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free fall ellipse or parabola?

1

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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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edited 18 mins ago

Aaron Stevens

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asked 38 mins ago

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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

edited 18 mins ago

Aaron Stevens

13.7k42250

asked 38 mins ago

user56930user56930

174

$endgroup$

add a comment | 

$begingroup$

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

edited 18 mins ago

Aaron Stevens

13.7k42250

asked 38 mins ago

user56930user56930

174

$endgroup$

share|cite|improve this question

edited 18 mins ago

Aaron Stevens

13.7k42250

asked 38 mins ago

user56930user56930

174

share|cite|improve this question

edited 18 mins ago

Aaron Stevens

13.7k42250

asked 38 mins ago

user56930user56930

174

edited 18 mins ago

Aaron Stevens

13.7k42250

edited 18 mins ago

Aaron Stevens

13.7k42250

asked 38 mins ago

user56930user56930

174

asked 38 mins ago

user56930user56930

174

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

answered 31 mins ago

Cort AmmonCort Ammon

24k34779

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3

$begingroup$

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

answered 31 mins ago

Cort AmmonCort Ammon

24k34779

$endgroup$

add a comment | 

3

$begingroup$

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

answered 31 mins ago

Cort AmmonCort Ammon

24k34779

$endgroup$

add a comment | 

$begingroup$

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

answered 31 mins ago

Cort AmmonCort Ammon

24k34779

$endgroup$

share|cite|improve this answer

answered 31 mins ago

Cort AmmonCort Ammon

24k34779

answered 31 mins ago

Cort AmmonCort Ammon

24k34779

answered 31 mins ago

Cort AmmonCort Ammon

24k34779