# The diameter of the sun makes an angle of .53 degrees from earth. How many minutes does it take for the sun to move one solar diameter in an overhead sky? ( Remember that it takes 24 hours or 1440 minutes for the sun to move through 360 degrees).

How does your answer compare with the time it takes for the sun to disappear once its lower edge meets the horizon at sunset?

Does defreaction affect the answer?

.53/360 = x/1440

x=2.12

The sun's angular speedin the sky is 360 degrees/1440 min = 0.250 degrees/min

The time to move one solar diameter is

T = (0.53 deg)/(0.25 deg/min) = 2.1 min

Make your own observation of the actual sunset time. Refraction is one reason for a discrepancy. So is the angle that the sun's path makes with the horizon. It sets obliquely, not perpendicular to the horizon at temperate and polar latitudes.

3 years ago

8 months ago

## Well, isn't the sun just a master of disappearing acts? It takes 2.1 minutes for the sun to move one solar diameter in an overhead sky. But when it comes to sunset, things get a bit trickier. You see, refraction and the angle of the sun's path with the horizon like to throw a wrench in our calculations.

You might want to make your own observation of the actual sunset time because it can vary. The sun doesn't always set in a straight line down, especially at temperate and polar latitudes. It likes to set obliquely, giving us a run for our money.

And let's not forget about refraction, that notorious troublemaker. Refraction can cause the sun to appear higher in the sky than it actually is, delaying the moment it disappears. So, while it may take 2.1 minutes for the sun to move one solar diameter overhead, it could take a bit longer for it to bid us adieu at sunset. The sun just loves to keep us on our toes!

8 months ago

## The time it takes for the sun to move one solar diameter in an overhead sky is approximately 2.1 minutes.

The time it takes for the sun to disappear once its lower edge meets the horizon at sunset can vary and is influenced by factors like refraction and the angle of the sun's path with the horizon. It does not necessarily match the time required for the sun to move one solar diameter.

Refraction can affect the answer because it causes the apparent position of the sun to be slightly higher in the sky than its actual position. This means that the sun may appear to move slightly slower than its true angular speed, which can result in a longer time for it to move one solar diameter or for it to disappear at sunset.

8 months ago

## To calculate how many minutes it takes for the sun to move one solar diameter in an overhead sky, you can use the following steps:

1. Start with the given information that the diameter of the sun makes an angle of 0.53 degrees from Earth.

2. Recognize that it takes 24 hours, or 1440 minutes, for the sun to move through 360 degrees.

3. Set up a proportion to find the unknown time (let's call it 'x') it takes for the sun to move one solar diameter.

4. The proportion can be set up as: 0.53 degrees / 360 degrees = x minutes / 1440 minutes.

5. Cross multiply and solve for 'x': 0.53 * 1440 = 360 * x.

6. Simplify the equation: 763.2 = 360 * x.

7. Divide both sides by 360 to isolate 'x': x = 763.2 / 360.

8. Calculate the value of 'x': x โ 2.12 minutes.

Therefore, it takes approximately 2.12 minutes for the sun to move one solar diameter in an overhead sky.

Regarding the time it takes for the sun to disappear once its lower edge meets the horizon at sunset, this process is influenced by several factors such as refraction and the angle of the sun's path with the horizon. Refraction refers to the bending of light as it passes through the Earth's atmosphere, which can affect the apparent position of the sun. Additionally, the sun's path sets obliquely, not perpendicular to the horizon at temperate and polar latitudes, causing it to take longer to disappear completely.

To determine the actual sunset time, it is recommended to make your own observation as the specific conditions can vary depending on the location, weather, and time of year.