free fall ellipse or parabola?












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Herbert Spencer somewhere says that the parabola of a ballistic object is actually a portion of an ellipse that is indistinguishable from a parabola--is that true? It would seem plausible since satellite orbits are ellipses and artillery trajectories are interrupted orbits.






newtonian-mechanics gravity orbital-motion projectile free-fall







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    4












    $begingroup$


    Herbert Spencer somewhere says that the parabola of a ballistic object is actually a portion of an ellipse that is indistinguishable from a parabola--is that true? It would seem plausible since satellite orbits are ellipses and artillery trajectories are interrupted orbits.






    newtonian-mechanics gravity orbital-motion projectile free-fall







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$


      Herbert Spencer somewhere says that the parabola of a ballistic object is actually a portion of an ellipse that is indistinguishable from a parabola--is that true? It would seem plausible since satellite orbits are ellipses and artillery trajectories are interrupted orbits.






      newtonian-mechanics gravity orbital-motion projectile free-fall







      share|cite|improve this question











      $endgroup$




      Herbert Spencer somewhere says that the parabola of a ballistic object is actually a portion of an ellipse that is indistinguishable from a parabola--is that true? It would seem plausible since satellite orbits are ellipses and artillery trajectories are interrupted orbits.






      newtonian-mechanics gravity orbital-motion projectile free-fall




      newtonian-mechanics gravity orbital-motion projectile free-fall






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      edited 2 hours ago









      Aaron Stevens

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      asked 2 hours ago









      user56930user56930

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          1 Answer
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          The difference between the two cases is the direction of the gravity vector. If gravity is pulling towards a point (as we see in orbital mechanics), ballistic objects follow an elliptical (or sometimes hyperbolic) path. If, however, gravity points in a constant direction (as we often assume in terrestrial physics problems: it pulls "down"), we get a parabolic trajectory.



          On the timescales of these trajectories that we call parabolic, the difference in direction of gravity from start to end of the flight is so tremendously minimal, that we can treat it as a perturbation from the "down" vector and then ignore it entirely. This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect.



          At orbital velocities, the effect is so non-trivial that we don't even try to model it as a "down" vector plus a perturbation. We just model the vector pointing towards the center of the gravitational body.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
            $endgroup$
            – NLambert
            25 mins ago












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          $begingroup$

          The difference between the two cases is the direction of the gravity vector. If gravity is pulling towards a point (as we see in orbital mechanics), ballistic objects follow an elliptical (or sometimes hyperbolic) path. If, however, gravity points in a constant direction (as we often assume in terrestrial physics problems: it pulls "down"), we get a parabolic trajectory.



          On the timescales of these trajectories that we call parabolic, the difference in direction of gravity from start to end of the flight is so tremendously minimal, that we can treat it as a perturbation from the "down" vector and then ignore it entirely. This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect.



          At orbital velocities, the effect is so non-trivial that we don't even try to model it as a "down" vector plus a perturbation. We just model the vector pointing towards the center of the gravitational body.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
            $endgroup$
            – NLambert
            25 mins ago
















          8












          $begingroup$

          The difference between the two cases is the direction of the gravity vector. If gravity is pulling towards a point (as we see in orbital mechanics), ballistic objects follow an elliptical (or sometimes hyperbolic) path. If, however, gravity points in a constant direction (as we often assume in terrestrial physics problems: it pulls "down"), we get a parabolic trajectory.



          On the timescales of these trajectories that we call parabolic, the difference in direction of gravity from start to end of the flight is so tremendously minimal, that we can treat it as a perturbation from the "down" vector and then ignore it entirely. This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect.



          At orbital velocities, the effect is so non-trivial that we don't even try to model it as a "down" vector plus a perturbation. We just model the vector pointing towards the center of the gravitational body.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
            $endgroup$
            – NLambert
            25 mins ago














          8












          8








          8





          $begingroup$

          The difference between the two cases is the direction of the gravity vector. If gravity is pulling towards a point (as we see in orbital mechanics), ballistic objects follow an elliptical (or sometimes hyperbolic) path. If, however, gravity points in a constant direction (as we often assume in terrestrial physics problems: it pulls "down"), we get a parabolic trajectory.



          On the timescales of these trajectories that we call parabolic, the difference in direction of gravity from start to end of the flight is so tremendously minimal, that we can treat it as a perturbation from the "down" vector and then ignore it entirely. This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect.



          At orbital velocities, the effect is so non-trivial that we don't even try to model it as a "down" vector plus a perturbation. We just model the vector pointing towards the center of the gravitational body.






          share|cite|improve this answer









          $endgroup$



          The difference between the two cases is the direction of the gravity vector. If gravity is pulling towards a point (as we see in orbital mechanics), ballistic objects follow an elliptical (or sometimes hyperbolic) path. If, however, gravity points in a constant direction (as we often assume in terrestrial physics problems: it pulls "down"), we get a parabolic trajectory.



          On the timescales of these trajectories that we call parabolic, the difference in direction of gravity from start to end of the flight is so tremendously minimal, that we can treat it as a perturbation from the "down" vector and then ignore it entirely. This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect.



          At orbital velocities, the effect is so non-trivial that we don't even try to model it as a "down" vector plus a perturbation. We just model the vector pointing towards the center of the gravitational body.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          Cort AmmonCort Ammon

          24.1k34779




          24.1k34779












          • $begingroup$
            For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
            $endgroup$
            – NLambert
            25 mins ago


















          • $begingroup$
            For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
            $endgroup$
            – NLambert
            25 mins ago
















          $begingroup$
          For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
          $endgroup$
          – NLambert
          25 mins ago




          $begingroup$
          For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
          $endgroup$
          – NLambert
          25 mins ago


















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