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#### EvilDrGanymede

##### Guest

Orbit. . distance(Mkm). .Year Length. . Example

0. . . . 30. . . . . . . . . 0.10. . . . -

1. . . . 60. . . . . . . . . 0.25. . . . Mercury

2. . . . 105. . . . . . . . . 0.6. . . . Venus

3. . . . 150. . . . . . . . . 1.0. . . . Earth

4. . . . 240. . . . . . . . . 2.0. . . . Mars

5. . . . 420. . . . . . . . . 4.7. . . . Asteroids

6. . . . 780. . . . . . . . 11.9. . . . Jupiter

7. . . .1500. . . . . . . . 31.8. . . . Saturn

8. . . .3000. . . . . . . . 89.8. . . . Uranus

(8.5. . 4500. . . . . . . . 165.0. . . . Neptune)

9. . . .5800. . . . . . . . 241.4. . . . Pluto

... (the rest are fine)

20. . .9467280. . . . . 15,920,000

Note that Orbit 20 here corresponds to a distance of 1.0 light year (0rbit 19 is 0.62 ly, Orbit 18 is 0.31 ly, Orbit 17 is 0.15 ly, etc).

As noted previously, they haven't factored in the other important factor in determining the orbital period - the mass of the star. This is because they don't actually *give* you the mass of the star in T20 (if you have a copy of CT Book 6/CT Reprints, stellar masses can be found therein).

IF you have the stellar masses available, you'll need to do the following:

Step 1) Express the mass of the primary as a decimal fraction of the mass of the sun (e.g. an M0 V star has a mass of 0.489 Sol). Label this number "MSol".

Step 2) Find the square root of [1 / (MSol)]

The number calculated in Step 3 is the factor by which you need to multiply all the year lengths shown above to take into account the different mass of the primary.

Thus, an M0 V star, massing 0.489 Sol, would yield a factor of x1.43. A planet orbiting it in Orbit 3 (1 AU, 150 Mkm) would complete one orbit in 1.43 years. A planet orbiting in Orbit 5 (2.8 AU, 420 Mkm) would complete one orbit in (4.7 * 1.43 =) 6.72 years.

That should make everything clearer I hope.

[EDIT: Stupid editor ignores multiple spaces! Fixed table now to make it more readable!]