Official degrees of earth’s rotation per day












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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







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    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.






    earth rotation







    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.






      earth rotation







      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




      earth rotation






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      share|improve this question











      share|improve this question




      share|improve this question










      asked 9 hours ago









      AutodidactAutodidact

      1145




      1145






















          2 Answers
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          8












          $begingroup$

          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
            8 hours ago





















          4












          $begingroup$


          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°.






          share|improve this answer









          $endgroup$













          • $begingroup$
            @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
            26 mins ago













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          2 Answers
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          2 Answers
          2






          active

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          active

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          active

          oldest

          votes









          8












          $begingroup$

          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
            8 hours ago


















          8












          $begingroup$

          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
            8 hours ago
















          8












          8








          8





          $begingroup$

          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 8 hours ago









          BillDOeBillDOe

          901410




          901410












          • $begingroup$
            “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
            8 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
            8 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
          8 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
          8 hours ago













          4












          $begingroup$


          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°.






          share|improve this answer









          $endgroup$













          • $begingroup$
            @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
            26 mins ago


















          4












          $begingroup$


          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°.






          share|improve this answer









          $endgroup$













          • $begingroup$
            @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
            26 mins ago
















          4












          4








          4





          $begingroup$


          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°.






          share|improve this answer









          $endgroup$




          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°.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 4 hours ago









          uhohuhoh

          6,22821763




          6,22821763












          • $begingroup$
            @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
            26 mins ago




















          • $begingroup$
            @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
            26 mins ago


















          $begingroup$
          @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
          26 mins ago






          $begingroup$
          @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
          26 mins ago




















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