Now, wait until the planet is at the winter solstice (12:00am). a) How many hours of daylight are there on the winter solstice with the inclination at 45 degrees? Use the same criteria for the length of day that you used for the previous question.
sunlight-hours = (angle factor)*(hours of daylight)
b) Use the above formula and the results from the earlier questions to calculate the number of sunlight-hours of daylight that one receives on the winter solstice for a planet tilted by 45 degrees. Show your work for full credit.
a) To calculate the number of hours of daylight on the winter solstice with a 45-degree inclination, we first need to determine the angle factor. The angle factor represents the proportion of daylight received at a given angle compared to the daylight received at the maximum peak angle.
The angle factor can be calculated using the formula:
angle factor = cos(90° - absolute value of inclination angle).
In this case, the inclination angle is 45 degrees. The absolute value of 45 degrees is still 45 degrees since it is already positive.
So, angle factor = cos(90° - 45°) = cos(45°) ≈ 0.7071.
Next, we multiply the angle factor by the number of hours of daylight obtained from the previous question.
Let's assume the number of hours of daylight obtained from the earlier question is X hours.
Therefore, hours of daylight on the winter solstice = angle factor * X = 0.7071 * X.
b) To calculate the number of sunlight-hours of daylight received on the winter solstice for a planet tilted by 45 degrees, we can plug in the value of X (obtained from the earlier question) into the formula determined in part (a).
For example, if the earlier question provided X = 10 hours of daylight, then the number of sunlight-hours of daylight received on the winter solstice for a 45-degree inclined planet would be:
sunlight-hours = angle factor * X
= 0.7071 * 10
≈ 7.071 hours of sunlight.