1
$\begingroup$

The new theories in the field demonstrate the possibility of non-Keplerian orbits.

https://www.pourlascience.fr/sd/spatial/de-nouvelles-orbites-geostationnaires-10636.php

enter image description here

I am interested in calculating orbits above the parallels of 45 and 80 degrees. A large disk placed on these orbits should always be above an area. Have geostationary behavior.

enter image description here

$\endgroup$
  • 1
    $\begingroup$ Orbits like your top two are impossible. In simple terms, the satellite must orbit the center of mass of the Earth. Consider reading through this article: earthobservatory.nasa.gov/features/OrbitsCatalog $\endgroup$ – Organic Marble Oct 5 '19 at 1:34
  • $\begingroup$ The new theories in the field demonstrate the possibility of non-Keplerian orbits. $\endgroup$ – Ion Corbu Oct 5 '19 at 2:27
  • $\begingroup$ I edited the post and added the photo and link. $\endgroup$ – Ion Corbu Oct 5 '19 at 2:28
  • 2
    $\begingroup$ The article says it might be possible to displace the satellites by 5 to 30 miles. At the distance of a geostationary satellite, that offset won't get you anywhere close to 45 or 80 degrees latitude. $\endgroup$ – JohnHoltz Oct 5 '19 at 3:27
  • 1
    $\begingroup$ A satellite moves on its orbit without engine thrust. If thrust is neccessary for displacement, it is no orbital path. There is no thrust for the orbits of solar system planets and moons $\endgroup$ – Uwe Oct 5 '19 at 22:06
4
$\begingroup$

The cited article is talking about a 1984 idea by Robert Forward. (See this English summary). In it, he proposed that solar pressure could be used to move an geostationary satellite a few 10’s of kilometers north or south of the usual geostationary orbit. These are more generally called “statites”.

There’s really nothing here that points to much larger offsets from an equatorial orbit.

$\endgroup$
2
$\begingroup$

note: The question was modified while I was writing this answer, so I will have update(d) this (with a new) answer with more math shortly finally, here!


So let's think about what an orbit is. One way to think about it is a combination of moving forward while falling down. At each moment the object keeps moving forward, but it also accelerates towards the Earth because of the Earth's gravitational attraction.

In which direction is it always accelerating?

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.

There are no other free orbits than those. The only way to make circles like that is to constantly fight gravity with powerful propulsion, and there's no known technology that can sustain that right now for longer than several minutes.

That would be a spaceflight equivalent of Franky Zapata's Flyboard.

Franky Zapata and Flyboard from CNN

From CNN's French inventor makes 'beautiful' flight across Channel on hoverboard, linked in What is it exactly about flying a Flyboard across the English channel that made Zapata's thighs burn?

There might be some ideas of using pressure from sunlight on a reflective film only a few atoms thick so as to be extremely low mass, but it's not something you could really build today.

$\endgroup$
  • 1
    $\begingroup$ Well said! Cool 'flyboard' too. $\endgroup$ – Organic Marble Oct 5 '19 at 2:39
  • $\begingroup$ Have you seen the new post edit? With non-Keplerie orbits? The orbit remains high but less than the equatorial one. The attraction to the center of the Earth is counteracted by altitude, speed and trajectory in the form of a flattened figure 8. The altitude corrections do not require too many edges. $\endgroup$ – Ion Corbu Oct 5 '19 at 2:49
  • $\begingroup$ @IonCorbu oh I see, that was added while I was writing. Okay I'll have a look now and then update my answer, thanks! $\endgroup$ – uhoh Oct 5 '19 at 2:54
  • $\begingroup$ Here is apparently most of the same information in English: phys.org/news/2010-07-scots-space-year-old-theory.html . It includes cites to 2 papers: "-- R L Forward, Light-levitated geostationary cylindrical orbits using perforated light sails, can be found in the Journal of the Astronautical Sciences, Vol.32, Apr-June, pp.221-226, 1984. -- S Baig and C R McInnes, Light-Levitated Geostationary Cylindrical Orbits are Feasible, Journal of Guidance, Control and Dynamics, Vol. 33, No. 3, pp. 782-793, 2010. -- A pre-print of the new paper is available at: strathprints.strath.ac.uk/18865/" $\endgroup$ – bit chaser Oct 5 '19 at 21:43
  • $\begingroup$ That preprint url doesn't resolve for me, but pureportal.strath.ac.uk/en/publications/… does and the full paper is at pure.strath.ac.uk/ws/portalfiles/portal/67305638/… $\endgroup$ – bit chaser Oct 5 '19 at 21:51
1
$\begingroup$

If the aim is to hover a solar powered ion satellite the basic numbers come out as:

Weight for 1KW of solar panels = 5.33kg

Force of gravity at at 400km up = 8.6 Newtons per Kg

Mass 5.33 kg so that = 46 Newtons of gravitational force

Thrust of a 0 mass Xenon engine per KW = 0.034 Newtons

So pretty clearly it will not be possible to make something that hovers over the poles on electric power, even with unlimited Xenon available and a magical 0 mass thruster. Suspect there may be better power per KG choices for solar panels than used above but they would need to be 1000 times better to make this math work even for a magical thruster.

The thrust force needed will be sine of the latitude you are hovering over so 80 degrees would need 45 Newtons and 45 degrees would need 32 Newtons per KW of power.

For non polar locations you also need to work with the Earth's shadow, cutting our available power by half for a single platform, by rather more for a fixed ring due shade from adjacent panels, less as orbit height increases.

Where the concept starts to work is at 0.04 degrees where the numbers above equal but only get a couple of Km of displacement from equatorial.

The other method where this might start to work for high latitudes is for very high orbits where the force of gravity from earth is reduced, but that will place your satellite at distances greater than the moons orbit above a pole, defeating most of the purpose of being geostationary, and still not doing much for 45 degrees latitude.

$\endgroup$
1
$\begingroup$

Okay, I had said that I would post another answer based on your clarification, but then I got temporarily distracted. Here it goes now.

Here is just what you've asked for; How to calculate the radii of high orbits above the parallels of 45 and 80 degrees and then some even higher ones!

Just fyi I have confirmed by a simple 3D orbit simulation that this works. I calculate the real acceleration, then zero-out the z-component in order to simulate the thrust. Since the magnitude of the thrust is not part of the question, there was no need to calculate it explicitly, but it will be sizable, and a realistic conventional spacecraft couldn't maintain it very long.

We will assume you have some "vertical" acceleration due to thrust $a_T$ produced by some unconventional thrust mechanism pushing your spacecraft "up and away" from the plane of the equator, that exactly cancels the "downward" component of Earth's gravitational acceleration. This leaves only a horizontal acceleration $a_H$.

enter image description here

Every unconventional orbit is defined by the axial distance $R$ and the height above the equatorial plane $h$. The distance to the center of the earth $r=\sqrt{R^2 + h^2}$ is used to calculate the radial acceleration

$$a = \frac{GM}{r^2} = \frac{GM}{R^2 + h^2}$$

and the horizontal acceleration is (by use of similar triangles)

$$a_H = \frac{GM}{R^2 + h^2} \times \frac{R}{\sqrt{R^2 + h^2}} = \frac{GM \ R}{(R^2+h^2)^{3/2}}.$$

For a circular orbit the acceleration is $v^2/r$, so we can write

$$v^2 = a_H R = \frac{GM \ R^2}{(R^2+h^2)^{3/2}}.$$

Also for a circular orbit, the relationship between the radius, velocity and the period is

$$T = \frac{2 \pi R}{v}$$

$$v^2 = \frac{4 \pi^2R^2}{T^2}$$

If we set the two expressions for v^2 equal, we get the surprisingly simple result:

$$R^2 + h^2 = \left( \frac{GM \ T^2}{4 \pi^2} \right)^{2/3}$$

Try it! The standard gravitational parameter $GM$ for the Earth is 3.986+14 m^3/s^2. Put in 86164 seconds for T (one sidereal day, 23h 56m 4s) and start with a normal GEO orbit with $h=0$. You should get 42,164 kilometers (42,164,000 meters).

Now, $h$ is given by

$$h = r_E \sin(lat)$$

and let's just use the average radius of the Earth of 6371 km.

Considering that the GEO orbit is so far from Earth, this is only a tiny variation. Let's do some crazy orbits way above the North Pole as well.

lat(degs)    h (km)       R (km)      v (km/s)
---------    ------      -------      --------
    0            0       42,164        3.075
   45        4,505       41,922        3.057
   80        6,274       41,694        3.040
   --       10,000       40,961        2.987
   --       20,000       37,119        2.707
   --       30,000       29,628        2.161
   --       40,000       13,335        0.972
$\endgroup$
  • 1
    $\begingroup$ Cool stuff! Now if we can just devise a bootstrap-lifter! $\endgroup$ – Organic Marble Oct 12 '19 at 22:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.