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In the not too distant future Earth will enter its golden age of industrialization and construction. Pulling raw materials from other planets (not from Earth), humans begin building mega structures around the planet. Huge rings, tethered space stations, multiple moons...

But scientists and activists around the world have begun to worry that the nonstop construction might be changing Earth's orbit. No longer can environmental and ecological change be blamed on the massive shadows and regular eclipses. Something more seems to be going on...

Given that:

  • All constructs are no more distant than the Moon. 75% of the constructs are in geosynchronous Earth orbit. 10% are in low Earth orbit. The rest are at or near the lunar orbit.
  • The mass of the Earth is 5.972 × 1024 kg. The mass of the constructs are in addition to this.
  • How or why the constructs are built is not relevant to the question. How the constructs remain in place is also not relevant to the question. Assume Clarkean Magic got them there and keeps them there.

Question: How much mass can be added to the volume defined by the average orbit of the Moon before the orbit of the Earth changes by 3% in either velocity or distance from the sun?

Why 3%? It comes from my college days learning to be an engineer. Back then, 3% was considered a rule-of-thumb "noise threshold" for any measurement. Think of it this way, if you're trying to propagate TTL (0V–5V) signals, then the circuit should never under any circumstances propagate a transition if the incoming signal is +/- 0.15V or less. Yes, the TTL standard is quite a bit more accommodating than this, but we're talking about the noise threshold, not the TTL spec limits. Cheers.

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  • $\begingroup$ Which values of distance and velocity are you taking as baseline for calculating the 3%? $\endgroup$
    – L.Dutch ♦
    Commented Apr 30 at 14:39
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    $\begingroup$ I'd be more worried about perturbing other planets destructively before you reach that threshold, unless my intuition is off. $\endgroup$
    – Escaped dental patient.
    Commented Apr 30 at 14:59
  • $\begingroup$ I don't know how to answer your question, @L.Dutch. $\endgroup$
    – JBH
    Commented Apr 30 at 15:42
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    $\begingroup$ Is this actually a worldbuilding question? This looks like a straight physics question with some worldbuilding flavored distractions that we are supposed to ignore. Although physics SE would close it as homework like until you changed it to a conceptual, rather than computational, question. $\endgroup$
    – g s
    Commented Apr 30 at 19:01
  • $\begingroup$ The question is ill-defined. As Monty Wild answered, the orbit itself is almost independent of the Earth mass, but it doesn't take into account change of mass. How do you get all that extra mass to Earth? The way it would most likely be done involves lots of gravitational assists, and those could most certainly alter the orbit of Earth substantially. But how much exactly depends on how you do it. It would be conceivable to gather the material from different parts of the solar system in a way that the side effects of the gr. assists cancel out. $\endgroup$
    – leftaroundabout
    Commented Apr 30 at 23:31

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This is a matter of two-body orbital mechanics.

$$T=2\pi\sqrt{\frac{a^3}{G(M_1+M_2+M_3)}}$$

where:

  • $T$ = orbit time
  • $G$ = Gravitational Constant (6.6743E-11)
  • $a$ = Semi Major Axis (1.49598023E11)
  • $M_1$ = Mass of Sun (1.989E30 kg)
  • $M_2$ = Mass of Earth (5.972E24 kg)
  • $M_3$ = Mass of Earth-orbital infrastructure (variable)

If we calculate T ($T_0$) with $M_3$ = 0 we get 3.1554E7 seconds.

To get the 3% difference, we can calculate T ($T_1$) with $M_3$ > 0 such that $\frac{T_0}{T_1}$ = 1.03.

I plugged the numbers into Excel, and got $M_3$ = 1.211E29 kg. This is 6% of the mass of the sun, or 20,279 times the mass of the earth.

I'm assuming that the orbital distance remains constant here, that the clarkean magic used to get all this mass into Earth's orbit does so by putting it into Earth's orbit, bit by bit, at Earth's speed at the time.

Solving for orbital distance with the increased mass, with orbital time remaining constant, the orbital distance increases by 3% at $M_3$ = 1.84E29kg, which is 9% of the sun's mass or 30,850 times the mass of earth.

So, taking the lesser of the two calculated masses, you'd have to add 20,279 times the earth's mass, or 6% of the Sun's mass, to get a 3% significant difference in Earth's orbital time.

This is an awful lot of mass... so I think it's safe... for a while, at least.

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    $\begingroup$ Or, alternatively, use plain reasoning without mathematics: since we are bringing the new mass from wherever so that its speed relative to Earth is just about zero, at least on the average, it means that the new mass is already in the same orbit around the Sun as Earth is. $\endgroup$
    – AlexP
    Commented Apr 30 at 15:15
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    $\begingroup$ @AlexP This is true "instantaneously", but not over time: If we somehow brought a bunch of mass to "dock" with the earth at an instant without changing its velocity, the orbit would still change, because it would change the relative positioning of the earth-sun center of mass. Because the earth-sun gravitational force is larger, it would result in a lowering of the apogee and average distance, and a shorter year. $\endgroup$
    – Mario Carneiro
    Commented May 1 at 2:13
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    $\begingroup$ @MarioCarneiro: Yes, of course, but in a first order approximation the Sun is infinitely heavier than Earth. To move the center of mass appreciably the Earth would have to gain sufficient mass as to be comparable to the Sun. That's an awful lot of mass. The point being that as long as the added mass remains comparable to the original mass of the Earth it doesn't matter. $\endgroup$
    – AlexP
    Commented May 1 at 6:55
  • $\begingroup$ There's arguably a factor of two error here. If the mass is added instantaneously (which, as far as we know since it's magic, it will be), the orbit will become more eccentric. With only a 1.5% change in average orbital distance, the distance from the sun will vary by 3% a quarter of a year later. Not that I think this changes the overall picture very much... $\endgroup$
    – addaon
    Commented May 2 at 22:27

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