The sun appears in the sky and for a Period of time each day during the summer?

Ronald 7

Gosh !! So it does

Bookbinder

This isn't a question, it's a statement. What kind of replies are you expecting? You're new to this website, aren't you? You're just pushing buttons to see what will happen. Go away and learn how to use the website, then come back and ask an intelligent question.

Anonymous

No the Sun does not appear, it is there all the time, the Earth orbits about the Sun, and is itself rotating, so constant and consistent sections of the Earth have visibility of the Sun (climatic conditions permitting).

Pearl L

i think thats true

puzzled

yes

David

Do you have a question, troll?

Nick

What season does the sun appear in the sky for a Period each day?

Poseidon

The Sun appears in the sky 24/7. Summer, fall, winter and spring. It is seen during the daylight hours in every part of the world but it is the spinning of the Earth which makes it disappear, but it appears in other parts of the Earth.

quantumclaustrophobe

Depends where on Earth you are, but - yes that's generally true.

poldi2

Yes it does. And each day during the spring, fall, and winter too (not just the summer).

John P

Indeed in moderate latitudes (not extreme north or south) the Sun appears every day throughout the year. The further north or south you go from the equator, the more the day lengths vary from winter to summer. You really need to get help in expressing yourself well in English.

Gerald

when the paper kid slings it onto the doorstep of the lager louts of England still dead to the world

Mopes

It is there always but blocked by the position of the earth you are at as the earth rotates. Astronauts go thru dark when they are on the opposite side of earth from the sun position.

Tom S

Yes, we call that period "daytime", it happens in the winter too unless you are in the arctic or antarctic.

Raymond

Here is a (relatively) simple equation to calculate the height (H) of the Sun above the horizon: sin(H) = sin(LAT)sin(DEC) + cos(LAT)cos(DEC)cos(PA) where LAT is the latitude of the observer (that would be you) DEC is the declination of the Sun -- its angular distance from the celestial equator. It changes smoothly from day to day (there are tables on the Web). PA is the "polar angle" of the Sun -- the angle between the meridian and the line between the Sun and the pole. For the Sun, PA changes by 15 degrees per hour. When PA = 0, the Sun is at the observer's meridian (and the Sun is at its highest for that day). Take the equation backwards. In marine navigation, sunrise happens when the calculated height of the Sun is exactly 0 (meaning that the centre of the Sun is on the theoretical horizon -- 90 degrees from the zenith). If you are a non-moving observer (your LAT is fixed). For a single day, you can consider DEC to be fixed for that day (take the DEC near local noon). If H=0, then sin(H) = sin(0) = 0 The original equation Sin(H) = sin(LAT)sin(DEC) + cos(LAT)cos(DEC)cos(PA) becomes 0 = sin(LAT)sin(DEC) + cos(LAT)cos(DEC)cos(PA) flip it around (the equal sign does not care which way we go) sin(LAT)sin(DEC) + cos(LAT)cos(DEC)cos(PA) = 0 move the sines to the right cos(LAT)cos(DEC)cos(PA) = -sin(LAT)sin(DEC) divide both sides by the cosines: cos(PA) = -sin(LAT)sin(DEC) / cos(LAT)cosDEC) we know that tan = sin/cos for each angle (for LAT and for DEC) cos(PA) - tan(LAT)tan(DEC) Using your latitude and the declination of the Sun for a particular day, you can find the Polar Angle of the Sun at time of sunrise (or sunset, same calculation). We know that "half-the-day" (sunrise to solar noon) is the time it takes from this angle PA to go to zero AND we know the angle PA changes by exactly 15 degrees per hour. --- Sounds complicated? Here is an example. Let's say you are at latitude 40 N (called +40 in nautical astronomy); that could be in the northern part of Indianapolis, for example. Let's pick "April 4, 2019" as the date (simply because my declination table fell open on that date) The mid-day DEC is around 5.5 N (+5.5 -- signs are important, because of the "tangent" function) How long is the theoretical day, on that date? at sunrise: cos(PA) = -tan(40)tan(5.5) = -0.0807961... take the arccos (the "inverse the cosine" function, and you get the angle PA = 94.634... degrees. Make sure your calculator is in degrees, and don't forget that annoying minus sign. We divide this angle by 15 degrees to get the number of hours from sunrise to solar noon: 94.634... / 15 = 6.309 hours By symmetry, the time from noon to sunset can be taken as being the same. Total duration sunrise to sunset = 2 * 6.309 = 12.618 hours = 12h37m Same place (LAT = +40) but on June 1st (DEC = +22) cos(PA) = -tan(40)tan(22) = -0.33902 PA = 109.817 Total day = 2 * 109.817/15 = 14.642 hours = 14h38.5m --- In reality, the people will see an image of the Sun that is raised (by 1 degree) by atmospheric refraction AND they will also consider sunrise (or sunset) to be when the top of the disk barely peeks above their local horizon. This will add roughly 5 to 10 minutes to the actual length-of-day, but the difference is approximately the same on both dates. The 2 hour difference in the length of day remains true.

PAMELA

the sun is always in the sky, in the summer we rotate nearer to it.