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The Forbes article New York Architects Plan Enormous Skyscraper Hanging From An Asteroid In Space describes a sky scraper suspended from an asteroid in a geosynchronous (but not geostationary) orbit that does not quite touch the Earth. The orbit is inclined with respect to the Earth's equator, and slightly elliptical. This is the recipe for a figure-eight ground track, and several answers here describe them in more detail (I'll try to find links to a few, if you know a good one, please add it.)

The architecture firm announcing the concept is Clouds AO and the building is currently referred to as Analemma Tower, and both of those links are worth exploring.

Question: Because the orbit is elliptical, will it only approach the Earth's surface for "boarding" at periapsis in the southern hemisphere, or could you enter this building when it was over New York City as well?

note 1: As pointed out nicely by @sweber, the path shown can not be the ground track of a compact object in orbit. It goes up to about 41 North in latitude but only down to about 18 South. But the building is something more like a a 2D pendulum with Coriolis(?), so need to look closer at what the plot represents.

note 2: Remember that the earth is oblate by many kilometers!


below: All images are from the Clouds AO website for the Analemma Tower concept, first one is cropped from the second one.

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below: from Google Street View - anyone who doesn't believe that things from space touch down in NYC hasn't been there, or at least hasn't seen MIB.

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    $\begingroup$ Assuming humanity ever figures out the engineering challenges involved and decides it makes good business sense to build this absurdist monument to the stereotype of architects not understanding physics -- it depends. What is the eccentricity? And how do you get into a building that never stops moving? $\endgroup$ – Tristan Mar 29 '17 at 15:23
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    $\begingroup$ I found that this hanging tower should go substantially below 8000 m height above the ground. But what about the speed differential between the building and the air due to jet streams and the movement over the ground track? Moving more than 6000 km in 12 hours is a speed of some 500 km/h. $\endgroup$ – Uwe Mar 29 '17 at 16:41
  • $\begingroup$ @Tristan I think that how to get on is discussed in the link. I think this is not a final design - architecture firms often show general concept designs as well as proposed designs for specific projects. Also I suspect that plenty of architects understand plenty of physics. Let's apply some orbital mechanics here and see what happens. $\endgroup$ – uhoh Mar 30 '17 at 5:06
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – called2voyage Mar 31 '17 at 16:56
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Due to the nature of elliptical orbits, and given that perigee is in the south, the building will fly fast and low in the south and slow and high in the north.

Let's calculate it.

In an elliptical orbit, the earth sits in one of the focal points and the eccentricity $e$ is the distance between earth and the geometrical center of the ellipse. The eccentricity and the semi-major axis $a$ give the closest distance between earth and satellite as $(a-e)$ and the farthest as $(a+e)$. Interestingly, $a$ is the same for all satellites with same period, so we can just take $a=42\,000km$ from the geostationary ones.

enter image description here

The height of the upper "half" of the eight-shaped track is $(a+e)\sin(inclination)$ and that of the lower "half" $(a-e)\sin(inclination)$. From your small drawing with the label "ground trace", the upper "half" is about 3 times larger than the lower half. So:

$$\frac{a+e}{a-e}=3$$ $$a+e=3a-3e$$ $$e=a/2=21\,000km$$

Perigee is $a-e=21\,000km$, apogee is $a+e=62\,000km$. This means the satellite / building will be 42000km farther away when over New York than when somewhere over south America. Even if the upper "half" would only be 10% higher than the lower, the altitude of the building would change by 2000km. For comparison: The ISS flies at 400km.

Another comparison: Here is a 2D-plot showing the size of the earth as well as the size and shape of several orbits:

enter image description here


EDIT:

There's something odd here. The orbit is in the same plane as the center of the earth. Due to this, the most southern latitude of the track equals the most northern one. If the ground track hits New York at 40°N, the most southern position will also be at 40°S, i.e. somewhere in the middle of chile, not be border to Peru at 20°S.
So the tracks shown are NOT ground tracks.
If one looks from far, far away, a satellite seems to fly much higher into the north due to its distance. But something geosynchronous would fly higher than the north pole... The shown track could be the orbit of the building, but even then, the shape of the curve doesn't fit. I'll think about this.

Ah, and if we neglect that we can't build something like this... The building and the asteroid form a nice pendulum, and it definitely will swing heavily.

But in general, I'd say this is one of the strange news which pop up everywhere at the moment due to the special date these days. Nevertheless, one could try the math.

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  • $\begingroup$ I did a quick calculation and noted the results in the question, the eccentricity is about 0.04. $\endgroup$ – uhoh Mar 30 '17 at 2:38
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    $\begingroup$ @uhoh: There are two types of eccentricity, I used the "distance" type as shown in the sketch. Yet, there's something terribly wrong with the images from the question, see edit. $\endgroup$ – sweber Mar 30 '17 at 5:40
  • $\begingroup$ OK re 'distance-type', very convenient notation. You are 100% right Good catch! There are no tricks of map projection that I can imagine to explain their drawing if it represents the ground-track of a compact body in orbit. Any plane that passes through the center of a sphere will pass equally close to each pole (equally North and South from the equator). But yep - there is no reason to think the building doesn't swing. What's the motion of a pendulum hanging from a body in an elliptical orbit? :) This is a really interesting problem! (and I'd better delete my eccentricity estimate) $\endgroup$ – uhoh Mar 30 '17 at 7:21
  • $\begingroup$ If the ground path of a body in orbit should be symmetric to the equator, why should the path of a pendulum hanging from that body be asymmetric to the equator? $\endgroup$ – Uwe Mar 31 '17 at 14:14

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