Rotating in orbit?
$begingroup$
Reading this question (Is the cupola, on the inside of the ISS, cold or warm to the touch?) prompted me to wonder about an object in orbit's orientation as it orbits its host object. For example, consider a hollow tube that is in orbit around the earth and it is perfectly perpendicular to the earth. Spaceman Spiff is immediately beneath the cylinder, and is able to look through the cylinder and observe a specific star.
As he and the tube orbit around the earth (assuming no external influences other than the earth's gravity), would the tube remain pointed at the star (hence the tube changing from perpendicular orientation in relation to the earth to parallel as it gets to 1/4 around its orbit) and Spiff would continue to see the same star in his view? Or, would the cylinder remain perpendicular to the earth, and Spiff would see a continually changing field of stars during his orbit, until he gets back to his starting position, when the star would come back into view as he completes his orbit?
Most spacecraft, I believe, are designed and controlled very carefaully to maintain a specific orientation with the earth, but I don't know what would happen if there was no spacecraft control mechanism or external forces acting on it.
orbital-mechanics
asked 8 hours ago
MilwrdfanMilwrdfan
653210
$endgroup$
add a comment |
$begingroup$
Reading this question (Is the cupola, on the inside of the ISS, cold or warm to the touch?) prompted me to wonder about an object in orbit's orientation as it orbits its host object. For example, consider a hollow tube that is in orbit around the earth and it is perfectly perpendicular to the earth. Spaceman Spiff is immediately beneath the cylinder, and is able to look through the cylinder and observe a specific star.
As he and the tube orbit around the earth (assuming no external influences other than the earth's gravity), would the tube remain pointed at the star (hence the tube changing from perpendicular orientation in relation to the earth to parallel as it gets to 1/4 around its orbit) and Spiff would continue to see the same star in his view? Or, would the cylinder remain perpendicular to the earth, and Spiff would see a continually changing field of stars during his orbit, until he gets back to his starting position, when the star would come back into view as he completes his orbit?
Most spacecraft, I believe, are designed and controlled very carefaully to maintain a specific orientation with the earth, but I don't know what would happen if there was no spacecraft control mechanism or external forces acting on it.
orbital-mechanics
asked 8 hours ago
MilwrdfanMilwrdfan
653210
$endgroup$
$begingroup$
I don't think this is necessarily a duplicate but it has good information about what happens to long thin objects (like your tube) in orbit. space.stackexchange.com/questions/17816/…
$endgroup$
– Organic Marble
8 hours ago
$begingroup$
If Spiff were rotating once about his axis for every orbit around the Earth, in the same direction as his orbit then the Earth would appear very nearly stationary below him (depending on eccentricity and uniformity of gravitational field) and the star would appear in the tube once per obit. If, on the other hand Spiff had no rotation, then the star would remain visible in the tube (except when obscured by the Earth. Any other states of angular momentum would have different results.
$endgroup$
– JCRM
7 hours ago
$begingroup$
(assuming no external influences other than the earth's gravity) Can we theoretically balance a perfectly symmetrical pencil on its one-atom tip? - The uncertainty principle says, no. But if you spun it...
$endgroup$
– Mazura
27 mins ago
add a comment |
$begingroup$
Reading this question (Is the cupola, on the inside of the ISS, cold or warm to the touch?) prompted me to wonder about an object in orbit's orientation as it orbits its host object. For example, consider a hollow tube that is in orbit around the earth and it is perfectly perpendicular to the earth. Spaceman Spiff is immediately beneath the cylinder, and is able to look through the cylinder and observe a specific star.
As he and the tube orbit around the earth (assuming no external influences other than the earth's gravity), would the tube remain pointed at the star (hence the tube changing from perpendicular orientation in relation to the earth to parallel as it gets to 1/4 around its orbit) and Spiff would continue to see the same star in his view? Or, would the cylinder remain perpendicular to the earth, and Spiff would see a continually changing field of stars during his orbit, until he gets back to his starting position, when the star would come back into view as he completes his orbit?
Most spacecraft, I believe, are designed and controlled very carefaully to maintain a specific orientation with the earth, but I don't know what would happen if there was no spacecraft control mechanism or external forces acting on it.
orbital-mechanics
asked 8 hours ago
MilwrdfanMilwrdfan
653210
$endgroup$
Reading this question (Is the cupola, on the inside of the ISS, cold or warm to the touch?) prompted me to wonder about an object in orbit's orientation as it orbits its host object. For example, consider a hollow tube that is in orbit around the earth and it is perfectly perpendicular to the earth. Spaceman Spiff is immediately beneath the cylinder, and is able to look through the cylinder and observe a specific star.
As he and the tube orbit around the earth (assuming no external influences other than the earth's gravity), would the tube remain pointed at the star (hence the tube changing from perpendicular orientation in relation to the earth to parallel as it gets to 1/4 around its orbit) and Spiff would continue to see the same star in his view? Or, would the cylinder remain perpendicular to the earth, and Spiff would see a continually changing field of stars during his orbit, until he gets back to his starting position, when the star would come back into view as he completes his orbit?
Most spacecraft, I believe, are designed and controlled very carefaully to maintain a specific orientation with the earth, but I don't know what would happen if there was no spacecraft control mechanism or external forces acting on it.
orbital-mechanics
orbital-mechanics
asked 8 hours ago
MilwrdfanMilwrdfan
653210
asked 8 hours ago
MilwrdfanMilwrdfan
653210
asked 8 hours ago
MilwrdfanMilwrdfan
653210
asked 8 hours ago
MilwrdfanMilwrdfan
653210
asked 8 hours ago
MilwrdfanMilwrdfan
653210
653210
$begingroup$
I don't think this is necessarily a duplicate but it has good information about what happens to long thin objects (like your tube) in orbit. space.stackexchange.com/questions/17816/…
$endgroup$
– Organic Marble
8 hours ago
$begingroup$
If Spiff were rotating once about his axis for every orbit around the Earth, in the same direction as his orbit then the Earth would appear very nearly stationary below him (depending on eccentricity and uniformity of gravitational field) and the star would appear in the tube once per obit. If, on the other hand Spiff had no rotation, then the star would remain visible in the tube (except when obscured by the Earth. Any other states of angular momentum would have different results.
$endgroup$
– JCRM
7 hours ago
$begingroup$
(assuming no external influences other than the earth's gravity) Can we theoretically balance a perfectly symmetrical pencil on its one-atom tip? - The uncertainty principle says, no. But if you spun it...
$endgroup$
– Mazura
27 mins ago
add a comment |
$begingroup$
I don't think this is necessarily a duplicate but it has good information about what happens to long thin objects (like your tube) in orbit. space.stackexchange.com/questions/17816/…
$endgroup$
– Organic Marble
8 hours ago
$begingroup$
If Spiff were rotating once about his axis for every orbit around the Earth, in the same direction as his orbit then the Earth would appear very nearly stationary below him (depending on eccentricity and uniformity of gravitational field) and the star would appear in the tube once per obit. If, on the other hand Spiff had no rotation, then the star would remain visible in the tube (except when obscured by the Earth. Any other states of angular momentum would have different results.
$endgroup$
– JCRM
7 hours ago
$begingroup$
(assuming no external influences other than the earth's gravity) Can we theoretically balance a perfectly symmetrical pencil on its one-atom tip? - The uncertainty principle says, no. But if you spun it...
$endgroup$
– Mazura
27 mins ago
$begingroup$
I don't think this is necessarily a duplicate but it has good information about what happens to long thin objects (like your tube) in orbit. space.stackexchange.com/questions/17816/…
$endgroup$
– Organic Marble
8 hours ago
$begingroup$
I don't think this is necessarily a duplicate but it has good information about what happens to long thin objects (like your tube) in orbit. space.stackexchange.com/questions/17816/…
$endgroup$
– Organic Marble
8 hours ago
$begingroup$
If Spiff were rotating once about his axis for every orbit around the Earth, in the same direction as his orbit then the Earth would appear very nearly stationary below him (depending on eccentricity and uniformity of gravitational field) and the star would appear in the tube once per obit. If, on the other hand Spiff had no rotation, then the star would remain visible in the tube (except when obscured by the Earth. Any other states of angular momentum would have different results.
$endgroup$
– JCRM
7 hours ago
$begingroup$
If Spiff were rotating once about his axis for every orbit around the Earth, in the same direction as his orbit then the Earth would appear very nearly stationary below him (depending on eccentricity and uniformity of gravitational field) and the star would appear in the tube once per obit. If, on the other hand Spiff had no rotation, then the star would remain visible in the tube (except when obscured by the Earth. Any other states of angular momentum would have different results.
$endgroup$
– JCRM
7 hours ago
$begingroup$
(assuming no external influences other than the earth's gravity) Can we theoretically balance a perfectly symmetrical pencil on its one-atom tip? - The uncertainty principle says, no. But if you spun it...
$endgroup$
– Mazura
27 mins ago
$begingroup$
(assuming no external influences other than the earth's gravity) Can we theoretically balance a perfectly symmetrical pencil on its one-atom tip? - The uncertainty principle says, no. But if you spun it...
$endgroup$
– Mazura
27 mins ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It depends.
If the cylinder's rotation was carefully stopped, then nothing else acted on it, it would hold its orientation and Spiff would see the same star through a complete orbit, while the tube went from Earth-surface-perpendicular to parallel and back.
If the cylinder's rotation was carefully matched to its orbit, so that it turned through 360 degrees over the course of a single revolution around the Earth, then nothing else acted on it, it would remain Earth-surface-perpendicular throughout.
As @Organic Marble's linked QA describes, long skinny objects in orbit are subject to tidal gradient forces, so if the tube is long enough, it will slowly get pulled into an always-perpendicular-to-Earth rotating orientation.
In low Earth orbit, there's also a very very small amount of atmosphere that the tube would be moving through; this would tend to have the opposite effect, making it tumble, or (if one end of it is denser/heavier than the other) making it tend to stabilize in an earth-parallel orientation.
The relative strength of the tidal and atmospheric effects will vary with the size and mass of the tube and orbital altitude, and I'm too lazy to work through a test case to see which one dominates. :)
answered 7 hours ago
Russell BorogoveRussell Borogove
83.9k2281362
$endgroup$
$begingroup$
Won't Spiff orbit faster than the tube? Also, "it would hold its orientation..." should be qualified "begin to rotate immediately due to tidal effects" Plug in some numbers for the long tube to see how quickly it would drift away from the star's position.
$endgroup$
– uhoh
38 mins 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: "508"
};
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%2fspace.stackexchange.com%2fquestions%2f33586%2frotating-in-orbit%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It depends.
If the cylinder's rotation was carefully stopped, then nothing else acted on it, it would hold its orientation and Spiff would see the same star through a complete orbit, while the tube went from Earth-surface-perpendicular to parallel and back.
If the cylinder's rotation was carefully matched to its orbit, so that it turned through 360 degrees over the course of a single revolution around the Earth, then nothing else acted on it, it would remain Earth-surface-perpendicular throughout.
As @Organic Marble's linked QA describes, long skinny objects in orbit are subject to tidal gradient forces, so if the tube is long enough, it will slowly get pulled into an always-perpendicular-to-Earth rotating orientation.
In low Earth orbit, there's also a very very small amount of atmosphere that the tube would be moving through; this would tend to have the opposite effect, making it tumble, or (if one end of it is denser/heavier than the other) making it tend to stabilize in an earth-parallel orientation.
The relative strength of the tidal and atmospheric effects will vary with the size and mass of the tube and orbital altitude, and I'm too lazy to work through a test case to see which one dominates. :)
answered 7 hours ago
Russell BorogoveRussell Borogove
83.9k2281362
$endgroup$
$begingroup$
Won't Spiff orbit faster than the tube? Also, "it would hold its orientation..." should be qualified "begin to rotate immediately due to tidal effects" Plug in some numbers for the long tube to see how quickly it would drift away from the star's position.
$endgroup$
– uhoh
38 mins ago
add a comment |
$begingroup$
It depends.
If the cylinder's rotation was carefully stopped, then nothing else acted on it, it would hold its orientation and Spiff would see the same star through a complete orbit, while the tube went from Earth-surface-perpendicular to parallel and back.
If the cylinder's rotation was carefully matched to its orbit, so that it turned through 360 degrees over the course of a single revolution around the Earth, then nothing else acted on it, it would remain Earth-surface-perpendicular throughout.
As @Organic Marble's linked QA describes, long skinny objects in orbit are subject to tidal gradient forces, so if the tube is long enough, it will slowly get pulled into an always-perpendicular-to-Earth rotating orientation.
In low Earth orbit, there's also a very very small amount of atmosphere that the tube would be moving through; this would tend to have the opposite effect, making it tumble, or (if one end of it is denser/heavier than the other) making it tend to stabilize in an earth-parallel orientation.
The relative strength of the tidal and atmospheric effects will vary with the size and mass of the tube and orbital altitude, and I'm too lazy to work through a test case to see which one dominates. :)
answered 7 hours ago
Russell BorogoveRussell Borogove
83.9k2281362
$endgroup$
$begingroup$
Won't Spiff orbit faster than the tube? Also, "it would hold its orientation..." should be qualified "begin to rotate immediately due to tidal effects" Plug in some numbers for the long tube to see how quickly it would drift away from the star's position.
$endgroup$
– uhoh
38 mins ago
add a comment |
$begingroup$
It depends.
If the cylinder's rotation was carefully stopped, then nothing else acted on it, it would hold its orientation and Spiff would see the same star through a complete orbit, while the tube went from Earth-surface-perpendicular to parallel and back.
If the cylinder's rotation was carefully matched to its orbit, so that it turned through 360 degrees over the course of a single revolution around the Earth, then nothing else acted on it, it would remain Earth-surface-perpendicular throughout.
As @Organic Marble's linked QA describes, long skinny objects in orbit are subject to tidal gradient forces, so if the tube is long enough, it will slowly get pulled into an always-perpendicular-to-Earth rotating orientation.
In low Earth orbit, there's also a very very small amount of atmosphere that the tube would be moving through; this would tend to have the opposite effect, making it tumble, or (if one end of it is denser/heavier than the other) making it tend to stabilize in an earth-parallel orientation.
The relative strength of the tidal and atmospheric effects will vary with the size and mass of the tube and orbital altitude, and I'm too lazy to work through a test case to see which one dominates. :)
answered 7 hours ago
Russell BorogoveRussell Borogove
83.9k2281362
$endgroup$
It depends.
If the cylinder's rotation was carefully stopped, then nothing else acted on it, it would hold its orientation and Spiff would see the same star through a complete orbit, while the tube went from Earth-surface-perpendicular to parallel and back.
If the cylinder's rotation was carefully matched to its orbit, so that it turned through 360 degrees over the course of a single revolution around the Earth, then nothing else acted on it, it would remain Earth-surface-perpendicular throughout.
As @Organic Marble's linked QA describes, long skinny objects in orbit are subject to tidal gradient forces, so if the tube is long enough, it will slowly get pulled into an always-perpendicular-to-Earth rotating orientation.
In low Earth orbit, there's also a very very small amount of atmosphere that the tube would be moving through; this would tend to have the opposite effect, making it tumble, or (if one end of it is denser/heavier than the other) making it tend to stabilize in an earth-parallel orientation.
The relative strength of the tidal and atmospheric effects will vary with the size and mass of the tube and orbital altitude, and I'm too lazy to work through a test case to see which one dominates. :)
answered 7 hours ago
Russell BorogoveRussell Borogove
83.9k2281362
answered 7 hours ago
Russell BorogoveRussell Borogove
83.9k2281362
answered 7 hours ago
Russell BorogoveRussell Borogove
83.9k2281362
answered 7 hours ago
Russell BorogoveRussell Borogove
83.9k2281362
83.9k2281362
$begingroup$
Won't Spiff orbit faster than the tube? Also, "it would hold its orientation..." should be qualified "begin to rotate immediately due to tidal effects" Plug in some numbers for the long tube to see how quickly it would drift away from the star's position.
$endgroup$
– uhoh
38 mins ago
add a comment |
$begingroup$
Won't Spiff orbit faster than the tube? Also, "it would hold its orientation..." should be qualified "begin to rotate immediately due to tidal effects" Plug in some numbers for the long tube to see how quickly it would drift away from the star's position.
$endgroup$
– uhoh
38 mins ago
$begingroup$
Won't Spiff orbit faster than the tube? Also, "it would hold its orientation..." should be qualified "begin to rotate immediately due to tidal effects" Plug in some numbers for the long tube to see how quickly it would drift away from the star's position.
$endgroup$
– uhoh
38 mins ago
$begingroup$
Won't Spiff orbit faster than the tube? Also, "it would hold its orientation..." should be qualified "begin to rotate immediately due to tidal effects" Plug in some numbers for the long tube to see how quickly it would drift away from the star's position.
$endgroup$
– uhoh
38 mins ago
add a comment |
Thanks for contributing an answer to Space Exploration 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%2fspace.stackexchange.com%2fquestions%2f33586%2frotating-in-orbit%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
$begingroup$
I don't think this is necessarily a duplicate but it has good information about what happens to long thin objects (like your tube) in orbit. space.stackexchange.com/questions/17816/…
$endgroup$
– Organic Marble
8 hours ago
$begingroup$
If Spiff were rotating once about his axis for every orbit around the Earth, in the same direction as his orbit then the Earth would appear very nearly stationary below him (depending on eccentricity and uniformity of gravitational field) and the star would appear in the tube once per obit. If, on the other hand Spiff had no rotation, then the star would remain visible in the tube (except when obscured by the Earth. Any other states of angular momentum would have different results.
$endgroup$
– JCRM
7 hours ago
$begingroup$
(assuming no external influences other than the earth's gravity) Can we theoretically balance a perfectly symmetrical pencil on its one-atom tip? - The uncertainty principle says, no. But if you spun it...
$endgroup$
– Mazura
27 mins ago