Is Earth an inertial reference frame?











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Is earth considered as inertial frame? i was confused because we learned about coriolis effect. We know that earth spins therefore coriolis effect should take place . But does it have minimal effect for motion of balls etc when they move with respect to the ground?










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  • you can calculate it's size, e.g, for a BMG 50 cal (853 m/s muzzle velocity) shot at its max effective range (1,800m) and tell us.
    – JEB
    Dec 2 at 0:05










  • This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
    – Geremia
    Dec 2 at 4:31






  • 2




    That the Earth's surface is not an inertial frame is nicely demonstrated by the Foucault pendulum.
    – Ruslan
    Dec 2 at 15:51






  • 1




    It depends on the physics problem you look at whether the Earth can be considered an inertial frame of reference or not.
    – jjack
    Dec 2 at 18:14










  • To emphasize @jjack's point, it's notoriously hard work to get a Foucault pendulum to work properly because the non-inertial effects, while present, are very small.
    – dmckee
    Dec 2 at 19:28















up vote
6
down vote

favorite
1












Is earth considered as inertial frame? i was confused because we learned about coriolis effect. We know that earth spins therefore coriolis effect should take place . But does it have minimal effect for motion of balls etc when they move with respect to the ground?










share|cite|improve this question
























  • you can calculate it's size, e.g, for a BMG 50 cal (853 m/s muzzle velocity) shot at its max effective range (1,800m) and tell us.
    – JEB
    Dec 2 at 0:05










  • This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
    – Geremia
    Dec 2 at 4:31






  • 2




    That the Earth's surface is not an inertial frame is nicely demonstrated by the Foucault pendulum.
    – Ruslan
    Dec 2 at 15:51






  • 1




    It depends on the physics problem you look at whether the Earth can be considered an inertial frame of reference or not.
    – jjack
    Dec 2 at 18:14










  • To emphasize @jjack's point, it's notoriously hard work to get a Foucault pendulum to work properly because the non-inertial effects, while present, are very small.
    – dmckee
    Dec 2 at 19:28













up vote
6
down vote

favorite
1









up vote
6
down vote

favorite
1






1





Is earth considered as inertial frame? i was confused because we learned about coriolis effect. We know that earth spins therefore coriolis effect should take place . But does it have minimal effect for motion of balls etc when they move with respect to the ground?










share|cite|improve this question















Is earth considered as inertial frame? i was confused because we learned about coriolis effect. We know that earth spins therefore coriolis effect should take place . But does it have minimal effect for motion of balls etc when they move with respect to the ground?







newtonian-mechanics reference-frames inertial-frames earth coriolis-effect






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edited Dec 2 at 2:25









Qmechanic

101k121821135




101k121821135










asked Dec 1 at 23:46









ado sar

1738




1738












  • you can calculate it's size, e.g, for a BMG 50 cal (853 m/s muzzle velocity) shot at its max effective range (1,800m) and tell us.
    – JEB
    Dec 2 at 0:05










  • This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
    – Geremia
    Dec 2 at 4:31






  • 2




    That the Earth's surface is not an inertial frame is nicely demonstrated by the Foucault pendulum.
    – Ruslan
    Dec 2 at 15:51






  • 1




    It depends on the physics problem you look at whether the Earth can be considered an inertial frame of reference or not.
    – jjack
    Dec 2 at 18:14










  • To emphasize @jjack's point, it's notoriously hard work to get a Foucault pendulum to work properly because the non-inertial effects, while present, are very small.
    – dmckee
    Dec 2 at 19:28


















  • you can calculate it's size, e.g, for a BMG 50 cal (853 m/s muzzle velocity) shot at its max effective range (1,800m) and tell us.
    – JEB
    Dec 2 at 0:05










  • This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
    – Geremia
    Dec 2 at 4:31






  • 2




    That the Earth's surface is not an inertial frame is nicely demonstrated by the Foucault pendulum.
    – Ruslan
    Dec 2 at 15:51






  • 1




    It depends on the physics problem you look at whether the Earth can be considered an inertial frame of reference or not.
    – jjack
    Dec 2 at 18:14










  • To emphasize @jjack's point, it's notoriously hard work to get a Foucault pendulum to work properly because the non-inertial effects, while present, are very small.
    – dmckee
    Dec 2 at 19:28
















you can calculate it's size, e.g, for a BMG 50 cal (853 m/s muzzle velocity) shot at its max effective range (1,800m) and tell us.
– JEB
Dec 2 at 0:05




you can calculate it's size, e.g, for a BMG 50 cal (853 m/s muzzle velocity) shot at its max effective range (1,800m) and tell us.
– JEB
Dec 2 at 0:05












This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
– Geremia
Dec 2 at 4:31




This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
– Geremia
Dec 2 at 4:31




2




2




That the Earth's surface is not an inertial frame is nicely demonstrated by the Foucault pendulum.
– Ruslan
Dec 2 at 15:51




That the Earth's surface is not an inertial frame is nicely demonstrated by the Foucault pendulum.
– Ruslan
Dec 2 at 15:51




1




1




It depends on the physics problem you look at whether the Earth can be considered an inertial frame of reference or not.
– jjack
Dec 2 at 18:14




It depends on the physics problem you look at whether the Earth can be considered an inertial frame of reference or not.
– jjack
Dec 2 at 18:14












To emphasize @jjack's point, it's notoriously hard work to get a Foucault pendulum to work properly because the non-inertial effects, while present, are very small.
– dmckee
Dec 2 at 19:28




To emphasize @jjack's point, it's notoriously hard work to get a Foucault pendulum to work properly because the non-inertial effects, while present, are very small.
– dmckee
Dec 2 at 19:28










4 Answers
4






active

oldest

votes

















up vote
12
down vote













The surface of the Earth is not, rigorously speaking, an inertial frame of reference. Objects at rest relative to Earth's surface are actually subject to a series of inertial effects, like the ficticious forces (Coriolis, centrifugal etc.) because of Earth's rotation, precession and other kinds of acceleration.



When solving physics problems, however, we usually take the Earth frame as being inertial. This is because the inertial effects are minuscule for most of our day-to-day experiences and experiments. For example, objects in the Equator are the ones subject to the strongest centrifugal force and it is only about $3 times10^{-3}$ or $0.3%$ of their weight.



So for the most part, if an experiment is short enough and happens in a small enough region, the surface of Earth can indeed be approximated to an inertial frame of reference since the effects on the experiment's results are very, very tiny.



This of course has exceptions, as cited in njspeer's answer.



If however by "Earth" you mean the reference frame in Earth's center, it is an inertial frame according to General Relativity (GR), since observers in free fall are inertial in GR. The Earth actually does have some proper acceleration due to external forces such as radiation pressure, but these are also minuscule effects.






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  • 2




    The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
    – David
    Dec 2 at 8:23












  • I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
    – João Vítor G. Lima
    Dec 2 at 12:52


















up vote
9
down vote













Because the earth is rotating, it is never strictly an inertial reference frame. However, because the effects are small in many situations, it can often be approximated as one. When to use Coriolis forces will have to be determined on a case-by-case basis. E.g. ballistic problems that cover large distances will most certainly require Coriolis-force corrections, and pendulums that swings for a long time would also require Coriolis-force corrections. For a block sliding down an inclined plane, or a spring on a mass, or a vibrating string, you should not need to take it into consideration.






share|cite|improve this answer






























    up vote
    3
    down vote













    Mach would say that non-inertial effects are due to the relative motion between the earth and the rest of the universe.



    See:
    • Assis, André K. T. Relational Mechanics and Implementation of Mach’s Principle with Weber’s Gravitational Force. Montréal: Apeiron, 2014.






    share|cite|improve this answer






























      up vote
      0
      down vote













      I know it's being a little pedantic, but I would say "earth" is a thing, not a reference frame. You could define an inertial reference frame that contains the earth.



      But suppose you mean a reference frame defined with reference to the earth: Z is perpendicular to the ground where you are standing, X and Y are parallel to the ground and perpendicular to each other.



      If you stand at rest in this reference frame, you may notice you feel acceleration (gravity). It is therefore not an inertial reference frame.






      share|cite|improve this answer





















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        4 Answers
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        active

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        4 Answers
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        up vote
        12
        down vote













        The surface of the Earth is not, rigorously speaking, an inertial frame of reference. Objects at rest relative to Earth's surface are actually subject to a series of inertial effects, like the ficticious forces (Coriolis, centrifugal etc.) because of Earth's rotation, precession and other kinds of acceleration.



        When solving physics problems, however, we usually take the Earth frame as being inertial. This is because the inertial effects are minuscule for most of our day-to-day experiences and experiments. For example, objects in the Equator are the ones subject to the strongest centrifugal force and it is only about $3 times10^{-3}$ or $0.3%$ of their weight.



        So for the most part, if an experiment is short enough and happens in a small enough region, the surface of Earth can indeed be approximated to an inertial frame of reference since the effects on the experiment's results are very, very tiny.



        This of course has exceptions, as cited in njspeer's answer.



        If however by "Earth" you mean the reference frame in Earth's center, it is an inertial frame according to General Relativity (GR), since observers in free fall are inertial in GR. The Earth actually does have some proper acceleration due to external forces such as radiation pressure, but these are also minuscule effects.






        share|cite|improve this answer



















        • 2




          The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
          – David
          Dec 2 at 8:23












        • I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
          – João Vítor G. Lima
          Dec 2 at 12:52















        up vote
        12
        down vote













        The surface of the Earth is not, rigorously speaking, an inertial frame of reference. Objects at rest relative to Earth's surface are actually subject to a series of inertial effects, like the ficticious forces (Coriolis, centrifugal etc.) because of Earth's rotation, precession and other kinds of acceleration.



        When solving physics problems, however, we usually take the Earth frame as being inertial. This is because the inertial effects are minuscule for most of our day-to-day experiences and experiments. For example, objects in the Equator are the ones subject to the strongest centrifugal force and it is only about $3 times10^{-3}$ or $0.3%$ of their weight.



        So for the most part, if an experiment is short enough and happens in a small enough region, the surface of Earth can indeed be approximated to an inertial frame of reference since the effects on the experiment's results are very, very tiny.



        This of course has exceptions, as cited in njspeer's answer.



        If however by "Earth" you mean the reference frame in Earth's center, it is an inertial frame according to General Relativity (GR), since observers in free fall are inertial in GR. The Earth actually does have some proper acceleration due to external forces such as radiation pressure, but these are also minuscule effects.






        share|cite|improve this answer



















        • 2




          The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
          – David
          Dec 2 at 8:23












        • I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
          – João Vítor G. Lima
          Dec 2 at 12:52













        up vote
        12
        down vote










        up vote
        12
        down vote









        The surface of the Earth is not, rigorously speaking, an inertial frame of reference. Objects at rest relative to Earth's surface are actually subject to a series of inertial effects, like the ficticious forces (Coriolis, centrifugal etc.) because of Earth's rotation, precession and other kinds of acceleration.



        When solving physics problems, however, we usually take the Earth frame as being inertial. This is because the inertial effects are minuscule for most of our day-to-day experiences and experiments. For example, objects in the Equator are the ones subject to the strongest centrifugal force and it is only about $3 times10^{-3}$ or $0.3%$ of their weight.



        So for the most part, if an experiment is short enough and happens in a small enough region, the surface of Earth can indeed be approximated to an inertial frame of reference since the effects on the experiment's results are very, very tiny.



        This of course has exceptions, as cited in njspeer's answer.



        If however by "Earth" you mean the reference frame in Earth's center, it is an inertial frame according to General Relativity (GR), since observers in free fall are inertial in GR. The Earth actually does have some proper acceleration due to external forces such as radiation pressure, but these are also minuscule effects.






        share|cite|improve this answer














        The surface of the Earth is not, rigorously speaking, an inertial frame of reference. Objects at rest relative to Earth's surface are actually subject to a series of inertial effects, like the ficticious forces (Coriolis, centrifugal etc.) because of Earth's rotation, precession and other kinds of acceleration.



        When solving physics problems, however, we usually take the Earth frame as being inertial. This is because the inertial effects are minuscule for most of our day-to-day experiences and experiments. For example, objects in the Equator are the ones subject to the strongest centrifugal force and it is only about $3 times10^{-3}$ or $0.3%$ of their weight.



        So for the most part, if an experiment is short enough and happens in a small enough region, the surface of Earth can indeed be approximated to an inertial frame of reference since the effects on the experiment's results are very, very tiny.



        This of course has exceptions, as cited in njspeer's answer.



        If however by "Earth" you mean the reference frame in Earth's center, it is an inertial frame according to General Relativity (GR), since observers in free fall are inertial in GR. The Earth actually does have some proper acceleration due to external forces such as radiation pressure, but these are also minuscule effects.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 2 at 12:54

























        answered Dec 2 at 0:10









        João Vítor G. Lima

        853219




        853219








        • 2




          The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
          – David
          Dec 2 at 8:23












        • I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
          – João Vítor G. Lima
          Dec 2 at 12:52














        • 2




          The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
          – David
          Dec 2 at 8:23












        • I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
          – João Vítor G. Lima
          Dec 2 at 12:52








        2




        2




        The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
        – David
        Dec 2 at 8:23






        The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
        – David
        Dec 2 at 8:23














        I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
        – João Vítor G. Lima
        Dec 2 at 12:52




        I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
        – João Vítor G. Lima
        Dec 2 at 12:52










        up vote
        9
        down vote













        Because the earth is rotating, it is never strictly an inertial reference frame. However, because the effects are small in many situations, it can often be approximated as one. When to use Coriolis forces will have to be determined on a case-by-case basis. E.g. ballistic problems that cover large distances will most certainly require Coriolis-force corrections, and pendulums that swings for a long time would also require Coriolis-force corrections. For a block sliding down an inclined plane, or a spring on a mass, or a vibrating string, you should not need to take it into consideration.






        share|cite|improve this answer



























          up vote
          9
          down vote













          Because the earth is rotating, it is never strictly an inertial reference frame. However, because the effects are small in many situations, it can often be approximated as one. When to use Coriolis forces will have to be determined on a case-by-case basis. E.g. ballistic problems that cover large distances will most certainly require Coriolis-force corrections, and pendulums that swings for a long time would also require Coriolis-force corrections. For a block sliding down an inclined plane, or a spring on a mass, or a vibrating string, you should not need to take it into consideration.






          share|cite|improve this answer

























            up vote
            9
            down vote










            up vote
            9
            down vote









            Because the earth is rotating, it is never strictly an inertial reference frame. However, because the effects are small in many situations, it can often be approximated as one. When to use Coriolis forces will have to be determined on a case-by-case basis. E.g. ballistic problems that cover large distances will most certainly require Coriolis-force corrections, and pendulums that swings for a long time would also require Coriolis-force corrections. For a block sliding down an inclined plane, or a spring on a mass, or a vibrating string, you should not need to take it into consideration.






            share|cite|improve this answer














            Because the earth is rotating, it is never strictly an inertial reference frame. However, because the effects are small in many situations, it can often be approximated as one. When to use Coriolis forces will have to be determined on a case-by-case basis. E.g. ballistic problems that cover large distances will most certainly require Coriolis-force corrections, and pendulums that swings for a long time would also require Coriolis-force corrections. For a block sliding down an inclined plane, or a spring on a mass, or a vibrating string, you should not need to take it into consideration.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 2 at 3:20

























            answered Dec 2 at 0:01









            njspeer

            4374




            4374






















                up vote
                3
                down vote













                Mach would say that non-inertial effects are due to the relative motion between the earth and the rest of the universe.



                See:
                • Assis, André K. T. Relational Mechanics and Implementation of Mach’s Principle with Weber’s Gravitational Force. Montréal: Apeiron, 2014.






                share|cite|improve this answer



























                  up vote
                  3
                  down vote













                  Mach would say that non-inertial effects are due to the relative motion between the earth and the rest of the universe.



                  See:
                  • Assis, André K. T. Relational Mechanics and Implementation of Mach’s Principle with Weber’s Gravitational Force. Montréal: Apeiron, 2014.






                  share|cite|improve this answer

























                    up vote
                    3
                    down vote










                    up vote
                    3
                    down vote









                    Mach would say that non-inertial effects are due to the relative motion between the earth and the rest of the universe.



                    See:
                    • Assis, André K. T. Relational Mechanics and Implementation of Mach’s Principle with Weber’s Gravitational Force. Montréal: Apeiron, 2014.






                    share|cite|improve this answer














                    Mach would say that non-inertial effects are due to the relative motion between the earth and the rest of the universe.



                    See:
                    • Assis, André K. T. Relational Mechanics and Implementation of Mach’s Principle with Weber’s Gravitational Force. Montréal: Apeiron, 2014.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 4 at 16:46

























                    answered Dec 2 at 4:35









                    Geremia

                    1,1802927




                    1,1802927






















                        up vote
                        0
                        down vote













                        I know it's being a little pedantic, but I would say "earth" is a thing, not a reference frame. You could define an inertial reference frame that contains the earth.



                        But suppose you mean a reference frame defined with reference to the earth: Z is perpendicular to the ground where you are standing, X and Y are parallel to the ground and perpendicular to each other.



                        If you stand at rest in this reference frame, you may notice you feel acceleration (gravity). It is therefore not an inertial reference frame.






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          I know it's being a little pedantic, but I would say "earth" is a thing, not a reference frame. You could define an inertial reference frame that contains the earth.



                          But suppose you mean a reference frame defined with reference to the earth: Z is perpendicular to the ground where you are standing, X and Y are parallel to the ground and perpendicular to each other.



                          If you stand at rest in this reference frame, you may notice you feel acceleration (gravity). It is therefore not an inertial reference frame.






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            I know it's being a little pedantic, but I would say "earth" is a thing, not a reference frame. You could define an inertial reference frame that contains the earth.



                            But suppose you mean a reference frame defined with reference to the earth: Z is perpendicular to the ground where you are standing, X and Y are parallel to the ground and perpendicular to each other.



                            If you stand at rest in this reference frame, you may notice you feel acceleration (gravity). It is therefore not an inertial reference frame.






                            share|cite|improve this answer












                            I know it's being a little pedantic, but I would say "earth" is a thing, not a reference frame. You could define an inertial reference frame that contains the earth.



                            But suppose you mean a reference frame defined with reference to the earth: Z is perpendicular to the ground where you are standing, X and Y are parallel to the ground and perpendicular to each other.



                            If you stand at rest in this reference frame, you may notice you feel acceleration (gravity). It is therefore not an inertial reference frame.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 2 at 15:51









                            Owen

                            27729




                            27729






























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