18 réponses à ce sujet

[swk3000]

I've been wondering about this for a long time: what the hell is a Keplerian Ratio? Every planet you look at (in both games) has this listed, but I can't seem to figure out what it means. The only thing I can find is an equation: K = T squared divided by r cubed. However, the site in question is a college university site, and fails to explain what the variables mean (presumably because they've already been explained in either the book or the class). What exactly is the significance of a Keplerian Ratio?

Also, the equation is written out because I can't figure out how to do Superscript and Subscript on these boards.

Modifié par swk3000, 19 septembre 2010 - 08:11 .

[Whatever42]

Wikipedia knows all:

The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

It simply describes the orbit of the planet around the star.

Source: http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion

Modifié par Whatever666343431431654324, 19 septembre 2010 - 08:23 .

[swk3000]

Are you able to translate that into common English?

[Pacifien]

If planet A is four times further away from the star than planet B, it's traveling a distance that's four times longer than planet B. This allows you to determine how long it takes for the planet to then complete its orbit.

Modifié par Pacifien, 19 septembre 2010 - 08:30 .

[swk3000]

Pacifien wrote...

If planet A is four times further away from the star than planet B, it's traveling a distance that's four times longer than planet B. This allows you to determine how long it takes for the planet to then complete its orbit.

Okay. That makes perfect sense. What seems kind of odd to me is BioWare's decision to include it. But it's their game, so it's their right.

Thanks for the replies.

[Fiery Phoenix]

Whatever666343431431654324 wrote...

Wikipedia knows all:

The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

Qualitatively, this means that the closer a planet is to a star, the shorter its orbital period. As such, the closer a planet is to a star, the faster the speed at which it orbits that star (due to increased gravitational effects).


The Keplerian ratio is the cube of the orbital distance of the planet divided by the square of the planet's orbital period. For instance, planet Endor lies at a distance of 0.7 AU from its host star and has an orbital period of 0.9 Earth-years. We can find the Keplerian ratio for Endor by (.7)^3 / (.9)^2 = 0.423 (answer).

Hope this helps (astronomy minor in college).

Modifié par FieryPhoenix7, 19 septembre 2010 - 09:03 .

[Whatever42]

Actually, its rather funny. Kepler discovered that planetary orbits were not actually circular but elliptical and his three laws model that. Yet, as far as I can tell with the perspective, in ME, all the orbits are drawn in nice circles.

[Fiery Phoenix]

Whatever666343431431654324 wrote...

Actually, its rather funny. Kepler discovered that planetary orbits were not actually circular but elliptical and his three laws model that. Yet, as far as I can tell with the perspective, in ME, all the orbits are drawn in nice circles.

It's for the sake of simplicity. There is no such thing as a perfectly circular orbit. Planetary orbits have a tendency to be elliptical, even if it's by the slightest bit. Take our own Earth as an example; the orbit is almost circular.

This is the same reason the words "average"  and "mean" are often accompanied with a planet's orbital period/velocity/distance, since it's a variable due to the not-exactly-circular orbit.

Modifié par FieryPhoenix7, 19 septembre 2010 - 08:56 .

[Pacifien]

Actually wish the systems had a bit of third dimension to them, without all the planets and asteroid belts on the exact same plane. Rotating the systems a bit, seeing them from different perspectives. Probably too small of a detail to add. But definitely wish all the planets had a lot more detail. Even Klendagon in ME2 is a featureless sphere when we know it's supposed to have a massive scar across its southern hemisphere.

[Yxiomel]

The mass effect ratios are weird. They should all be the same (thats the point of a constant of proportionality after all!) in a given system, but (according to wikia at least) they are not. Also, the ratio is a dimensionful quantity, but quoted without units on the star map. (OK, I assume they are in natural units (ie Au^3/years^2)

Modifié par Yxiomel, 19 septembre 2010 - 09:05 .

[swk3000]

FieryPhoenix7 wrote...

Whatever666343431431654324 wrote...

Wikipedia knows all:

The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

Qualitatively, this means that the closer a planet is to a star, the shorter its orbital period. As such, the closer a planet is to a star, the faster the speed at which it orbits that star (due to increased gravitational effects).


The Keplerian ratio is the cube of the orbital distance of the planet divided by the square of the planet's orbital period. For instance, planet Endor lies at a distance of 0.7 AU from its host star and has an orbital period of 0.9 Earth-years. We can find the Keplerian ratio for Endor by (.7)^3 / (.9)^2 = 0.423 (answer).

Hope this helps (astronomy minor in college).

Kind of a random question: is the distance in question taken at the closest point it's orbit takes it to it's sun, or the farthest point?

[Yxiomel]

The semi-major axis is (half) the long axis.

Modifié par Yxiomel, 19 septembre 2010 - 09:08 .

[Whatever42]

swk3000 wrote...

Kind of a random question: is the distance in question taken at the closest point it's orbit takes it to it's sun, or the farthest point?

At the farthest point. 

P.S. Actually, that is incorrect. Sorry. Yes, the centre of the ellipse to the long edge.  The sun, however ,might not be at the centre. 

Modifié par Whatever666343431431654324, 19 septembre 2010 - 09:09 .

[swk3000]

Alright. Not quite sure why I'm asking, but it's nice to get all the little details. Even if I'm not going to use them.

[Fiery Phoenix]

swk3000 wrote...

Kind of a random question: is the distance in question taken at the closest point it's orbit takes it to it's sun, or the farthest point?

It's the mean distance.

[Yxiomel]

FieryPhoenix7 wrote...

swk3000 wrote...

Kind of a random question: is the distance in question taken at the closest point it's orbit takes it to it's sun, or the farthest point?

It's the mean distance.

No it's not! It's the semi-major axis (center to the furtherst point). I only did astronomy for a year at uni, but I'm certain of this.

[swk3000]

Whatever666343431431654324 wrote...

swk3000 wrote...

Kind of a random question: is the distance in question taken at the closest point it's orbit takes it to it's sun, or the farthest point?

At the farthest point. 

P.S. Actually, that is incorrect. Sorry. Yes, the centre of the ellipse to the long edge.  The sun, however ,might not be at the centre. 

Ah. Interesting.

[Yxiomel]

Good point - requires clarification: an ellipse has 2 foci; the sun is at one of the these. The center is half way betweent eh 2 foci, and the distance in question, the semi-major axis, is the line from the center to the furthest edge (or half of longest diameter, if you prefer).

[Fiery Phoenix]

Yxiomel wrote...

FieryPhoenix7 wrote...

swk3000 wrote...

Kind of a random question: is the distance in question taken at the closest point it's orbit takes it to it's sun, or the farthest point?

It's the mean distance.

No it's not! It's the semi-major axis (center to the furtherst point). I only did astronomy for a year at uni, but I'm certain of this.

I was referring to my own example.

We can use the mean value of a planet's orbital radius to give approximate and initial results, but not in an in-depth study of some sort. You're correct, in other words.