Official degrees of earth’s rotation per dayWhat effect does the Earth's rotation have on plate tectonics?Why does this graph for sunlight intensity on land has a steeper slope during sunrise as compared to sunset?Cancelling out earth rotation speed, Altazimuth mountDoes a week represent something in astronomy?Ambiguity in Earth's “Tilt”Measurement precision of celestial eventsWhy is right ascension measured on a 24 hour scale rather than a 23 hours and 56 minutes scale?How can I calculate how far through the day the prime meridian is of different planetsModeling planet rotations (time of day, obliqueness, etc.)Field rotation / Parallactic angle question
How to answer questions about my characters?
Is a lawful good "antagonist" effective?
SQL Server Primary Login Restrictions
Can anyone tell me why this program fails?
Professor being mistaken for a grad student
Replacing Windows 7 security updates with anti-virus?
Am I not good enough for you?
What is the greatest age difference between a married couple in Tanach?
Provisioning profile doesn't include the application-identifier and keychain-access-groups entitlements
It's a yearly task, alright
What is IP squat space
Why does Deadpool say "You're welcome, Canada," after shooting Ryan Reynolds in the end credits?
Instead of Universal Basic Income, why not Universal Basic NEEDS?
Schematic conventions for different supply rails
Life insurance that covers only simultaneous/dual deaths
Making a sword in the stone, in a medieval world without magic
An Accountant Seeks the Help of a Mathematician
Can elves maintain concentration in a trance?
Good allowance savings plan?
Be in awe of my brilliance!
What are the possible solutions of the given equation?
Pinhole Camera with Instant Film
Possible Leak In Concrete
How do I hide Chekhov's Gun?
Official degrees of earth’s rotation per day
What effect does the Earth's rotation have on plate tectonics?Why does this graph for sunlight intensity on land has a steeper slope during sunrise as compared to sunset?Cancelling out earth rotation speed, Altazimuth mountDoes a week represent something in astronomy?Ambiguity in Earth's “Tilt”Measurement precision of celestial eventsWhy is right ascension measured on a 24 hour scale rather than a 23 hours and 56 minutes scale?How can I calculate how far through the day the prime meridian is of different planetsModeling planet rotations (time of day, obliqueness, etc.)Field rotation / Parallactic angle question
$begingroup$
What is the official degree to one decimal point please, of the earth’s rotation in one single day. Can it be confirmed that it is exactly 360.0 degrees using official data? Thank you in advance.
earth rotation
asked 8 hours ago
AutodidactAutodidact
1145
$endgroup$
add a comment |
$begingroup$
What is the official degree to one decimal point please, of the earth’s rotation in one single day. Can it be confirmed that it is exactly 360.0 degrees using official data? Thank you in advance.
earth rotation
asked 8 hours ago
AutodidactAutodidact
1145
$endgroup$
add a comment |
$begingroup$
What is the official degree to one decimal point please, of the earth’s rotation in one single day. Can it be confirmed that it is exactly 360.0 degrees using official data? Thank you in advance.
earth rotation
asked 8 hours ago
AutodidactAutodidact
1145
$endgroup$
What is the official degree to one decimal point please, of the earth’s rotation in one single day. Can it be confirmed that it is exactly 360.0 degrees using official data? Thank you in advance.
earth rotation
earth rotation
asked 8 hours ago
AutodidactAutodidact
1145
asked 8 hours ago
AutodidactAutodidact
1145
asked 8 hours ago
AutodidactAutodidact
1145
asked 8 hours ago
AutodidactAutodidact
1145
asked 8 hours ago
AutodidactAutodidact
1145
1145
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is a bit more complicated than it seems. First off, the definition of a day that matters to us earthlings is the average amount of time from one solar noon to the next (or alternatively, the time it takes for the Sun to appear above the same meridian from day to day); it is called a solar day. The sidereal day, which is the time it takes for some given distant star to appear above the same meridian from day to day, is not the one that really matters to us; this is also the amount of time it takes for the Earth to rotate 360 degrees.
While the Earth is rotating on its axis, it is also travelling along its orbit. In about the amount of time it takes to complete one revolution, it has also travelled about one degree along its orbital path so that in order for the Sun to appear above the same meridian, the Earth has to rotate about 361 degrees.
But then near perihelion (its closest approach to the Sun, which is around January) it's travelling even faster, so it has to rotate more than 361 degrees. Near aphelion it's travelling slower, so the Earth has to rotate less than 361 degrees.
As to your actual question, given the complexity of Earth's orbital variations, I'm not sure it's answerable.
answered 7 hours ago
BillDOeBillDOe
891410
$endgroup$
$begingroup$
“The true solar day tends to be longer near perihelion when the Sun apparently moves along the ecliptic through a greater angle than usual, taking about 10 seconds longer to do so. Conversely, it is about 10 seconds shorter near aphelion. It is about 20 seconds longer near a solstice when the projection of the Sun's apparent motion along the ecliptic onto the celestial equator causes the Sun to move through a greater angle than usual. Conversely, near an equinox the projection onto the equator is shorter by about 20 seconds” I’m trying to figure out how the perihelion can affect it that much
$endgroup$
– Autodidact
6 hours ago
add a comment |
$begingroup$
Can it be confirmed that it is exactly 360.0 degrees using official data?
TL;DR: No, it can not. Instead it can be confirmed to be 361.0 degrees.
To my knowledge:
The Earth's rotation period is very close to 23 hours, 56 minutes, 4.1 seconds or 86164.1 sec. That's called a sidereal day
A day is defined as 24 hours, or 86400.0 sec.
So in one day it turns
$$360° times frac86400.086164.1 approx 360.986°$$
Rounded "to one decimal point please" that's 361.0°.
Using "Official data":
NASA Earth Fact Sheet: https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html
Sidereal rotation period (hrs) 23.9345
Length of day (hrs) 24.0000
$$360° times frac24.000023.9345 approx 360.985°$$
Rounded "to one decimal point please" again, that's still 361.0°.
answered 3 hours ago
uhohuhoh
6,21821763
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "514"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fastronomy.stackexchange.com%2fquestions%2f29986%2fofficial-degrees-of-earth-s-rotation-per-day%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is a bit more complicated than it seems. First off, the definition of a day that matters to us earthlings is the average amount of time from one solar noon to the next (or alternatively, the time it takes for the Sun to appear above the same meridian from day to day); it is called a solar day. The sidereal day, which is the time it takes for some given distant star to appear above the same meridian from day to day, is not the one that really matters to us; this is also the amount of time it takes for the Earth to rotate 360 degrees.
While the Earth is rotating on its axis, it is also travelling along its orbit. In about the amount of time it takes to complete one revolution, it has also travelled about one degree along its orbital path so that in order for the Sun to appear above the same meridian, the Earth has to rotate about 361 degrees.
But then near perihelion (its closest approach to the Sun, which is around January) it's travelling even faster, so it has to rotate more than 361 degrees. Near aphelion it's travelling slower, so the Earth has to rotate less than 361 degrees.
As to your actual question, given the complexity of Earth's orbital variations, I'm not sure it's answerable.
answered 7 hours ago
BillDOeBillDOe
891410
$endgroup$
$begingroup$
“The true solar day tends to be longer near perihelion when the Sun apparently moves along the ecliptic through a greater angle than usual, taking about 10 seconds longer to do so. Conversely, it is about 10 seconds shorter near aphelion. It is about 20 seconds longer near a solstice when the projection of the Sun's apparent motion along the ecliptic onto the celestial equator causes the Sun to move through a greater angle than usual. Conversely, near an equinox the projection onto the equator is shorter by about 20 seconds” I’m trying to figure out how the perihelion can affect it that much
$endgroup$
– Autodidact
6 hours ago
add a comment |
$begingroup$
This is a bit more complicated than it seems. First off, the definition of a day that matters to us earthlings is the average amount of time from one solar noon to the next (or alternatively, the time it takes for the Sun to appear above the same meridian from day to day); it is called a solar day. The sidereal day, which is the time it takes for some given distant star to appear above the same meridian from day to day, is not the one that really matters to us; this is also the amount of time it takes for the Earth to rotate 360 degrees.
While the Earth is rotating on its axis, it is also travelling along its orbit. In about the amount of time it takes to complete one revolution, it has also travelled about one degree along its orbital path so that in order for the Sun to appear above the same meridian, the Earth has to rotate about 361 degrees.
But then near perihelion (its closest approach to the Sun, which is around January) it's travelling even faster, so it has to rotate more than 361 degrees. Near aphelion it's travelling slower, so the Earth has to rotate less than 361 degrees.
As to your actual question, given the complexity of Earth's orbital variations, I'm not sure it's answerable.
answered 7 hours ago
BillDOeBillDOe
891410
$endgroup$
$begingroup$
“The true solar day tends to be longer near perihelion when the Sun apparently moves along the ecliptic through a greater angle than usual, taking about 10 seconds longer to do so. Conversely, it is about 10 seconds shorter near aphelion. It is about 20 seconds longer near a solstice when the projection of the Sun's apparent motion along the ecliptic onto the celestial equator causes the Sun to move through a greater angle than usual. Conversely, near an equinox the projection onto the equator is shorter by about 20 seconds” I’m trying to figure out how the perihelion can affect it that much
$endgroup$
– Autodidact
6 hours ago
add a comment |
$begingroup$
This is a bit more complicated than it seems. First off, the definition of a day that matters to us earthlings is the average amount of time from one solar noon to the next (or alternatively, the time it takes for the Sun to appear above the same meridian from day to day); it is called a solar day. The sidereal day, which is the time it takes for some given distant star to appear above the same meridian from day to day, is not the one that really matters to us; this is also the amount of time it takes for the Earth to rotate 360 degrees.
While the Earth is rotating on its axis, it is also travelling along its orbit. In about the amount of time it takes to complete one revolution, it has also travelled about one degree along its orbital path so that in order for the Sun to appear above the same meridian, the Earth has to rotate about 361 degrees.
But then near perihelion (its closest approach to the Sun, which is around January) it's travelling even faster, so it has to rotate more than 361 degrees. Near aphelion it's travelling slower, so the Earth has to rotate less than 361 degrees.
As to your actual question, given the complexity of Earth's orbital variations, I'm not sure it's answerable.
answered 7 hours ago
BillDOeBillDOe
891410
$endgroup$
This is a bit more complicated than it seems. First off, the definition of a day that matters to us earthlings is the average amount of time from one solar noon to the next (or alternatively, the time it takes for the Sun to appear above the same meridian from day to day); it is called a solar day. The sidereal day, which is the time it takes for some given distant star to appear above the same meridian from day to day, is not the one that really matters to us; this is also the amount of time it takes for the Earth to rotate 360 degrees.
While the Earth is rotating on its axis, it is also travelling along its orbit. In about the amount of time it takes to complete one revolution, it has also travelled about one degree along its orbital path so that in order for the Sun to appear above the same meridian, the Earth has to rotate about 361 degrees.
But then near perihelion (its closest approach to the Sun, which is around January) it's travelling even faster, so it has to rotate more than 361 degrees. Near aphelion it's travelling slower, so the Earth has to rotate less than 361 degrees.
As to your actual question, given the complexity of Earth's orbital variations, I'm not sure it's answerable.
answered 7 hours ago
BillDOeBillDOe
891410
answered 7 hours ago
BillDOeBillDOe
891410
answered 7 hours ago
BillDOeBillDOe
891410
answered 7 hours ago
BillDOeBillDOe
891410
891410
$begingroup$
“The true solar day tends to be longer near perihelion when the Sun apparently moves along the ecliptic through a greater angle than usual, taking about 10 seconds longer to do so. Conversely, it is about 10 seconds shorter near aphelion. It is about 20 seconds longer near a solstice when the projection of the Sun's apparent motion along the ecliptic onto the celestial equator causes the Sun to move through a greater angle than usual. Conversely, near an equinox the projection onto the equator is shorter by about 20 seconds” I’m trying to figure out how the perihelion can affect it that much
$endgroup$
– Autodidact
6 hours ago
add a comment |
$begingroup$
“The true solar day tends to be longer near perihelion when the Sun apparently moves along the ecliptic through a greater angle than usual, taking about 10 seconds longer to do so. Conversely, it is about 10 seconds shorter near aphelion. It is about 20 seconds longer near a solstice when the projection of the Sun's apparent motion along the ecliptic onto the celestial equator causes the Sun to move through a greater angle than usual. Conversely, near an equinox the projection onto the equator is shorter by about 20 seconds” I’m trying to figure out how the perihelion can affect it that much
$endgroup$
– Autodidact
6 hours ago
$begingroup$
“The true solar day tends to be longer near perihelion when the Sun apparently moves along the ecliptic through a greater angle than usual, taking about 10 seconds longer to do so. Conversely, it is about 10 seconds shorter near aphelion. It is about 20 seconds longer near a solstice when the projection of the Sun's apparent motion along the ecliptic onto the celestial equator causes the Sun to move through a greater angle than usual. Conversely, near an equinox the projection onto the equator is shorter by about 20 seconds” I’m trying to figure out how the perihelion can affect it that much
$endgroup$
– Autodidact
6 hours ago
$begingroup$
“The true solar day tends to be longer near perihelion when the Sun apparently moves along the ecliptic through a greater angle than usual, taking about 10 seconds longer to do so. Conversely, it is about 10 seconds shorter near aphelion. It is about 20 seconds longer near a solstice when the projection of the Sun's apparent motion along the ecliptic onto the celestial equator causes the Sun to move through a greater angle than usual. Conversely, near an equinox the projection onto the equator is shorter by about 20 seconds” I’m trying to figure out how the perihelion can affect it that much
$endgroup$
– Autodidact
6 hours ago
add a comment |
$begingroup$
Can it be confirmed that it is exactly 360.0 degrees using official data?
TL;DR: No, it can not. Instead it can be confirmed to be 361.0 degrees.
To my knowledge:
The Earth's rotation period is very close to 23 hours, 56 minutes, 4.1 seconds or 86164.1 sec. That's called a sidereal day
A day is defined as 24 hours, or 86400.0 sec.
So in one day it turns
$$360° times frac86400.086164.1 approx 360.986°$$
Rounded "to one decimal point please" that's 361.0°.
Using "Official data":
NASA Earth Fact Sheet: https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html
Sidereal rotation period (hrs) 23.9345
Length of day (hrs) 24.0000
$$360° times frac24.000023.9345 approx 360.985°$$
Rounded "to one decimal point please" again, that's still 361.0°.
answered 3 hours ago
uhohuhoh
6,21821763
$endgroup$
add a comment |
$begingroup$
Can it be confirmed that it is exactly 360.0 degrees using official data?
TL;DR: No, it can not. Instead it can be confirmed to be 361.0 degrees.
To my knowledge:
The Earth's rotation period is very close to 23 hours, 56 minutes, 4.1 seconds or 86164.1 sec. That's called a sidereal day
A day is defined as 24 hours, or 86400.0 sec.
So in one day it turns
$$360° times frac86400.086164.1 approx 360.986°$$
Rounded "to one decimal point please" that's 361.0°.
Using "Official data":
NASA Earth Fact Sheet: https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html
Sidereal rotation period (hrs) 23.9345
Length of day (hrs) 24.0000
$$360° times frac24.000023.9345 approx 360.985°$$
Rounded "to one decimal point please" again, that's still 361.0°.
answered 3 hours ago
uhohuhoh
6,21821763
$endgroup$
add a comment |
$begingroup$
Can it be confirmed that it is exactly 360.0 degrees using official data?
TL;DR: No, it can not. Instead it can be confirmed to be 361.0 degrees.
To my knowledge:
The Earth's rotation period is very close to 23 hours, 56 minutes, 4.1 seconds or 86164.1 sec. That's called a sidereal day
A day is defined as 24 hours, or 86400.0 sec.
So in one day it turns
$$360° times frac86400.086164.1 approx 360.986°$$
Rounded "to one decimal point please" that's 361.0°.
Using "Official data":
NASA Earth Fact Sheet: https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html
Sidereal rotation period (hrs) 23.9345
Length of day (hrs) 24.0000
$$360° times frac24.000023.9345 approx 360.985°$$
Rounded "to one decimal point please" again, that's still 361.0°.
answered 3 hours ago
uhohuhoh
6,21821763
$endgroup$
Can it be confirmed that it is exactly 360.0 degrees using official data?
TL;DR: No, it can not. Instead it can be confirmed to be 361.0 degrees.
To my knowledge:
The Earth's rotation period is very close to 23 hours, 56 minutes, 4.1 seconds or 86164.1 sec. That's called a sidereal day
A day is defined as 24 hours, or 86400.0 sec.
So in one day it turns
$$360° times frac86400.086164.1 approx 360.986°$$
Rounded "to one decimal point please" that's 361.0°.
Using "Official data":
NASA Earth Fact Sheet: https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html
Sidereal rotation period (hrs) 23.9345
Length of day (hrs) 24.0000
$$360° times frac24.000023.9345 approx 360.985°$$
Rounded "to one decimal point please" again, that's still 361.0°.
answered 3 hours ago
uhohuhoh
6,21821763
answered 3 hours ago
uhohuhoh
6,21821763
answered 3 hours ago
uhohuhoh
6,21821763
answered 3 hours ago
uhohuhoh
6,21821763
6,21821763
add a comment |
add a comment |
Thanks for contributing an answer to Astronomy Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fastronomy.stackexchange.com%2fquestions%2f29986%2fofficial-degrees-of-earth-s-rotation-per-day%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
