In the future, in a galaxy far far away... I have a star just like the sun. I was wondering, how many planets could I fit into the Goldilocks zone (aka the region where water is liquid) around this star?

Note - I understand that these conditions are unlikely, but if it helps you to know this system can be manufactured by some futuristic alien race. They just need to calculate how many planets (and where they go).

The star is like our sun in every way. The planets have to be 1.25-.75 Gs. They can be as close to the sun as you want, but areas at the poles need to have liquid water. They can also be far away, but areas at their equators have to have liquid water. Assume that these planets are the only objects within their sun's sphere of influence, and that they don't have moons. (To make everyone's lives easier, no binary planets.)

Given these restrictions, how many planets can I have around this sun?

There are several reasons this question is not a duplicate.This differs from this question because I specify the size and heat of my star, I asked for no binary pairs of planets, no gas giants, and no moons. I also have only one star. The answers in the other question talk about how many are possible including binary planets. Not only is my question different in these ways, but I ask for hard science answers. I want to see your proof that this is possible.

  • $\begingroup$ @DaaaahWhoosh I don't think it's a duplicate, but definitely related. I don't want any gas giants, and my question is a little more specific. $\endgroup$ – Xandar The Zenon Feb 10 '16 at 14:41
  • $\begingroup$ I would also consider that if you terraform some planets that number could go up. Alien tech can explain some issues away $\endgroup$ – AndreiROM Feb 10 '16 at 15:11
  • $\begingroup$ @XandarTheZenon if you won't allow gas giants, you should make that restriction part of the question. Otherwise it reads like a duplicate of ^^ $\endgroup$ – nitsua60 Feb 10 '16 at 15:44
  • $\begingroup$ The answers for the linked question don't all use gas giants, so I think that might make it a dupe. $\endgroup$ – HDE 226868 Feb 10 '16 at 15:59
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    $\begingroup$ Two of the answers on the other question violate the conditions of my question, and the third would fall apart. That means that that question doesn't have an answer that answers my question. $\endgroup$ – Xandar The Zenon Feb 10 '16 at 18:15

For a Sol luminosity star:

Possibly up to 4

Two per Orbit

Based upon this answer, it looks like the maximum is 2 planets sharing the same approximate orbit. Their orbits are horseshoe shaped and they will alternate with one on the inside trace while the other is on the outside track, just like Saturn's Janus and Epimetheus moons

Two viable Orbits

If Mars were about 4x more massive, it most likely would have retained the bulk of its atmosphere. If it had, then it would be a cold place but it would likely be able to support life. Mars is borderline on whether it's inside or outside the Goldilocks Zone. If it were outside, moving it a little closer would likely not affect Earth dramatically.

Why Lagrange Points Buy you Nothing

Lagrange Points

L1, L2, & L3 are unstable

Three of the five Lagrange points are unstable (specifically they are statically stable but dynamically unstable). If you put objects in those locations, they will drift away from the point. Spacecraft in these locations require propulsion adjustments to remain in the location.

A planet, with no means of propulsion, cannot use these points for a stable configuration and maintain their location.

L4 & L5 don't help either

Lagrange points impose strict requirements on the masses of the bodies in the system. The central object (Sun) must be significantly more massive than the orbiting body (Planet) which must be significantly more massive than the object in the Lagrange point. The general case is: $$M_{\text{Sun}} \gg M_{\text{Planet}} \gg m$$

For our case, it looks like $$M_{\text{Sun}} > 25 \times M_{\text{Planet}}$$


$$M_{\text{Planet}} > 10 \times m_{\text{Lagrange}}$$

Both must be true for the configuration to remain stable.

This means anything you put in those points are too small to retain an atmosphere.

Why Klemperer Rosettes Buy you Nothing

Klemperer (or other) Rosettes suffer from the same problem as the Lagrange points, they are statically stable but dynamically unstable. If you place objects in those locations, they'll drift out of the desired location and lead to collisions and or ejections of the objects in the system.

Incidentally a 6 object Rosette is more stable than other configurations, because each object falls into another object's L4 or L5 location. It is still unstable, it is just more stable than the other configurations (objects will stay in their locations longer).

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  • $\begingroup$ +1 for the simple (yet important) explanation of why dynamic stability != static stability $\endgroup$ – nitsua60 Feb 10 '16 at 15:56
  • $\begingroup$ I'm somewhat sad about this thorough debunking, as the concept of a whole chain of habitable planets around the same star (as described by Peter F Hamilton) is one of my favourite pieces of SF imagery. Oh well, nobody ever accused him of writing particularly 'hard' SF! $\endgroup$ – Toadfish Feb 11 '16 at 14:04

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