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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













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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.










share|improve this question









$endgroup$
















    1












    $begingroup$


    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.










    share|improve this question









    $endgroup$














      1












      1








      1





      $begingroup$


      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.










      share|improve this question









      $endgroup$




      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






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 4 hours ago









      AutodidactAutodidact

      1145




      1145




















          1 Answer
          1






          active

          oldest

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          3












          $begingroup$

          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.






          share|improve this answer









          $endgroup$












          • $begingroup$
            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











          Your Answer





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          1 Answer
          1






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          active

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          active

          oldest

          votes









          3












          $begingroup$

          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.






          share|improve this answer









          $endgroup$












          • $begingroup$
            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
















          3












          $begingroup$

          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.






          share|improve this answer









          $endgroup$












          • $begingroup$
            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














          3












          3








          3





          $begingroup$

          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.






          share|improve this answer









          $endgroup$



          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.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          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

















          • $begingroup$
            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


















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