Official degrees of earth’s rotation per day
$begingroup$
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 yesterday
AutodidactAutodidact
1748
$endgroup$
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 yesterday
AutodidactAutodidact
1748
$endgroup$
7
$begingroup$
Official according to what or whom?
$endgroup$
– jpmc26
18 hours ago
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 yesterday
AutodidactAutodidact
1748
$endgroup$
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 yesterday
AutodidactAutodidact
1748
asked yesterday
AutodidactAutodidact
1748
asked yesterday
AutodidactAutodidact
1748
asked yesterday
AutodidactAutodidact
1748
asked yesterday
AutodidactAutodidact
1748
1748
7
$begingroup$
Official according to what or whom?
$endgroup$
– jpmc26
18 hours ago
add a comment |
7
$begingroup$
Official according to what or whom?
$endgroup$
– jpmc26
18 hours ago
7
7
$begingroup$
Official according to what or whom?
$endgroup$
– jpmc26
18 hours ago
$begingroup$
Official according to what or whom?
$endgroup$
– jpmc26
18 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
First, we need to decide which definition of "day" to employ. There are several types of days:
Apparent solar day: the time between two successive culminations of the Sun (apparent Noon) from an fixed Earth-based observer.
Mean solar day: an more uniform, averaged solar day without seasonal variations.
Stellar/Sidereal day: the time needed for the Earth to rotate once relative to the stars.
SI day: a unit of time containing exactly 86,400 SI seconds defined by caesium atoms.
Since we generally refer to the traditional day/night cycle when we say "day", this means a form of solar time. To get a more averaged value, let's use the mean solar day.
The current formula linking the Earth Rotation Angle (ERA) to the modern approximation of mean solar time, UT1 (basically the Earth's clock following the mean day/night cycle), is by definition :
$$ERA = 2π(0.7790572732640 + 1.00273781191135448 T_u) text{ radians}$$
Where Tu is the Julian UT1 Date - 2451545.0
So according to this formula, a (UT1) day is 1.00273781191135448 Earth rotations, which multiplied by 360° is about 360.98561°. However, the Earth's rotation and revolution are not constant, and are always changing at somewhat unpredictable rates, so the angle is not perfect, but the changes are very slow. So this is more a modern approximation rather than an exact value. Rounded to one decimal place, this gives you 361.0°, a figure that will likely remain true for at least several millenia.
If you want to know the amount of Earth rotation for every SI day, you're in luck: it is possible to consult reports of the Earth's orientation (rotation and polar motion) thanks to the IERS. Values are tabulated for each 0h of Terrestrial Time every day in the IERS publications, allowing to derive the angle that Earth has rotated every 86,400 SI seconds, allowing to scientifically monitor variations in Earth's rotation compared to a very constant unit of time realized by atomic clocks. Nowadays, the Earth's rotation is measured and reported thanks to radio telescopes and VLBI observing distant objects in the universe. Official reports of its orientation, the Earth Orientation Parameters are published on the IERS Website.
As an example, here is a graph showing the amount of Earth rotation every 86,400 seconds constructed with IERS data of the last year:
As we can see, there are several seasonal, periodic and unpredictable variations in Earth's rotation.
answered yesterday
FSimardGISFSimardGIS
70619
$endgroup$
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 yesterday
BillDOeBillDOe
1,011411
$endgroup$
$begingroup$
“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
yesterday
1
$begingroup$
@Autodidact: At perihelion we're about 3.3% closer to the sun than at aphelion, so by Kepler's second law the earth's linear speed relative to the sun is about 3.3% faster. However, being closer to the sun also means that for each kilometer we move, the angle to the sun will change by about 3.3% more. These effects add together, so the apparent speed of the sun among the fixed stars is about 6.6% faster at perihelion. This makes the difference between the solar and sidereal day 6.6% larger which indeed works out to around 15 seconds. (Ignoring the effects of axial tilt).
$endgroup$
– Henning Makholm
20 hours ago
$begingroup$
(There's still a difference of a few seconds from the source you're quoting, but I'll chalk that up to rounding. Remember that "about 10 seconds" has only one significant digit!)
$endgroup$
– Henning Makholm
20 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
$$360° times frac{86400.0}{86164.1} approx 360.986°$$
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
$$360° times frac{24.0000}{23.9345} approx 360.985°$$
Rounded "to one decimal point please" again, that's still 361.0°.
answered yesterday
uhohuhoh
6,33821764
$endgroup$
$begingroup$
@Jasper you're talking about a solar day not a sidereal day. The question asks about the rotation of the Earth which has to be considered in an inertial frame, and that's what a sidereal day is. A day is exactly 24 hours, or 86400 seconds. Solar days aren't really used in time measurement anymore because of the effects you've mentioned.
$endgroup$
– uhoh
yesterday
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
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First, we need to decide which definition of "day" to employ. There are several types of days:
Apparent solar day: the time between two successive culminations of the Sun (apparent Noon) from an fixed Earth-based observer.
Mean solar day: an more uniform, averaged solar day without seasonal variations.
Stellar/Sidereal day: the time needed for the Earth to rotate once relative to the stars.
SI day: a unit of time containing exactly 86,400 SI seconds defined by caesium atoms.
Since we generally refer to the traditional day/night cycle when we say "day", this means a form of solar time. To get a more averaged value, let's use the mean solar day.
The current formula linking the Earth Rotation Angle (ERA) to the modern approximation of mean solar time, UT1 (basically the Earth's clock following the mean day/night cycle), is by definition :
$$ERA = 2π(0.7790572732640 + 1.00273781191135448 T_u) text{ radians}$$
Where Tu is the Julian UT1 Date - 2451545.0
So according to this formula, a (UT1) day is 1.00273781191135448 Earth rotations, which multiplied by 360° is about 360.98561°. However, the Earth's rotation and revolution are not constant, and are always changing at somewhat unpredictable rates, so the angle is not perfect, but the changes are very slow. So this is more a modern approximation rather than an exact value. Rounded to one decimal place, this gives you 361.0°, a figure that will likely remain true for at least several millenia.
If you want to know the amount of Earth rotation for every SI day, you're in luck: it is possible to consult reports of the Earth's orientation (rotation and polar motion) thanks to the IERS. Values are tabulated for each 0h of Terrestrial Time every day in the IERS publications, allowing to derive the angle that Earth has rotated every 86,400 SI seconds, allowing to scientifically monitor variations in Earth's rotation compared to a very constant unit of time realized by atomic clocks. Nowadays, the Earth's rotation is measured and reported thanks to radio telescopes and VLBI observing distant objects in the universe. Official reports of its orientation, the Earth Orientation Parameters are published on the IERS Website.
As an example, here is a graph showing the amount of Earth rotation every 86,400 seconds constructed with IERS data of the last year:
As we can see, there are several seasonal, periodic and unpredictable variations in Earth's rotation.
answered yesterday
FSimardGISFSimardGIS
70619
$endgroup$
add a comment |
$begingroup$
First, we need to decide which definition of "day" to employ. There are several types of days:
Apparent solar day: the time between two successive culminations of the Sun (apparent Noon) from an fixed Earth-based observer.
Mean solar day: an more uniform, averaged solar day without seasonal variations.
Stellar/Sidereal day: the time needed for the Earth to rotate once relative to the stars.
SI day: a unit of time containing exactly 86,400 SI seconds defined by caesium atoms.
Since we generally refer to the traditional day/night cycle when we say "day", this means a form of solar time. To get a more averaged value, let's use the mean solar day.
The current formula linking the Earth Rotation Angle (ERA) to the modern approximation of mean solar time, UT1 (basically the Earth's clock following the mean day/night cycle), is by definition :
$$ERA = 2π(0.7790572732640 + 1.00273781191135448 T_u) text{ radians}$$
Where Tu is the Julian UT1 Date - 2451545.0
So according to this formula, a (UT1) day is 1.00273781191135448 Earth rotations, which multiplied by 360° is about 360.98561°. However, the Earth's rotation and revolution are not constant, and are always changing at somewhat unpredictable rates, so the angle is not perfect, but the changes are very slow. So this is more a modern approximation rather than an exact value. Rounded to one decimal place, this gives you 361.0°, a figure that will likely remain true for at least several millenia.
If you want to know the amount of Earth rotation for every SI day, you're in luck: it is possible to consult reports of the Earth's orientation (rotation and polar motion) thanks to the IERS. Values are tabulated for each 0h of Terrestrial Time every day in the IERS publications, allowing to derive the angle that Earth has rotated every 86,400 SI seconds, allowing to scientifically monitor variations in Earth's rotation compared to a very constant unit of time realized by atomic clocks. Nowadays, the Earth's rotation is measured and reported thanks to radio telescopes and VLBI observing distant objects in the universe. Official reports of its orientation, the Earth Orientation Parameters are published on the IERS Website.
As an example, here is a graph showing the amount of Earth rotation every 86,400 seconds constructed with IERS data of the last year:
As we can see, there are several seasonal, periodic and unpredictable variations in Earth's rotation.
answered yesterday
FSimardGISFSimardGIS
70619
$endgroup$
add a comment |
$begingroup$
First, we need to decide which definition of "day" to employ. There are several types of days:
Apparent solar day: the time between two successive culminations of the Sun (apparent Noon) from an fixed Earth-based observer.
Mean solar day: an more uniform, averaged solar day without seasonal variations.
Stellar/Sidereal day: the time needed for the Earth to rotate once relative to the stars.
SI day: a unit of time containing exactly 86,400 SI seconds defined by caesium atoms.
Since we generally refer to the traditional day/night cycle when we say "day", this means a form of solar time. To get a more averaged value, let's use the mean solar day.
The current formula linking the Earth Rotation Angle (ERA) to the modern approximation of mean solar time, UT1 (basically the Earth's clock following the mean day/night cycle), is by definition :
$$ERA = 2π(0.7790572732640 + 1.00273781191135448 T_u) text{ radians}$$
Where Tu is the Julian UT1 Date - 2451545.0
So according to this formula, a (UT1) day is 1.00273781191135448 Earth rotations, which multiplied by 360° is about 360.98561°. However, the Earth's rotation and revolution are not constant, and are always changing at somewhat unpredictable rates, so the angle is not perfect, but the changes are very slow. So this is more a modern approximation rather than an exact value. Rounded to one decimal place, this gives you 361.0°, a figure that will likely remain true for at least several millenia.
If you want to know the amount of Earth rotation for every SI day, you're in luck: it is possible to consult reports of the Earth's orientation (rotation and polar motion) thanks to the IERS. Values are tabulated for each 0h of Terrestrial Time every day in the IERS publications, allowing to derive the angle that Earth has rotated every 86,400 SI seconds, allowing to scientifically monitor variations in Earth's rotation compared to a very constant unit of time realized by atomic clocks. Nowadays, the Earth's rotation is measured and reported thanks to radio telescopes and VLBI observing distant objects in the universe. Official reports of its orientation, the Earth Orientation Parameters are published on the IERS Website.
As an example, here is a graph showing the amount of Earth rotation every 86,400 seconds constructed with IERS data of the last year:
As we can see, there are several seasonal, periodic and unpredictable variations in Earth's rotation.
answered yesterday
FSimardGISFSimardGIS
70619
$endgroup$
First, we need to decide which definition of "day" to employ. There are several types of days:
Apparent solar day: the time between two successive culminations of the Sun (apparent Noon) from an fixed Earth-based observer.
Mean solar day: an more uniform, averaged solar day without seasonal variations.
Stellar/Sidereal day: the time needed for the Earth to rotate once relative to the stars.
SI day: a unit of time containing exactly 86,400 SI seconds defined by caesium atoms.
Since we generally refer to the traditional day/night cycle when we say "day", this means a form of solar time. To get a more averaged value, let's use the mean solar day.
The current formula linking the Earth Rotation Angle (ERA) to the modern approximation of mean solar time, UT1 (basically the Earth's clock following the mean day/night cycle), is by definition :
$$ERA = 2π(0.7790572732640 + 1.00273781191135448 T_u) text{ radians}$$
Where Tu is the Julian UT1 Date - 2451545.0
So according to this formula, a (UT1) day is 1.00273781191135448 Earth rotations, which multiplied by 360° is about 360.98561°. However, the Earth's rotation and revolution are not constant, and are always changing at somewhat unpredictable rates, so the angle is not perfect, but the changes are very slow. So this is more a modern approximation rather than an exact value. Rounded to one decimal place, this gives you 361.0°, a figure that will likely remain true for at least several millenia.
If you want to know the amount of Earth rotation for every SI day, you're in luck: it is possible to consult reports of the Earth's orientation (rotation and polar motion) thanks to the IERS. Values are tabulated for each 0h of Terrestrial Time every day in the IERS publications, allowing to derive the angle that Earth has rotated every 86,400 SI seconds, allowing to scientifically monitor variations in Earth's rotation compared to a very constant unit of time realized by atomic clocks. Nowadays, the Earth's rotation is measured and reported thanks to radio telescopes and VLBI observing distant objects in the universe. Official reports of its orientation, the Earth Orientation Parameters are published on the IERS Website.
As an example, here is a graph showing the amount of Earth rotation every 86,400 seconds constructed with IERS data of the last year:
As we can see, there are several seasonal, periodic and unpredictable variations in Earth's rotation.
answered yesterday
FSimardGISFSimardGIS
70619
answered yesterday
FSimardGISFSimardGIS
70619
answered yesterday
FSimardGISFSimardGIS
70619
answered yesterday
FSimardGISFSimardGIS
70619
70619
add a comment |
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 yesterday
BillDOeBillDOe
1,011411
$endgroup$
$begingroup$
“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
yesterday
1
$begingroup$
@Autodidact: At perihelion we're about 3.3% closer to the sun than at aphelion, so by Kepler's second law the earth's linear speed relative to the sun is about 3.3% faster. However, being closer to the sun also means that for each kilometer we move, the angle to the sun will change by about 3.3% more. These effects add together, so the apparent speed of the sun among the fixed stars is about 6.6% faster at perihelion. This makes the difference between the solar and sidereal day 6.6% larger which indeed works out to around 15 seconds. (Ignoring the effects of axial tilt).
$endgroup$
– Henning Makholm
20 hours ago
$begingroup$
(There's still a difference of a few seconds from the source you're quoting, but I'll chalk that up to rounding. Remember that "about 10 seconds" has only one significant digit!)
$endgroup$
– Henning Makholm
20 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 yesterday
BillDOeBillDOe
1,011411
$endgroup$
$begingroup$
“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
yesterday
1
$begingroup$
@Autodidact: At perihelion we're about 3.3% closer to the sun than at aphelion, so by Kepler's second law the earth's linear speed relative to the sun is about 3.3% faster. However, being closer to the sun also means that for each kilometer we move, the angle to the sun will change by about 3.3% more. These effects add together, so the apparent speed of the sun among the fixed stars is about 6.6% faster at perihelion. This makes the difference between the solar and sidereal day 6.6% larger which indeed works out to around 15 seconds. (Ignoring the effects of axial tilt).
$endgroup$
– Henning Makholm
20 hours ago
$begingroup$
(There's still a difference of a few seconds from the source you're quoting, but I'll chalk that up to rounding. Remember that "about 10 seconds" has only one significant digit!)
$endgroup$
– Henning Makholm
20 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 yesterday
BillDOeBillDOe
1,011411
$endgroup$
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 yesterday
BillDOeBillDOe
1,011411
answered yesterday
BillDOeBillDOe
1,011411
answered yesterday
BillDOeBillDOe
1,011411
answered yesterday
BillDOeBillDOe
1,011411
1,011411
$begingroup$
“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
yesterday
1
$begingroup$
@Autodidact: At perihelion we're about 3.3% closer to the sun than at aphelion, so by Kepler's second law the earth's linear speed relative to the sun is about 3.3% faster. However, being closer to the sun also means that for each kilometer we move, the angle to the sun will change by about 3.3% more. These effects add together, so the apparent speed of the sun among the fixed stars is about 6.6% faster at perihelion. This makes the difference between the solar and sidereal day 6.6% larger which indeed works out to around 15 seconds. (Ignoring the effects of axial tilt).
$endgroup$
– Henning Makholm
20 hours ago
$begingroup$
(There's still a difference of a few seconds from the source you're quoting, but I'll chalk that up to rounding. Remember that "about 10 seconds" has only one significant digit!)
$endgroup$
– Henning Makholm
20 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
yesterday
1
$begingroup$
@Autodidact: At perihelion we're about 3.3% closer to the sun than at aphelion, so by Kepler's second law the earth's linear speed relative to the sun is about 3.3% faster. However, being closer to the sun also means that for each kilometer we move, the angle to the sun will change by about 3.3% more. These effects add together, so the apparent speed of the sun among the fixed stars is about 6.6% faster at perihelion. This makes the difference between the solar and sidereal day 6.6% larger which indeed works out to around 15 seconds. (Ignoring the effects of axial tilt).
$endgroup$
– Henning Makholm
20 hours ago
$begingroup$
(There's still a difference of a few seconds from the source you're quoting, but I'll chalk that up to rounding. Remember that "about 10 seconds" has only one significant digit!)
$endgroup$
– Henning Makholm
20 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
yesterday
$begingroup$
“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
yesterday
1
1
$begingroup$
@Autodidact: At perihelion we're about 3.3% closer to the sun than at aphelion, so by Kepler's second law the earth's linear speed relative to the sun is about 3.3% faster. However, being closer to the sun also means that for each kilometer we move, the angle to the sun will change by about 3.3% more. These effects add together, so the apparent speed of the sun among the fixed stars is about 6.6% faster at perihelion. This makes the difference between the solar and sidereal day 6.6% larger which indeed works out to around 15 seconds. (Ignoring the effects of axial tilt).
$endgroup$
– Henning Makholm
20 hours ago
$begingroup$
@Autodidact: At perihelion we're about 3.3% closer to the sun than at aphelion, so by Kepler's second law the earth's linear speed relative to the sun is about 3.3% faster. However, being closer to the sun also means that for each kilometer we move, the angle to the sun will change by about 3.3% more. These effects add together, so the apparent speed of the sun among the fixed stars is about 6.6% faster at perihelion. This makes the difference between the solar and sidereal day 6.6% larger which indeed works out to around 15 seconds. (Ignoring the effects of axial tilt).
$endgroup$
– Henning Makholm
20 hours ago
$begingroup$
(There's still a difference of a few seconds from the source you're quoting, but I'll chalk that up to rounding. Remember that "about 10 seconds" has only one significant digit!)
$endgroup$
– Henning Makholm
20 hours ago
$begingroup$
(There's still a difference of a few seconds from the source you're quoting, but I'll chalk that up to rounding. Remember that "about 10 seconds" has only one significant digit!)
$endgroup$
– Henning Makholm
20 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
$$360° times frac{86400.0}{86164.1} approx 360.986°$$
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
$$360° times frac{24.0000}{23.9345} approx 360.985°$$
Rounded "to one decimal point please" again, that's still 361.0°.
answered yesterday
uhohuhoh
6,33821764
$endgroup$
$begingroup$
@Jasper you're talking about a solar day not a sidereal day. The question asks about the rotation of the Earth which has to be considered in an inertial frame, and that's what a sidereal day is. A day is exactly 24 hours, or 86400 seconds. Solar days aren't really used in time measurement anymore because of the effects you've mentioned.
$endgroup$
– uhoh
yesterday
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
$$360° times frac{86400.0}{86164.1} approx 360.986°$$
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
$$360° times frac{24.0000}{23.9345} approx 360.985°$$
Rounded "to one decimal point please" again, that's still 361.0°.
answered yesterday
uhohuhoh
6,33821764
$endgroup$
$begingroup$
@Jasper you're talking about a solar day not a sidereal day. The question asks about the rotation of the Earth which has to be considered in an inertial frame, and that's what a sidereal day is. A day is exactly 24 hours, or 86400 seconds. Solar days aren't really used in time measurement anymore because of the effects you've mentioned.
$endgroup$
– uhoh
yesterday
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
$$360° times frac{86400.0}{86164.1} approx 360.986°$$
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
$$360° times frac{24.0000}{23.9345} approx 360.985°$$
Rounded "to one decimal point please" again, that's still 361.0°.
answered yesterday
uhohuhoh
6,33821764
$endgroup$
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
$$360° times frac{86400.0}{86164.1} approx 360.986°$$
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
$$360° times frac{24.0000}{23.9345} approx 360.985°$$
Rounded "to one decimal point please" again, that's still 361.0°.
answered yesterday
uhohuhoh
6,33821764
answered yesterday
uhohuhoh
6,33821764
answered yesterday
uhohuhoh
6,33821764
answered yesterday
uhohuhoh
6,33821764
6,33821764
$begingroup$
@Jasper you're talking about a solar day not a sidereal day. The question asks about the rotation of the Earth which has to be considered in an inertial frame, and that's what a sidereal day is. A day is exactly 24 hours, or 86400 seconds. Solar days aren't really used in time measurement anymore because of the effects you've mentioned.
$endgroup$
– uhoh
yesterday
add a comment |
$begingroup$
@Jasper you're talking about a solar day not a sidereal day. The question asks about the rotation of the Earth which has to be considered in an inertial frame, and that's what a sidereal day is. A day is exactly 24 hours, or 86400 seconds. Solar days aren't really used in time measurement anymore because of the effects you've mentioned.
$endgroup$
– uhoh
yesterday
$begingroup$
@Jasper you're talking about a solar day not a sidereal day. The question asks about the rotation of the Earth which has to be considered in an inertial frame, and that's what a sidereal day is. A day is exactly 24 hours, or 86400 seconds. Solar days aren't really used in time measurement anymore because of the effects you've mentioned.
$endgroup$
– uhoh
yesterday
$begingroup$
@Jasper you're talking about a solar day not a sidereal day. The question asks about the rotation of the Earth which has to be considered in an inertial frame, and that's what a sidereal day is. A day is exactly 24 hours, or 86400 seconds. Solar days aren't really used in time measurement anymore because of the effects you've mentioned.
$endgroup$
– uhoh
yesterday
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
7
$begingroup$
Official according to what or whom?
$endgroup$
– jpmc26
18 hours ago