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free fall ellipse or parabola?
The Next CEO of Stack OverflowCoriolis force in free fallContainer of liquid in free fallFree fall from spaceMimicking Free FallFree Fall with Air ResistanceFree fall around EarthFree fall of stonesVelocity of a body in free fallFree fall in a centrifugal space ship?The maths of free fall and near free fall
$begingroup$
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
edited 48 mins ago
Aaron Stevens
13.7k42250
asked 1 hour 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
edited 48 mins ago
Aaron Stevens
13.7k42250
asked 1 hour 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
edited 48 mins ago
Aaron Stevens
13.7k42250
asked 1 hour ago
user56930user56930
174
$endgroup$
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
edited 48 mins ago
Aaron Stevens
13.7k42250
asked 1 hour ago
user56930user56930
174
edited 48 mins ago
Aaron Stevens
13.7k42250
asked 1 hour ago
user56930user56930
174
edited 48 mins ago
Aaron Stevens
13.7k42250
edited 48 mins ago
Aaron Stevens
13.7k42250
edited 48 mins ago
Aaron Stevens
13.7k42250
13.7k42250
asked 1 hour ago
user56930user56930
174
asked 1 hour ago
user56930user56930
174
asked 1 hour ago
user56930user56930
174
174
add a comment |
add a comment |
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.
answered 1 hour ago
Cort AmmonCort Ammon
24k34779
$endgroup$
add a comment |
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1 Answer
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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.
answered 1 hour 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.
answered 1 hour 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.
answered 1 hour ago
Cort AmmonCort Ammon
24k34779
$endgroup$
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.
answered 1 hour ago
Cort AmmonCort Ammon
24k34779
answered 1 hour ago
Cort AmmonCort Ammon
24k34779
answered 1 hour ago
Cort AmmonCort Ammon
24k34779
answered 1 hour ago
Cort AmmonCort Ammon
24k34779
24k34779
add a comment |
add a comment |
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