Free fall ellipse or parabola?Can a very small portion of an ellipse be a parabola?Prove that orbits are conic sectionsDo the planets really orbit the Sun?What kind of effect would cosmic expansion have on planetary motion or keplerian orbit?Stability of solar systemSimple explanation of why object falls in same time it takes to go up?Meaning of the focus of an elliptical orbitHow did people get ellipses of Newton's equations of motion and gravitation?Calculating theta/anomaly for a hyperbolic orbit based on time stepIs spacetime in an ellipse around a massive object, or does it just slope down towards the massive object?
What exactly is ineptocracy?
how do we prove that a sum of two periods is still a period?
Pact of Blade Warlock with Dancing Blade
Am I breaking OOP practice with this architecture?
Machine learning testing data
Send out email when Apex Queueable fails and test it
Could neural networks be considered metaheuristics?
Is this draw by repetition?
OP Amp not amplifying audio signal
Mathematica command that allows it to read my intentions
Finitely generated matrix groups whose eigenvalues are all algebraic
Convert seconds to minutes
Should I tell management that I intend to leave due to bad software development practices?
How to travel to Japan while expressing milk?
Knowledge-based authentication using Domain-driven Design in C#
Do Iron Man suits sport waste management systems?
How does a dynamic QR code work?
How could sorcerers who are able to produce/manipulate almost all forms of energy communicate over large distances?
How to show a landlord what we have in savings?
Why didn't Boeing produce its own regional jet?
How to compare a string
How dangerous is XSS
Does the Idaho Potato Commission associate potato skins with healthy eating?
Can I hook these wires up to find the connection to a dead outlet?
Free fall ellipse or parabola?
Can a very small portion of an ellipse be a parabola?Prove that orbits are conic sectionsDo the planets really orbit the Sun?What kind of effect would cosmic expansion have on planetary motion or keplerian orbit?Stability of solar systemSimple explanation of why object falls in same time it takes to go up?Meaning of the focus of an elliptical orbitHow did people get ellipses of Newton's equations of motion and gravitation?Calculating theta/anomaly for a hyperbolic orbit based on time stepIs 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 newtonian-gravity orbital-motion projectile free-fall
edited yesterday
Qmechanic♦
107k121991233
asked yesterday
user56930user56930
14226
$endgroup$
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 newtonian-gravity orbital-motion projectile free-fall
edited yesterday
Qmechanic♦
107k121991233
asked yesterday
user56930user56930
14226
$endgroup$
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 newtonian-gravity orbital-motion projectile free-fall
edited yesterday
Qmechanic♦
107k121991233
asked yesterday
user56930user56930
14226
$endgroup$
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 newtonian-gravity orbital-motion projectile free-fall
newtonian-mechanics newtonian-gravity orbital-motion projectile free-fall
edited yesterday
Qmechanic♦
107k121991233
asked yesterday
user56930user56930
14226
edited yesterday
Qmechanic♦
107k121991233
asked yesterday
user56930user56930
14226
edited yesterday
Qmechanic♦
107k121991233
edited yesterday
Qmechanic♦
107k121991233
edited yesterday
Qmechanic♦
107k121991233
107k121991233
asked yesterday
user56930user56930
14226
asked yesterday
user56930user56930
14226
asked yesterday
user56930user56930
14226
14226
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 yesterday
Cort AmmonCort Ammon
24.3k34880
$endgroup$
4
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
yesterday
3
$begingroup$
"This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect" or it works as long as constant gravity field is a good approximation. I think this wording is better, because it's a direct relation, now for whatever reason it changes (speed, timescale, mass, etc.) the simplified model stops working. It also makes clear that the parabola is not a part of the ellipsis, but the approximation as well. (Notabene: Radial gravity model also stops working at certain conditions.)
$endgroup$
– luk32
yesterday
$begingroup$
For certain types of artillery and missiles we need to do elliptical calculations instead of parabolic because the distances covered means we can no longer assume that the Earth is flat. When firing beyond the horizon the error can be quite significant
$endgroup$
– slebetman
17 hours ago
add a comment |
$begingroup$
One easy way to tell the difference between a highly eccentric elliptical orbit and a true parabolic orbit is that an object in a parabolic orbit travels at its escape velocity exactly. In astronomy, such orbits are as rare as circular orbits, i.e. they don't exist. An object well below the escape velocity can be in an elliptical orbit that has an eccentricity very close to 1, making it look much like a parabolic orbit when only part of the curve is examined.
A relatively slow projectile on the surface of the Earth is actually a closed curve ellipse, and if the Earth got out of its way by shrinking to the size of basketball with the same gravitational field, the object would return to its original place in a long cigar shaped elliptical path.
As an aside, if an object is traveling faster than its escape velocity, it is in a hyperbolic orbit.
answered yesterday
Bill WattsBill Watts
38217
$endgroup$
2
$begingroup$
To expand a little more: the path of a Newtonian ballistic projectile under the influence of a single other mass is always a conic section. If the eccentricity is exactly 0, its orbit is a circle, if it is less than 1, the orbit is an ellipse, if it is exactly 1, the orbit is a parabola, and if it is greater than 1, the orbit is a hyperbola. However, the very concept of an "exact value" for a physical measurement is not definable (serious issues under Newton, and strictly not definable in QM). So true circular and parabolic orbits do not exist in reality.
$endgroup$
– Paul Sinclair
yesterday
$begingroup$
Because in the movie Hidden Figures the intriguing issue was going from ellipse math to parabola math. Thank you
$endgroup$
– user56930
yesterday
$begingroup$
What is said in this answer is, sure. But I don't think this is what the paraphrase in the question alluded to.
$endgroup$
– Arthur
10 hours ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "151"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f469780%2ffree-fall-ellipse-or-parabola%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 yesterday
Cort AmmonCort Ammon
24.3k34880
$endgroup$
4
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
yesterday
3
$begingroup$
"This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect" or it works as long as constant gravity field is a good approximation. I think this wording is better, because it's a direct relation, now for whatever reason it changes (speed, timescale, mass, etc.) the simplified model stops working. It also makes clear that the parabola is not a part of the ellipsis, but the approximation as well. (Notabene: Radial gravity model also stops working at certain conditions.)
$endgroup$
– luk32
yesterday
$begingroup$
For certain types of artillery and missiles we need to do elliptical calculations instead of parabolic because the distances covered means we can no longer assume that the Earth is flat. When firing beyond the horizon the error can be quite significant
$endgroup$
– slebetman
17 hours 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 yesterday
Cort AmmonCort Ammon
24.3k34880
$endgroup$
4
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
yesterday
3
$begingroup$
"This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect" or it works as long as constant gravity field is a good approximation. I think this wording is better, because it's a direct relation, now for whatever reason it changes (speed, timescale, mass, etc.) the simplified model stops working. It also makes clear that the parabola is not a part of the ellipsis, but the approximation as well. (Notabene: Radial gravity model also stops working at certain conditions.)
$endgroup$
– luk32
yesterday
$begingroup$
For certain types of artillery and missiles we need to do elliptical calculations instead of parabolic because the distances covered means we can no longer assume that the Earth is flat. When firing beyond the horizon the error can be quite significant
$endgroup$
– slebetman
17 hours 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 yesterday
Cort AmmonCort Ammon
24.3k34880
$endgroup$
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 yesterday
Cort AmmonCort Ammon
24.3k34880
answered yesterday
Cort AmmonCort Ammon
24.3k34880
answered yesterday
Cort AmmonCort Ammon
24.3k34880
answered yesterday
Cort AmmonCort Ammon
24.3k34880
24.3k34880
4
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
yesterday
3
$begingroup$
"This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect" or it works as long as constant gravity field is a good approximation. I think this wording is better, because it's a direct relation, now for whatever reason it changes (speed, timescale, mass, etc.) the simplified model stops working. It also makes clear that the parabola is not a part of the ellipsis, but the approximation as well. (Notabene: Radial gravity model also stops working at certain conditions.)
$endgroup$
– luk32
yesterday
$begingroup$
For certain types of artillery and missiles we need to do elliptical calculations instead of parabolic because the distances covered means we can no longer assume that the Earth is flat. When firing beyond the horizon the error can be quite significant
$endgroup$
– slebetman
17 hours ago
add a comment |
4
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
yesterday
3
$begingroup$
"This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect" or it works as long as constant gravity field is a good approximation. I think this wording is better, because it's a direct relation, now for whatever reason it changes (speed, timescale, mass, etc.) the simplified model stops working. It also makes clear that the parabola is not a part of the ellipsis, but the approximation as well. (Notabene: Radial gravity model also stops working at certain conditions.)
$endgroup$
– luk32
yesterday
$begingroup$
For certain types of artillery and missiles we need to do elliptical calculations instead of parabolic because the distances covered means we can no longer assume that the Earth is flat. When firing beyond the horizon the error can be quite significant
$endgroup$
– slebetman
17 hours ago
4
4
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
yesterday
$begingroup$
For the orbital model, the magnitude of the vector changes (as $1/r^2$) as well as the direction.
$endgroup$
– NLambert
yesterday
3
3
$begingroup$
"This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect" or it works as long as constant gravity field is a good approximation. I think this wording is better, because it's a direct relation, now for whatever reason it changes (speed, timescale, mass, etc.) the simplified model stops working. It also makes clear that the parabola is not a part of the ellipsis, but the approximation as well. (Notabene: Radial gravity model also stops working at certain conditions.)
$endgroup$
– luk32
yesterday
$begingroup$
"This works until the object is flying fast enough that the changing gravity vector starts to have a non-trivial effect" or it works as long as constant gravity field is a good approximation. I think this wording is better, because it's a direct relation, now for whatever reason it changes (speed, timescale, mass, etc.) the simplified model stops working. It also makes clear that the parabola is not a part of the ellipsis, but the approximation as well. (Notabene: Radial gravity model also stops working at certain conditions.)
$endgroup$
– luk32
yesterday
$begingroup$
For certain types of artillery and missiles we need to do elliptical calculations instead of parabolic because the distances covered means we can no longer assume that the Earth is flat. When firing beyond the horizon the error can be quite significant
$endgroup$
– slebetman
17 hours ago
$begingroup$
For certain types of artillery and missiles we need to do elliptical calculations instead of parabolic because the distances covered means we can no longer assume that the Earth is flat. When firing beyond the horizon the error can be quite significant
$endgroup$
– slebetman
17 hours ago
add a comment |
$begingroup$
One easy way to tell the difference between a highly eccentric elliptical orbit and a true parabolic orbit is that an object in a parabolic orbit travels at its escape velocity exactly. In astronomy, such orbits are as rare as circular orbits, i.e. they don't exist. An object well below the escape velocity can be in an elliptical orbit that has an eccentricity very close to 1, making it look much like a parabolic orbit when only part of the curve is examined.
A relatively slow projectile on the surface of the Earth is actually a closed curve ellipse, and if the Earth got out of its way by shrinking to the size of basketball with the same gravitational field, the object would return to its original place in a long cigar shaped elliptical path.
As an aside, if an object is traveling faster than its escape velocity, it is in a hyperbolic orbit.
answered yesterday
Bill WattsBill Watts
38217
$endgroup$
2
$begingroup$
To expand a little more: the path of a Newtonian ballistic projectile under the influence of a single other mass is always a conic section. If the eccentricity is exactly 0, its orbit is a circle, if it is less than 1, the orbit is an ellipse, if it is exactly 1, the orbit is a parabola, and if it is greater than 1, the orbit is a hyperbola. However, the very concept of an "exact value" for a physical measurement is not definable (serious issues under Newton, and strictly not definable in QM). So true circular and parabolic orbits do not exist in reality.
$endgroup$
– Paul Sinclair
yesterday
$begingroup$
Because in the movie Hidden Figures the intriguing issue was going from ellipse math to parabola math. Thank you
$endgroup$
– user56930
yesterday
$begingroup$
What is said in this answer is, sure. But I don't think this is what the paraphrase in the question alluded to.
$endgroup$
– Arthur
10 hours ago
add a comment |
$begingroup$
One easy way to tell the difference between a highly eccentric elliptical orbit and a true parabolic orbit is that an object in a parabolic orbit travels at its escape velocity exactly. In astronomy, such orbits are as rare as circular orbits, i.e. they don't exist. An object well below the escape velocity can be in an elliptical orbit that has an eccentricity very close to 1, making it look much like a parabolic orbit when only part of the curve is examined.
A relatively slow projectile on the surface of the Earth is actually a closed curve ellipse, and if the Earth got out of its way by shrinking to the size of basketball with the same gravitational field, the object would return to its original place in a long cigar shaped elliptical path.
As an aside, if an object is traveling faster than its escape velocity, it is in a hyperbolic orbit.
answered yesterday
Bill WattsBill Watts
38217
$endgroup$
2
$begingroup$
To expand a little more: the path of a Newtonian ballistic projectile under the influence of a single other mass is always a conic section. If the eccentricity is exactly 0, its orbit is a circle, if it is less than 1, the orbit is an ellipse, if it is exactly 1, the orbit is a parabola, and if it is greater than 1, the orbit is a hyperbola. However, the very concept of an "exact value" for a physical measurement is not definable (serious issues under Newton, and strictly not definable in QM). So true circular and parabolic orbits do not exist in reality.
$endgroup$
– Paul Sinclair
yesterday
$begingroup$
Because in the movie Hidden Figures the intriguing issue was going from ellipse math to parabola math. Thank you
$endgroup$
– user56930
yesterday
$begingroup$
What is said in this answer is, sure. But I don't think this is what the paraphrase in the question alluded to.
$endgroup$
– Arthur
10 hours ago
add a comment |
$begingroup$
One easy way to tell the difference between a highly eccentric elliptical orbit and a true parabolic orbit is that an object in a parabolic orbit travels at its escape velocity exactly. In astronomy, such orbits are as rare as circular orbits, i.e. they don't exist. An object well below the escape velocity can be in an elliptical orbit that has an eccentricity very close to 1, making it look much like a parabolic orbit when only part of the curve is examined.
A relatively slow projectile on the surface of the Earth is actually a closed curve ellipse, and if the Earth got out of its way by shrinking to the size of basketball with the same gravitational field, the object would return to its original place in a long cigar shaped elliptical path.
As an aside, if an object is traveling faster than its escape velocity, it is in a hyperbolic orbit.
answered yesterday
Bill WattsBill Watts
38217
$endgroup$
One easy way to tell the difference between a highly eccentric elliptical orbit and a true parabolic orbit is that an object in a parabolic orbit travels at its escape velocity exactly. In astronomy, such orbits are as rare as circular orbits, i.e. they don't exist. An object well below the escape velocity can be in an elliptical orbit that has an eccentricity very close to 1, making it look much like a parabolic orbit when only part of the curve is examined.
A relatively slow projectile on the surface of the Earth is actually a closed curve ellipse, and if the Earth got out of its way by shrinking to the size of basketball with the same gravitational field, the object would return to its original place in a long cigar shaped elliptical path.
As an aside, if an object is traveling faster than its escape velocity, it is in a hyperbolic orbit.
answered yesterday
Bill WattsBill Watts
38217
answered yesterday
Bill WattsBill Watts
38217
answered yesterday
Bill WattsBill Watts
38217
answered yesterday
Bill WattsBill Watts
38217
38217
2
$begingroup$
To expand a little more: the path of a Newtonian ballistic projectile under the influence of a single other mass is always a conic section. If the eccentricity is exactly 0, its orbit is a circle, if it is less than 1, the orbit is an ellipse, if it is exactly 1, the orbit is a parabola, and if it is greater than 1, the orbit is a hyperbola. However, the very concept of an "exact value" for a physical measurement is not definable (serious issues under Newton, and strictly not definable in QM). So true circular and parabolic orbits do not exist in reality.
$endgroup$
– Paul Sinclair
yesterday
$begingroup$
Because in the movie Hidden Figures the intriguing issue was going from ellipse math to parabola math. Thank you
$endgroup$
– user56930
yesterday
$begingroup$
What is said in this answer is, sure. But I don't think this is what the paraphrase in the question alluded to.
$endgroup$
– Arthur
10 hours ago
add a comment |
2
$begingroup$
To expand a little more: the path of a Newtonian ballistic projectile under the influence of a single other mass is always a conic section. If the eccentricity is exactly 0, its orbit is a circle, if it is less than 1, the orbit is an ellipse, if it is exactly 1, the orbit is a parabola, and if it is greater than 1, the orbit is a hyperbola. However, the very concept of an "exact value" for a physical measurement is not definable (serious issues under Newton, and strictly not definable in QM). So true circular and parabolic orbits do not exist in reality.
$endgroup$
– Paul Sinclair
yesterday
$begingroup$
Because in the movie Hidden Figures the intriguing issue was going from ellipse math to parabola math. Thank you
$endgroup$
– user56930
yesterday
$begingroup$
What is said in this answer is, sure. But I don't think this is what the paraphrase in the question alluded to.
$endgroup$
– Arthur
10 hours ago
2
2
$begingroup$
To expand a little more: the path of a Newtonian ballistic projectile under the influence of a single other mass is always a conic section. If the eccentricity is exactly 0, its orbit is a circle, if it is less than 1, the orbit is an ellipse, if it is exactly 1, the orbit is a parabola, and if it is greater than 1, the orbit is a hyperbola. However, the very concept of an "exact value" for a physical measurement is not definable (serious issues under Newton, and strictly not definable in QM). So true circular and parabolic orbits do not exist in reality.
$endgroup$
– Paul Sinclair
yesterday
$begingroup$
To expand a little more: the path of a Newtonian ballistic projectile under the influence of a single other mass is always a conic section. If the eccentricity is exactly 0, its orbit is a circle, if it is less than 1, the orbit is an ellipse, if it is exactly 1, the orbit is a parabola, and if it is greater than 1, the orbit is a hyperbola. However, the very concept of an "exact value" for a physical measurement is not definable (serious issues under Newton, and strictly not definable in QM). So true circular and parabolic orbits do not exist in reality.
$endgroup$
– Paul Sinclair
yesterday
$begingroup$
Because in the movie Hidden Figures the intriguing issue was going from ellipse math to parabola math. Thank you
$endgroup$
– user56930
yesterday
$begingroup$
Because in the movie Hidden Figures the intriguing issue was going from ellipse math to parabola math. Thank you
$endgroup$
– user56930
yesterday
$begingroup$
What is said in this answer is, sure. But I don't think this is what the paraphrase in the question alluded to.
$endgroup$
– Arthur
10 hours ago
$begingroup$
What is said in this answer is, sure. But I don't think this is what the paraphrase in the question alluded to.
$endgroup$
– Arthur
10 hours ago
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f469780%2ffree-fall-ellipse-or-parabola%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
