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In my story I have an orbital tower or space elevator on a world the size of Earth. Yes, it goes up to geosync orbit. The tower has a tethered asteroid -- how big can it be without exerting gravitational effects on the world?

I'm envisioning an asteroid about 100 kilometers in diameter- Would that be too big?

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    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Jan 14 at 20:20
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    $\begingroup$ 100km is nothing, gravitationally. But you will need one XXXXX of a tether to keep it under control. $\endgroup$
    – PcMan
    Jan 14 at 20:20
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    $\begingroup$ "It" being the tower and the asteroid, just the tower, or - most likely - just the asteroid? $\endgroup$
    – Joachim
    Jan 14 at 20:54
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    $\begingroup$ if you are asking about a 100km tall tower from surface up.... that is a very difficult engineering problem, practically too expensive. For objects in space size is not very important compared to how much mass. $\endgroup$ Jan 14 at 21:22
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    $\begingroup$ I don't understand how this would work. Being connected to a fixed location on the surface, the "tether" will have to hold the asteroid at a distance suitable for geostationary orbit, you'd need a tether with a huge length (35,786 km) far exceeding the height and weight of your building. It would be a bit.. wobbly $\endgroup$
    – Goodies
    Jan 14 at 21:57
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Trite answer: gravitational fields have effectively infinite reach. Even a fleck of sand in orbit technically has gravitational effects on the Earth.

Practical answer: a rock a mere 100km across will have an insignificant gravitational field and the chances are that only some scientists and a small number of engineers will notice.

Assuming your asteroid is conveniently spherical and homogenous, the strength of its gravitational field at some point above its surface can be described by $g = {GM \over r^2}$ where $G$ is the gravitational constant, $M$ is the mass of the asteroid and $r$ is the distance from its centre. Even if it were made of pure iridium (density 22.56g/cm3) it would weigh about 94 petatonnes, but at its surface the acceleration due to gravity would be only ~0.6m/s2, or 0.06 of Earth's surface gravity.

Now, you say that the tower has a tethered asteroid. This must necessarily be at or above the altitude of a geostationary orbit, otherwise it will just come crashing back down to Earth, pulled by its tether, as it won't be travelling at orbital velocities. This means it will be at least 35786km above the surface of the Earth.

Now, the force of gravity between two masses $m_1$ and $m_2$is $F = G{m_1m_2 \over r^2}$ where $m_1$ is the mass of your asteroid, and, say, $m_2$ is the mass of an 80kg human standing on the surface of the Earth directly below it. It will pull on you with a force of 0.4 millinewtons, which is about 2 million times less than the force with which the Earth pulls you down.

This is something of an extreme example, as finding an asteroid that big made of pure iridium is unlikely, and moving it into Earth orbit would be challenging as well as extremely risky. More plausible asteroids will be made of much less dense materials (probably rock or ice) and will generate a weaker gravitational field as a result.

Note on terminology: "orbital tower" seems like a curious thing to call something that reaches out to or past geostationary altitude. Traditionally, one might call this a "space elevator". I'd save "orbital tower" for things like space fountains or the tethers of an orbital ring that might perhaps reach orbital altitudes, but don't need to be tens of thousands of kilometres long.

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    $\begingroup$ Do you even have to care about homogeneity? As long as you can define a spherical closed surface between the two bodies you can treat them as point masses at the centre of mass IIRC. I think it’s a result of Gauss’s law? Been a while since I did any orbital mechanics though, so I might have made that up… $\endgroup$
    – Joe Bloggs
    Jan 15 at 22:45
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    $\begingroup$ @JoeBloggs i assume the spherical and homogenous assumptions are here to be able to calculate a mass. $\endgroup$
    – ths
    Jan 15 at 23:09
  • $\begingroup$ @ths Well now, if you’re going to go around the Internet making sense then… …I’m going to agree with you. That would make a lot of sense. $\endgroup$
    – Joe Bloggs
    Jan 16 at 0:50

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