Hillariously high orbital eccentricies

[Heavy Johnson]

So I find this:

This 378 earth-mass class-I giant has an orbital ecc of 0.9984! At aphelion it's at 51.36 au, and at perihelion a mere 0.04112 au (20.5 light-seconds).

 

[Caramel Clown]

Now you just have to wait for a month for some sweet sun proximity shots

 

[ShadowNUAR]

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0.9988

 

[GroG79]

What does this Chinese mean?

 

[[PS4] Filthymick420]

What does this Chinese mean?

It's not Chinese it is a Cyrillic alphabet language possibly Russian

 

[GroG79]

I was referring to "This 378 earth-mass class-I giant has an orbital ecc of 0.9984! At aphelion it's at 51.36 au, and at perihelion a mere 0.04112 au (20.5 light-seconds)."

 

[varonica]

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0.9988

This drove me to check for eccentric orbits us Elite Observatory, according that I have a body with an orbital eccentricity of 1.0, surely that would mean it just shoots off into the galaxy? Ahit seems anything over 0.99 counts as 1 for Elite Observatory, rounding I guess.

Well it appears that 0.9987 is my highest eccentricity;

 

[iain666]

I was referring to "This 378 earth-mass class-I giant has an orbital ecc of 0.9984! At aphelion it's at 51.36 au, and at perihelion a mere 0.04112 au (20.5 light-seconds)."

Eccentricity is a value between 0 and 1 that indicates how circular an orbit is, 0 being perfectly circular and 1 being an escape orbit - so 0.9984 is extremely non-circular (for Earth it's 0.0167, for example) and the planet goes from 20ls from the star at it's closes approach to 51.36AU from the star at it's farthest away (AU = Astronomical Unit = distance between Earth and Sol).

 

[Sapyx]

What does this Chinese mean?

It means the OP's planet is, in effect, a giant comet the size of Jupiter.

 

[Zieman]

So I find this:

This 378 earth-mass class-I giant has an orbital ecc of 0.9984! At aphelion it's at 51.36 au, and at perihelion a mere 0.04112 au (20.5 light-seconds).

Thanks for the transcript.

 

[Heavy Johnson]

This drove me to check for eccentric orbits us Elite Observatory, according that I have a body with an orbital eccentricity of 1.0, surely that would mean it just shoots off into the galaxy? Ahit seems anything over 0.99 counts as 1 for Elite Observatory, rounding I guess.

This is actually how I found mine. "1.0 ecc? what?"

 

[Eahlstan]

The body with the highest eccentricity I found is this one, sadly not as nice at yours:

 

[Orvidius]

Looks like my highest eccentricity is Blua Hypue TY-S d3-93 D 1 at 0.987886, so not quite as high. And I certainly didn't notice at the time. I'll have to take a look at it again.

 

[Eahlstan]

and the planet goes from 20ls from the star at it's closes approach to 51.36AU from the star at it's farthest away (AU = Astronomical Unit = distance between Earth and Sol).

To be more comparable: 53.16 AUs equal 25628.886 light seconds. That's 1250 times farther away than the nearest point in it's orbit.

 

[Northpin]

To be more comparable: 53.16 AUs equal 25628.886 light seconds. That's 1250 times farther away than the nearest point in it's orbit.

Gives a new meaning to "The Winter is Coming", isn't it?

 

[Straha Yeagar]

 

[iain666]

To be more comparable: 53.16 AUs equal 25628.886 light seconds. That's 1250 times farther away than the nearest point in it's orbit.

I just assume everybody knows the distance between earth and Sol, at least to "about 8 light minutes" accuracy.

I concede that this may be optimistic of me.

 

[Eahlstan]

I just assume everybody knows

I have long ago stopped assuming anything about anyone (or at least I try hard not to).
In that case i just thought that it's way more intuitive to use the same unit at both numbers, although I have to admit expressing 0.2 light seconds as 0.000400798 AUs would have been a funnier way to compare them.

 

[Heavy Johnson]

I just assume everybody knows the distance between earth and Sol, at least to "about 8 light minutes" accuracy.

I concede that this may be optimistic of me.

I have this problem too. "Well it's only about 0.63 au, or Venus".
blank look

 

[Heavy Johnson]

The other "sad" thing is that due to Kepler's law (equal distance in equal time), the chances of you finding one of these when close to the star (or other orbiting body) is tiny.