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free fall ellipse or parabola?
The Next CEO of Stack OverflowHohmann transfer orbit questionsPractical limits on the size of orbiting objects: could two pebbles orbit each otherCan a very small portion of an ellipse be a parabola?Are retrograde capture orbits “easier” than prograde capture orbits?How does acceleration affect the orbit?Prove that orbits are conic sectionsDo the planets really orbit the Sun?Meaning of the focus of an elliptical orbitHow did people get ellipses of Newton's equations of motion and gravitation?Is spacetime in an ellipse around a massive object, or does it just slope down towards the massive object?
$begingroup$
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 2 hours ago
Aaron Stevens
13.7k42250
asked 2 hours ago
user56930user56930
274
$endgroup$
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 2 hours ago
Aaron Stevens
13.7k42250
asked 2 hours ago
user56930user56930
274
$endgroup$
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 2 hours ago
Aaron Stevens
13.7k42250
asked 2 hours ago
user56930user56930
274
$endgroup$
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 2 hours ago
Aaron Stevens
13.7k42250
asked 2 hours ago
user56930user56930
274
edited 2 hours ago
Aaron Stevens
13.7k42250
asked 2 hours ago
user56930user56930
274
edited 2 hours ago
Aaron Stevens
13.7k42250
edited 2 hours ago
Aaron Stevens
13.7k42250
edited 2 hours ago
Aaron Stevens
13.7k42250
13.7k42250
asked 2 hours ago
user56930user56930
274
asked 2 hours ago
user56930user56930
274
asked 2 hours ago
user56930user56930
274
274
add a comment |
add a comment |
1 Answer
1
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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 2 hours ago
Cort AmmonCort Ammon
24k34779
$endgroup$
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
7 mins ago
add a comment |
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1 Answer
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1 Answer
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oldest
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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 2 hours ago
Cort AmmonCort Ammon
24k34779
$endgroup$
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
7 mins ago
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 2 hours ago
Cort AmmonCort Ammon
24k34779
$endgroup$
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
7 mins ago
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 2 hours 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 2 hours ago
Cort AmmonCort Ammon
24k34779
answered 2 hours ago
Cort AmmonCort Ammon
24k34779
answered 2 hours ago
Cort AmmonCort Ammon
24k34779
answered 2 hours ago
Cort AmmonCort Ammon
24k34779
24k34779
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
7 mins ago
add a comment |
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
7 mins ago
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
7 mins ago
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
7 mins ago
add a comment |
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