What orbital parameters could I give seven Mars/Earth sized planets so they have stable orbits around a G-type star like The Sun, they need to fit between 0.5 AU and 1.7 AU.

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    $\begingroup$ This is a complex question to answer. To make it more on topic for this site, can you rephrase it as "is it possible to have seven Mars/earth sized planets betwen 0.5 AU and 1.7 AU of our solar system?" The answer probably depends a lot on the presence of absence of Jupiter, in particular, so you might want to specify what other planets there are in this system. $\endgroup$ – kingledion Apr 25 '17 at 1:01
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    $\begingroup$ @kingledion Not so complex a question as all the planets in the TRAPPIST-1 system share the same 3:2 orbital resonance. I cncur that the absence of a Jupiter-mass planet is probably necessary. Since we don't know what other planets might be in the TRAPPIST-1 system, so let's keep it simple and pretend there's only the seven planets present until more planets are found there. Sensible, eh? $\endgroup$ – a4android Apr 25 '17 at 3:37
  • $\begingroup$ See here: planetplanet.net/2014/05/21/…. TRAPPIST-1 is a chain of resonances so space the planets as I do for giant planets. $\endgroup$ – Sean Raymond Jun 14 '18 at 7:41

So, to elaborate on the math I mentioned in my comment on a4android's answer:

Orbital resonances are based on orbital period, not orbital radius, so let's use the equation a4android mentioned to convert our target range (.5 AU to 1.7 AU) from radii into periods: $$P = \sqrt{R^3}$$ (I removed the /M since we're working with a one-solar-mass star.)

$$P_{min} = \sqrt{0.5^3} \approx 0.3536 $$

$$P_{max} = \sqrt{1.7^3} \approx 2.2165 $$

The most stable orbital resonances are in low integer ratios: 2/3, 3/5, etc. So the period of the innermost planet multiplied by the larger number and then divided by the smaller number gives you the period of the next planet out. Then take that period and repeat the process with the next ratio to find the period of the third planet, and so on.

With seven planets in our system, there are six such adjacent-planet ratios. If we want all six ratios to be equal, then we're multiplying our smallest period (P1) by the ratio (r) six times to get the outermost planet's period (P7). Thus: $$P_7 = P_1 \times {r^6}$$ Since we know the smallest P1 and largest P7 that will meet our radius requirements, we can plug those in and solve for rmax to get the greatest possible ratio that will let seven planets fit in the target range. \begin{align} 2.2165 & = 0.3536 \times {{r_{max}}^6}\\ r_{max} & = \sqrt[6]{\frac{2.2165}{0.3536}} \\ r_{max} & \approx 1.3579 \end{align} That, of course, is not a nice whole-number ratio. 4:3 looks like the largest ratio that is less than rmax without using overly-large integers (e.g 135:100), at 1.333...

If you want to vary the ratios between adjacent planets, you can check that they will work correctly by multiplying them all together and making sure the result is less than ~6.2692.

As a suggested solution, below is one possible set of orbital radii (in AU) and periods (in Earth years) that work for your question. This set was generated with R7 pegged to 1.70 AU, since that gave a center planet close to Earth's parameters.

\begin{array}{c|rr} Planet & Radius & Period\\ 1 & 0.54 & 0.39\\ 2 & 0.65 & 0.53\\ 3 & 0.79 & 0.70\\ 4 & 0.96 & 0.94\\ 5 & 1.16 & 1.25\\ 6 & 1.40 & 1.66\\ 7 & 1.70 & 2.22\\ \end{array}

  • $\begingroup$ Also, just have to say, this answer was my first time working with MathJax, and I had way too much fun playing with it, so I apologize if I overdid it a bit. ^_~ $\endgroup$ – Salda007 Apr 26 '17 at 8:23
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    $\begingroup$ I do not see anything excessive here, explaining how you determined correct ratios is an useful thing to include. Now the answer is useful to even people who need a solution for other parameters and has less magic numbers in general. Well done. $\endgroup$ – Ville Niemi Apr 26 '17 at 9:21
  • $\begingroup$ This is great! I'm going to assume a gas giant orbiting at Jupiter's distance, ~5 AU, would destabilize the system. $\endgroup$ – Stephanie Apr 28 '17 at 16:19
  • $\begingroup$ No idea! A bit under that (About 4.97 AU ;-) ) is actually in a 5:1 resonance with the 7th planet using the table I calculated, so, maybe not? $\endgroup$ – Salda007 Apr 29 '17 at 8:50
  • $\begingroup$ using x^\frac 3 2 would be preferably in the math part, but is almost illegible. generally, using x^\frac 1 N for the Nth root of something is preferably. $\endgroup$ – Trish Jun 13 '18 at 14:23

Essentially you use the same orbital parameters of the TRAPPIST-1 system and locate your seven Mars/Earth mass planets in the seven orbits. It's basically a matter of keeping the orbital resonances the same and scaling the distances from the primary star.

The key feature of the TRAPPIST-1 system are that they all share the same 3:2 orbital resonance. So this is how we build a TRAPIST-1 type of planetary system with seven planets.

Assume there is a one Earth-mass planet at one astronomical unit (AU) from the primary G-star, and an orbital period of one year (approx. 365 days, for pedants worried about what constitutes a planet's year).

This means the next planet outwards will have an orbital period such that every time it orbits the star twice our Earth-clone planet will have orbited it three times. This what constitutes a 3:2 orbital resonance. Therefore, its orbital period will be half of three years or eighteen (18) months.

It also means the next planet inwards towards the G-star will have an orbital period of eight (8) months. As this is one-third of two standard Earth years (all refers here to years are standard Earth years).

Following the orbital resonance relationships in this manner a worldbuilder can construct their own version of a TRAPPIST-1 system. At least, in terms of the planets and their orbital periods.

Just continue applying the same sets of 3:2 ratios and this will yield the complete of the planets' orbital periods.

The distances of the planets from the primary G-star can be calculated using this approximate equation derived from Kepler's Laws.

$$P = \sqrt{\frac{R^3}{M}}$$

where P = the orbital period, R = the radius of the planet's orbit (assumed circular for simplicity), and M = the mass of the primary star in solar masses.

Therefore, to calculate the orbital radius of the planets in your clone TRAPPIST-1 system, and the primary star's mass is one solar mass, rearrange the above equation

$$R = \sqrt[3]{P^2}$$

Most reasonable scientific calculators can do this calculation. Also, there are a slew of helpful online calculators.

Using the above information and equations you should be able to your own TRAPPIST-1 lookalike system with its own septet of Mars/Earth mass planets.

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    $\begingroup$ @Secespitus Many thanks for the edit especially the MathJaxing of the equations. Much appreciated. $\endgroup$ – a4android Apr 25 '17 at 7:35
  • $\begingroup$ Doing the math, you can only fit five planets between .5 AU and 1.7 AU with 3:2 resonances between adjacent planets, and if you peg one of the planets to Earth's orbit (1 AU, 1 Earth year), then you only actually get four. Similarly, using the actual TRAPPIST-1 resonances (8/5, 5/3, 3/2, 3/2, 4/3, 3/2), then you can only fit four in that range. However, seven planets in 4/3 resonances can fit in the target range, if the center planet orbits between .85 and .90 AU. No idea how stable that would be long-term, though. (Yay spreadsheets!) $\endgroup$ – Salda007 Apr 25 '17 at 9:19
  • $\begingroup$ Salda007 That's a beautiful piece of work. It gives the OP additional options for their system. Well done! Really delightful results. Orbital resonant systems are considered to have long-term stability. $\endgroup$ – a4android Apr 25 '17 at 10:54

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