Polar orbits

simonalexander2005

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Hey,

So I've been trying to figure out some stuff to make my story feasible. The premise is that a ship is in polar orbit around the earth. My questions are as follows:

1. How do I figure out how fast the ship could be moving, as viewed from the earth's surface (e.g. how long would it take for the ship's line-of-sight to move from the north pole to the south pole, or from Dresden to Brussels? What variables and equations do I need to do this? Or can someone give me some approximate values?)

2. How does orbit work? Would the ship be able to be in a polar orbit without applying any power of its own to maintain it? If so, would that change/simplify question 1.?

Any help on this stuff would be welcomed :)

Thanks,

Simon
 
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Drachen Jager

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1) it's all about altitude. There should be plenty of orbital calculators out there if you just google it.

2) You mean an orbit that crosses from north to south poles, right? Otherwise you need some remedial physics lessons to even contemplate writing the sort of fiction you appear to be attempting. You can't have an orbit that simply circles a pole.

Orbits, by their nature don't require power to maintain (perhaps a slight bit, but nothing of consequence) because it's merely centrifugal force competing against gravity, much like when you spin around holding the handle of a bucket. The bucket can go straight out, or even upside-down without spilling water, because the curvature of the path (and the tendency to travel in a straight line) forces it outward.
 

Russell Secord

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Unless you're talking about a geosynchronous orbit, it doesn't matter where your satellite/ship goes. As Drachen Jager points out, the speed of the ship will vary with the height of the orbit. Whether the planet is rotating beneath it doesn't affect the math in any significant way. That rotation, however, will affect the places where people on the surface can see the ship at any given time.

To pick a simple example, suppose the ship takes 24 hours to make one orbit. If it passes over Brussels on the first orbit, it will pass over Brussels on every orbit, because it takes Brussels 24 hours to rotate around the Earth's axis and return to that same point.
 

blacbird

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There are plenty of scientific satellites studying Earth in polar orbits. Those orbits will be as stable as equatorial orbits, and like the latter, dependent primarily on altitude. Shouldn't present any sort of technical problem.

To have a polar orbit that takes 24 hours per orbit, the ship would need to be at about the same distance from Earth as geosynchronic equatorial orbits are, which is a distance from the surface of Earth of a little more than 22,000 miles (~36,000 km). Presuming, of course, an essentially circular orbit.

caw
 
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wendymarlowe

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The math for how fast the ship has to go is pretty easy, too - just (height above the earth) * (pi). That gives you the speed per day (in whatever units you measure height in), and then you can break that down into hours or seconds or whatever you need. Of course, that assumes you're high enough you're not having to deal with drag from crashing back into the atmosphere or anything :p
 

simonalexander2005

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yeah so I guess I mean a geosynchronous polar orbit - so at some point the ship will cover every point on the earth's surface (like a GPS sat does). The question is, what is a realistic speed for the ship to be covering points on the surface - how long would it take to get from A to B, and what's the maths for working that out? Also obviously the route it takes in covering the surface of the earth would have to be logical - it can't jump around - so are there any rules about routes it has to take?
 

Schilcote

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(Disclaimer: Everything I know about space comes from playing Kerbal Space Program and watching Scott Manley's Youtube channel)

No such thing as a geosynchronus polar orbit; indeed, polar orbit is as far from geosynchronus orbit as you can get. :p

Part of what you're looking for is what's called the 'orbital period'; how long it takes the complete an orbit. Wikipedia's got an article on it, though the math is just a bit heavy. As long as two points are on the same longitude line (more or less), the time difference will be some fraction of that.

Now, if the spacecraft will not be passing over both locations in the same orbit, that's a little tougher. The Earth will rotate under the spacecraft's orbit- one full rotation every twenty-four hours- but there's no guarantee that the spacecraft will actually be over the right spot.

Your orbital period in LEO is about 100 minutes, and that's about as low as you can go (any lower and you start really rubbing against the atmosphere and need to burn your engines every so often; the ISS has to do this).

So, all in all, it's a pretty safe bet that your spaceship will be over whatever point you need it to be in a straight vertical line on the map as long as you have a leeway of about an hour and a half, and will be over whatever point you want it to be period at some point in a twenty-for hour period (as that's how long it takes the Earth to rotate fully).

Also, orbit has nothing to do with centrifugal force. Centrifugal force is actually just an effect of inerta; the object wants to go the way it's already going, and if it's connected to a spinning object then that won't be the way it's being pulled, so it gets 'pulled' away from the thing it's fixed to.

Orbit is literally trying to fall and missing. You just have to get going fast enough around the Earth that by the time you hit the ground, you're not over it anymore.

Also, you do in fact need to apply a bit of energy to stay in orbit. Certainly if you're really low, like the previously mentioned ISS, but even if you're far outside the atmosphere solar wind will push you away from the sun, eventually putting your perigee (lowest point of the orbit) inside the atmosphere and causing you to decelerate right into the surface.

The line that traces the satellite's position above the planet looks like this, by the way:


Of course, the size of the gaps will vary depending on your orbital period, as I mentioned.
 
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