Has any attempt been made to change the "plane" of the orbit (not altitude) of a satellite, across earth's diameter, continuously in small steps, from pole to pole for a satellite with zero inclination, and from East to west for a satellite with 90 degree inclination?. Will this help us in giving a 3D scan of earth surface?
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Organic Marble
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Niranjan
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As an "add-on", can we revolve the plane about an axis, which in itself can be made to rotate around the earth axis? – Niranjan Sep 05 '20 at 04:12
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5Why would you want to move East/West at a 90° inclination when you can just let the Earth rotate beneath you? – Edlothiad Sep 05 '20 at 13:20
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@Edlothiad, you are right. E-W movement is not required, just having a "stationary" satellite would suffice, but the other way round will help in capturing data which might remain hidden from view in E-W rotation. Thanks for pointing that. But our satellite would need to move from pole to pole, to capture details from those line of sights. – Niranjan Sep 05 '20 at 18:08
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1@Niranjan One satellite could easily cover 90-95& of the surface area, the missing parts are either less interesting or can be covered by alternative means. Keeping a satellite simple is preferred over making it multifunctional. Satellites with bonus features are a pain to design, launch and administer. – Mast Sep 06 '20 at 07:57
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Your comment on this answer has, I think, led me to understand what you are really asking about.
What I am saying is that, presuming the inclination to be ZERO (Plane of orbit parallel to the equatorial plane), the entire plane keeps shifting from pole to pole - parallel to the equatorial plane.
You want to move the orbit like this:
If that is the right interpretation, no, you can't do that.
The center of mass of the Earth must lie in the plane of the orbit.
Since the Earth is very close to spherical, the gravitational attractions from each part of it add up to an attraction towards the center of the Earth. Even if an object started on one of those circles above or below the equatorial plane, Earth's gravity would pull accelerate it in a plane that intersects the spacecraft's position and the center of the Earth. That would be the orbital plane, and all orbital planes around spherical bodies pass through the center of the body.
Quoted from this answer.
Organic Marble
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PERFECT INTERPRETATION. Two questions: 1 Why do we need to have the centroid in the plane of orbit? For gravitational force? Presuming that the new orbital plane is curved around the earth's surface, the satellite will be subjected to same gravitational pull. and Question no. 2 : How to "create a sketch (like you have done) and post it as a part of question / answer. If i would have known this, I would have posted almost same sketch as part of my question, which would have been, then, very simple to understand. Thanks. – Niranjan Sep 05 '20 at 12:29
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- You are right, a picture can really, really help in cases like this. I used the "Impress" app in Libreoffice, a free office suite https://www.libreoffice.org/ It's basically a Powerpoint clone. Sadly I had to use Powerpoint a lot in my work so it was easy for me to gen this up quickly. You add the picture to your question by clicking the little icon of a mountain. I am looking for a good explanation of 1). I think there's one here on the site but I am having trouble finding it.
– Organic Marble Sep 05 '20 at 12:35 -
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- Quoting @uhoh from their answer here: https://space.stackexchange.com/a/39206/6944 Since the Earth is very close to spherical, the gravitational attractions from each part of it add up to an attraction towards the center of the Earth. Even if an object started on one of those circles above or below the equatorial plane, Earth's gravity would pull accelerate it in a plane that intersects the spacecraft's position and the center of the Earth. That would be the orbital plane, and all orbital planes around spherical bodies pass through the center of the body. I will add this to my answer.
– Organic Marble Sep 05 '20 at 12:53 -
1I also suggest you read this article carefully: https://earthobservatory.nasa.gov/features/OrbitsCatalog – Organic Marble Sep 05 '20 at 12:53
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3@Niranjan: And the answer to your first question is "Because that's the way gravity works." – jamesqf Sep 05 '20 at 17:46
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@jamesqf, may be I need to study the "centroid" stuff. What I am assuming that gravitational force on an object, will be the same, so long as we maintain its altitude, regardless wheather its plane of orbit includes the centroid or not. Thats why I wanted to draw a sketch and explain my view point. Anyway, let me try and learn how to make a sketch, hope to come back on this shortly. Thanks. – Niranjan Sep 05 '20 at 18:15
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@Niranjan this may help: http://web.mit.edu/8.01t/www/materials/Presentations/Presentations_f10/Presentation_W14D1.pdf – Organic Marble Sep 05 '20 at 18:21
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1@Niranjan while the magnitude of the gravitational force will be the same as long as you maintain the altitude, the direction will vary. "wheather its plane of orbit includes the centroid or not." is a meaningless utterance because there is no such thing as a plane of orbit that does not include the centroid - in the illustration of this answer, the upper circle is not an orbit (which would happen "passively" given just momentum and gravity) it's a trajectory which can be maintained only with continuous thrust counteracting the part of gravity vector which is towards the equatorial plane. – Peteris Sep 05 '20 at 21:34
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@Niranjan: As simply as possible, I think you may be thinking about gravity as a scalar, when in fact it's a vector. If you think about the old Einsteinian rubber sheet model of gravity, a path around the centroid is flat, other paths go "uphill", and so require energy. – jamesqf Sep 05 '20 at 21:37
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@Petereris, your comment that "the upper circle in the illustration is not an orbit, but a trajectory, which needs to be "maintained" by spending energy does make sense. Can we therefore say that a trajectory in the plane which passes through the center of earth only can be called an "orbit"? If so, what I am looking for, would require the orbital plane to be rotated about the earth axis as well as about an axis which is perpendicular to earth's axis. Again, will it be correct to presume these rotations would be similar to mid course corrections - a minor push, & it will continue to rotate? – Niranjan Sep 07 '20 at 13:14
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Inclination changes are very expensive in terms of delta V/fuel. To do a 90 degree plane change you are thrusting against your current ~7 kilometer a second velocity to bring it to zero and also adding thrust to get a ~7 kilometer per second velocity in the new orbital plane. This is more than it took to get into orbit in the first place, and would require a truly substantial rocket.
As it happens you do not need to do this, a launch into high inclination polar orbit of the right period (height) can observe the whole of earths surface over time, with various missions building models of the earth with only the needed thrust to avoid unwanted inclination changes.
GremlinWranger
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@GremlinWrangler. Dear Gremlin, I think there is some confusion in the interpretation of my question, in first part of your answer. I DO NOT INTEND to CHANGE THE INCLINATION. What I am saying is that, presuming the inclination to be ZERO (Plane of orbit parallel to the equatorial plane), the entire plane keeps shifting from pole to pole - parallel to the equatorial plane. We may ignore altitude corrections required if any. On similar lines, a plane with 90 degree inclination moves from east to west or west to east . Both these motions would give us a more realistic 3D profile of earth surface. – Niranjan Sep 05 '20 at 11:56
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On further thinking, I think I need to clarify, that I understand inclination to be the angle between earth's equatorial plane, and the orbital plane of the satellite. In case "shifting" of the orbital plane - parallel to equatorial plane is also means changing the inclination, then your point about fuel consumption is equally valid and correct. Believing that changing inclination is just like "mid course corrections" (which is just a small push in the required quantity and direction, compensated in turn to reduce overshoots etc.) I thought it does not need much fuel Please comment. – Niranjan Sep 05 '20 at 12:21
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Although a little vexing, the GIF below and the answer it comes from shows that even fixed orbit will eventually pass over all places on a body that rotates underneath it that are at any latitude smaller than the inclination.
That being said you have to choose an altitude that spread them out evenly so that you don't have to wait a very long time.
That being said if you needed to be almost directly overhead of every point for say top-down imaging and it wasn't okay to cover a swath that was hundreds of kilometers wide by looking at some areas from says 10 or 20 degrees away from top-down for some reason, and you were in a big hurry, there are geometrical advantages under those constraints to first doing a low inclination orbit which would cover low latitudes more effectively in a short period of time then switching once to a nearly polar orbit to cover the higher latitudes more quickly.
As @GremlinWrangler's answer points out, this is a very demanding orbit change and would require your satellite to be more of a flying fuel tank than an Earth observation satellite, but if you absolutely had to scan the enter planet, viewing almost exactly top-down at each point, as absolutely fast as possible, then you could build a flying fuel tank and do it!
However you might consider just going to a higher altitude if possible. For example, from a 1000 km high polar orbit each 100 km wide swath would be within +/- 3 degrees of vertical, and each orbit will cross the equator twice. In 15 days you would have seen every location on Earth and viewed it within +/- 3 degrees of top town.
If you don't mind looking at 10 or 20 degrees, then a lower orbit would do what you need even faster!
See also answers to
- Analytical expression for the ground track of the International Space Station
- Trying to understand the two "holes" in this alt/az plot of the ISS
and from this answer to How do I determine the ground-track period of a LEO satellite?
uhoh
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From both the answers I understand that the kind of orbital plane changes I am looking at is a "fuel consuming" process which is avoidable, because for all practicality, and given the resolution of cameras available as of date, surface profile in the requisite details are available, and as such, this kind of maneuvers are not required. I can also see that you have been candid enough while answering my question, which is very much welcome. The last statement though, indicates that you are unnecessarily hurt and being sarcastic. Sorry for that. – Niranjan Sep 05 '20 at 12:09
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2@Niranjan I don't understand your comment. I'm not, and I don't see anything in my post that appears to be anything less than 100% cheerful and positive! If you do, can you help by quoting the specific sentence or phrase that you're referring to? Thanks! – uhoh Sep 05 '20 at 13:26
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Hi uhoh, as I said, the last statement... specifically "do what you need" is the exact phrase what made me think you got frustrated about my question, and hurt. May be because that is a typical reaction of we Indians, when we get frustrated, and give up all hope of changing things!! Anyway, I am happy that you were not hurt, neither sarcastic. Although I am a mechanical engineering degree holder, I am more into application engineering, and hence lack in specific words & knowledge in astrophysics. Hence such questions. But I am happy that my doubts are answered on such forums. Thanks. – Niranjan Sep 05 '20 at 18:00
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@Niranjan Oh I'm not aware of that phrase. For me "...a lower orbit would do what you need even faster!" only means that a lower orbit would provide what is required, in a shorter amount of time. – uhoh Sep 05 '20 at 21:25