angle that the sunlight strikes the surface.
In this case, the planet is tilted at 45 degrees and the sunlight arrives parallel to the surface. Since the light is parallel to the surface, it means that the angle between the light rays and the surface is 90 degrees.
When the angle between the light rays and the surface is 90 degrees, it means that all the light is hitting the surface perpendicularly. Therefore, the fraction of light that is perpendicular to the surface in this case is 100%.
So the correct answer is:
100%