Are high latitude orbits possible?

[Mutley]

Hi all,

I've been playing KSP for a few weeks now, and totally love it. The amount to learn is driving me mad, but I love it at the same time; hits me right in the OCD spot.

I've been learning about orbital mechanics, and trying to get a lot of principles straight in my head, but it is proving to be a challenge.

Here is my question:

If (for some reason), I wanted to put a satellite/probe/space station into a position such that it effectively had a short orbit around the north pole of Kerbin, is that possible? I mean, is that possible in the game, and is that possible in real life?

Clarification - I mean a high latitude orbit that always stays above, therefore parallel with, the equator (as a plane of reference), like a halo over the north pole (as Dave Kerbin put it below).

Must all orbits actually have orbits which intersect the equator? My thinking is that must be the case, but I don't know why, it just seems right - something do with the gravitational force exerted by the planet.

Can any one enlighten me (if you understand what I'm asking!)?

Thanks

Muts.

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"Mutley Aerospace Division - You'd be MAD not to." Edited September 15, 2014 by Mutley
Question answered! Corrected spelling, clarified meaning.

[Aethon]

Welcome Mutley!

Your question is a little confusing but I will say: All orbit must cross the equator at two points. Remember, gravity pulls toward the center of the parent body. I think you're asking about a polar orbit that looks like a yarmulke which is impossible.

[Xavven]

Hi all,

I've been playing KSP for a few weeks now, and totally love it. The amount to learn is driving me mad, but I love it at the same time; hits me right in the OCD spot.

I've been learning about orbital mechanics, and trying to get a lot of principles straight in my head, but it is proving to be a challenge. #

Here is my question:

If (for some reason), I wanted to put a satellite/probe/space station into a position such that it effectively had a short orbit around the north pole of Kerbin, is that possible? I mean, is that possible in the game, and is that possible in real life?

Must all orbits actually have orbits which intersect the equator? My thinking is that must be the case, but I don't know why, it just seems right - something do with the gravitational force exerted by the plant.

Can any one enlighten me (if you understand what I'm asking!)?

Thanks

Muts.

---

"Mutley Aerospace Division - You'd be MAD not to."

No, it's not possible. At least in patched conics system where N-body physics is not calculated, a satellite's orbital plane has to intersect the center of mass of the central body.

Although, I guess more accurately that would be the barycenter, but when you're talking about an artificial satellite with negligible mass compared to the planet or moon it's orbiting, the barycenter is pretty much the center of mass of the planet/moon.

Anyway...

The intuitive reasoning is that gravity is pulling the satellite towards the center of mass of the planet. If the satellite is at a high latitude, the center of mass of the planet is at a lower latitude.

[Dave Kerbin]

The simple answer is no, you can't. The part about orbits intersecting the equator isn't important, any line that divided Kerbin into two equal pieces would fill the same purpose.

To understand remember that an orbit is just falling while constantly missing the ground. The spacecraft is always being pulled down (where down is defined as a line between the ship and the center of Kerbin) but thanks to horizontal velocity the direction of down is constantly changing, translating the previous moments downward velocity into the new horizontal velocity. If you tried to orbit around one pole (like an angels halo over the 'head' of Kerbin) then the downward velocity would not be completely converted into horizontal velocity, so would continue to lose altitude until you hit the ground.

[Dave Kerbin]

At least in patched conics system where N-body physics is not calculated, a satellite's orbital plane has to intersect the center of mass of the central body.

Although, I guess more accurately that would be the barycenter, but when you're talking about an artificial satellite with negligible mass compared to the planet or moon it's orbiting, the barycenter is pretty much the center of mass of the planet/moon.

Yeah, I can see that. If you had some weird potato shaped asteroid (maybe one half is composed of a denser substance) then you might be able to have an orbit that would appear like a halo, though I think it would be a big stretch for all the numbers to fit together and not just look silly.

[capi3101]

Yeah, if you're asking what I think you're asking (say, a circular orbit that stays constantly around 80 degrees latitude), that's physically impossible - though an escape trajectory that coincides with that latitude is possible.

I'll point you to this section of a wikipedia article: http://en.wikipedia.org/wiki/Orbit#Understanding_orbits. Pay attention to the discussion of "Newton's Cannonball".

As for a polar orbit, those are possible, though if you want to launch straight into one you might want to aim a few degrees west of the direction you're going to compensate for the planet's rotation. You can correct any minor error later with a normal/anti-normal burn as usual if you've got the delta-v for it.

[mhoram]

Must all orbits actually have orbits which intersect the equator? My thinking is that must be the case, but I don't know why, it just seems right - something do with the gravitational force exerted by the plant.

Short answer: yes all orbits in this patched conics implementation that is based on Kepler celestial mechanics intersect the equator.

The Bit Longer Answer:

An object in KSP can be represented by two methods:

-Kepler Elements

-Cartesian Coordinates of a 3D-Position and a 3D-Velocity vector.

For Kepler Elements it is easily visible from the picture in the linked site that the orbit intersects the equator.

For Cartesian Coordinates consider the following:

The only force that is applied to the object in space is gravity from the central planet. This force induces an acceleration in the direction Obejct -> Planets Center Of Mass.

So the Velocity Vector changes in the direction of the Planets CoM.

Lets take the following:

- Position of the Object

- Planets CoM

- Velocity Vector of the Object.

These three together describe a well defined geometric plane in the 3D-space.

And since the Gravity-Force is a vector that is parallel to this plane, the gravity changes the velocity-vector, but also only within this plane.

This means that the movenent of the Object happens only within this plane.

The equator also defined a plane in 3D-space.

And the Planets Center of Mass happens to be in both planes, so both planes intersect.

Two planes that intersect have a common line that lies in both on them and the flightpath of the object intersects this line in two points.

So the object crosses the equator in two points.

QED ;-)

[Mutley]

Wow! What a fantastic bunch of replies, and a great welcome - thank you.

These explanations make perfect sense, and explain my intuited sense that orbits must cross the equator. I did create a drawing to explain what I meant, but I can't see a way of uploading it. Never mind, I think you all understood what I meant.

This is exactly what I meant:

if you're asking what I think you're asking (say, a circular orbit that stays constantly around 80 degrees latitude)

I have another question related to orbital mechanics, but I should post another question about that.

Thanks again !

[FleshJeb]

That was a really great question, Mutley.

If you're interested in high-latitude coverage, you might want to play with this:

http://en.wikipedia.org/wiki/Molniya_orbit

[TythosEternal]

For the record, any two non-parallel planes will intercept along a common nodal line. When these are orbital planes (including the equatorial plane) with a common focal point, an object orbiting in one plane will pass through the other plane twice--ascending and descending. This seems a simpler way to explain it, but maybe it's just me.

WRT the high-inclination question, it is entirely possible to launch (from any latitude) directly into a sufficiently-inclined orbit. Whether or not you want to pay the extra fuel cost is a different matter. There are a number of common highly-inclined inertial orbits, including SS (sun-synchronous) and the aforementioned Molniya (sometimes classified as a HEO, or highly-elliptical orbit, though it's not the only one).

There are also non-inertial solutions for sustaining a high-latitude position, in which a platform continually thrusts to avoid descending to or below the equator. Particularly at low altitudes though, these are extremely expensive as measures by fuel consumption. Nonetheless, it is possible in both real life and KSP (you just won't see it projected in your orbital map).

[Mutley]

For the record, any two non-parallel planes will intercept along a common nodal line. When these are orbital planes (including the equatorial plane) with a common focal point, an object orbiting in one plane will pass through the other plane twice--ascending and descending. This seems a simpler way to explain it, but maybe it's just me.

WRT the high-inclination question, it is entirely possible to launch (from any latitude) directly into a sufficiently-inclined orbit. Whether or not you want to pay the extra fuel cost is a different matter. There are a number of common highly-inclined inertial orbits, including SS (sun-synchronous) and the aforementioned Molniya (sometimes classified as a HEO, or highly-elliptical orbit, though it's not the only one).

There are also non-inertial solutions for sustaining a high-latitude position, in which a platform continually thrusts to avoid descending to or below the equator. Particularly at low altitudes though, these are extremely expensive as measures by fuel consumption. Nonetheless, it is possible in both real life and KSP (you just won't see it projected in your orbital map).

That is very interesting, with respect to what is possible in KSP, but what you won't see on the map. This is something I'm getting used to - that everything you see on the map is a projection in to the future based on a set of assumptions (velocity, acceleration, direction etc).

In the case of my original question (high latitude orbit, parallel to the equator is possibly a better way of putting it), this isn't somethng I would want to do, I don't think, but it just struck me if it was possible.

Thanks again.

[Mutley]

The simple answer is no, you can't. The part about orbits intersecting the equator isn't important, any line that divided Kerbin into two equal pieces would fill the same purpose.

To understand remember that an orbit is just falling while constantly missing the ground. The spacecraft is always being pulled down (where down is defined as a line between the ship and the center of Kerbin) but thanks to horizontal velocity the direction of down is constantly changing, translating the previous moments downward velocity into the new horizontal velocity. If you tried to orbit around one pole (like an angels halo over the 'head' of Kerbin) then the downward velocity would not be completely converted into horizontal velocity, so would continue to lose altitude until you hit the ground.

Good point, re the equator itself not being important - something else I hadn't considered until you said that, and is now obvious!