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There is a science fiction idea called terraforming. Assuming that it is possible to change the orbit of Venus, is there an orbit that would give Venus the same climate as the Earth (i.e. the same level of radiation from the sun), but which would have a zero risk of collision?

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    If you can move a planet, you can surely move it enough to keep it from colliding with another planet... – PearsonArtPhoto Jul 09 '17 at 14:13
  • @PearsonArtPhoto: Gravitating many-body systems are generically unstable, so it's probably not trivial to find an orbital setup that you can convince yourself is stable against collisions. Also, moving a planet is probably something you want to do before inhabiting it, not afterward. –  Jul 09 '17 at 23:02

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Our planet is right in the middle of the sweet spot or Goldilocks zone for orbital distance from the sun for habitable temperature. While the orbit of Venus is too close to the sun, the distance from the sun is not the main reason for the extreme heat. In fact, by studying Venus, scientists were able to better gauge the greenhouse affect of CO2 and thus extrapolate what could happen on earth.

The formula for approximating the mean surface temperature with albedo factor (surface reflectivity), greenhouse factor, and orbital distance included would be

$$T_p = T_s * \sqrt{R_s / 2R_o} \times (1 – A)^{0.25} \times (1 + G)^{0.25}$$ where
$T_p$ is the mean surface temperature of the planet
$T_s$ is the temperature of the Sun (approx. 6000 K)
$R_s$ is the radius of the Sun (taken as 700,000 km)
$R_o$ is the radius of the planet's orbit (in km)
$A$ is albedo factor (0.76 for Venus), and
$G$ is the greenhouse factor (120 for Venus)

According to Wikipedia, the mean surface temperature on the earth is 15 deg C (59 deg F, 288 deg K).
So if we re-arrange the above equation to figure out what $R_o$ needs to be to obtain a $T_p$ of 288 deg K, we have:

$$288 = 6000 \times \sqrt{350000 / R_o} \times 0.7 \times 3.317$$

$$R_o = 818,978,589 \text{ km or } 5.4745 \text{ AU}$$

To put things in perspective Mars is at 1.52 AU and Jupiter goes between 4.94 AU at perihelion and 5.454 at aphelion.

So we are talking just outside Jupiter's orbit, and too close to avoid collisions with its moons or diversion by its gravity.

If we revise the above calculation to not include the greenhouse and albedo factors we would have

$$T_p = T_s \times\sqrt{R_s / 2R_o}$$

and solving for $R_o$ would give us

$$288 = 6000 \times\sqrt{350000 / R_o}$$

$R_o$ = 151,909,722 km or 1.015 AU, right where Earth is!!!!

I would rather put it at about 2.5 to 3.0 AUs in the middle of the Kirkwood gap in the asteroid belt. Venus would still be too hot for us, but can possibly be terraformed by hitting it with a comet of water ice and then introducing specially engineered phytoplankton that would bloom into the water-filled and $CO_2$-dense planet.

The planet would turn green, and maybe in a few million years the $CO_2$ would be converted to Oxygen as on earth, the greenhouse factor would be much lower, so the surface temperature would lower to something more inhabitable. We would be able to tune the greenhouse effect to get the ideal temperature.

In my humble opinion, rather than trying to shift the orbit of Venus, it would be much easier relatively speaking to screen the planet from most of the sun's energy either by putting into the shadow of a large adjustable screen at ideal location between it and the Sun, and/or by putting the planet into permanent winter mode by sending an asteroid or comet into it causing a highly obscuring dust cloud in the atmosphere.

We can also focus on reforming an earth sized planet in the asteroid belt by clustering the asteroids back together. Then sending in a few comets and other objects from the Kuiper belt. We can custom engineer the planet from core up.

Nathan Tuggy
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    The mass of the combined asteroid belts is about 4% that of the moon. You're nowhere near what would be needed for a second earth sized planet. – Antzi Jul 11 '17 at 16:20
  • @Antzi What is your source? Wikipedia says "The total mass of the asteroid belt is significantly less than Pluto's, and approximately twice that of Pluto's moon Charon." I think that is much more than 4% of our moon. – 0tyranny0poverty Jul 11 '17 at 16:43
  • From https://en.wikipedia.org/wiki/Asteroid_belt

    The total mass of the asteroid belt is approximately 4% that of the Moon, or 22% that of Pluto, and roughly twice that of Pluto's moon Charon (whose diameter is 1200 km).

    – Antzi Jul 11 '17 at 16:53
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    Ok, so let's drag out 2 or more of Jupiter's moons into this region until we can accumulate an earth sized one. This would be easier than bringing Venus out there. And we can custom design the planet from core up. – 0tyranny0poverty Jul 11 '17 at 17:01
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Not in any reasonable way, if you want to have the same radiation from the sun then it has to be the same orbit. An L3 or L4 position would keep you in the same orbit but would not be stable. The exact opposite position from Earth in Earth's orbit is possible and would likely remain safe for a long time in human terms, say for thousands of years or more but would not be stable for long geologically ie millions of years. So someone would need to readjust it back to its original position or it would very much risk collision or slinging both planets into new and likely unfavorable orbits.

Shading Venus in its current orbit is the easiest and cheapest way to get it cooled down and begin the terraforming process. (Unless you don't mind blasting off huge amounts of atmosphere, and some of the surface, with asteroids which would likely be even cheaper still.)

Brooks Nelson
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