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Official degrees of earth’s rotation per day
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$begingroup$
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 4 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 4 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 4 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 4 hours ago
AutodidactAutodidact
1145
asked 4 hours ago
AutodidactAutodidact
1145
asked 4 hours ago
AutodidactAutodidact
1145
asked 4 hours ago
AutodidactAutodidact
1145
asked 4 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 3 hours ago
BillDOeBillDOe
851410
$endgroup$
$begingroup$
“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
3 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 10 mins ago
uhohuhoh
6,18821763
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
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$begingroup$
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 3 hours ago
BillDOeBillDOe
851410
$endgroup$
$begingroup$
“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
3 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 3 hours ago
BillDOeBillDOe
851410
$endgroup$
$begingroup$
“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
3 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 3 hours ago
BillDOeBillDOe
851410
$endgroup$
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 3 hours ago
BillDOeBillDOe
851410
answered 3 hours ago
BillDOeBillDOe
851410
answered 3 hours ago
BillDOeBillDOe
851410
answered 3 hours ago
BillDOeBillDOe
851410
851410
$begingroup$
“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
3 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
3 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
3 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
3 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 10 mins ago
uhohuhoh
6,18821763
$endgroup$
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
$$360° times frac{86400.0}{86164.1} approx 360.986°$$
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 10 mins ago
uhohuhoh
6,18821763
$endgroup$
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
$$360° times frac{86400.0}{86164.1} approx 360.986°$$
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 10 mins ago
uhohuhoh
6,18821763
$endgroup$
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 10 mins ago
uhohuhoh
6,18821763
answered 10 mins ago
uhohuhoh
6,18821763
answered 10 mins ago
uhohuhoh
6,18821763
answered 10 mins ago
uhohuhoh
6,18821763
6,18821763
add a comment |
add a comment |
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