Gas giant orbits a white dwarf very closely

Ozzie_J_Isaacs

Ozzie_J_Isaacs
In the Spoihee LM-V e2-372 system, my monitoring program sounded an alarm
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A gas giant orbits a white dwarf in 50min and this in a distance of only 0.87LS
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The fact that there is a second quite close white dwarf makes scanning a bit tricky
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But in the shadow of the gas giant I could approach
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And scan successfully
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something oval was the gas giant also still
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I have now the data of the planet and the two white dwarfs in English, had not thought yesterday to change my tool language.
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MiloGreenRanger

M
Morbad said:
Quick, someone with a pie calculate the orbital velocity.
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For an eccentric orbit, we use the vis-viva equation. The eccentricity of the orbit is given as 0.4205. The program lists the semimajor axis at 0.87 ls, which is 260,819,438 m. That means the periapsis will be 151,141,864 m and apoapsis will be 370,494,011 m.

v = sqrt( GM * (2/r - 1/a) )
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Where r is the distance, a is the semimajor axis. So the fastest velocity is 854.969 km/s, and the slowest velocity is 348.740 km/s.

If the orbit were circular, we could just take the circumference over the period, which we could verify like so:

T (in seconds) = 2 pi sqrt(a^3 / (GM))

M = 0.5859 solar masses = 1.16504x10^30 kg
a = 260,819,438 m
G = 6.67408 x 10^-11 m^3 / (kg s^2)
Click to expand...

So the period would match what the program also lists, 3001 seconds or 50 minutes.
For a circular orbit, the mean velocity would then be 2*pi*a / 3001 s = 546.076 km / s.

Edit: Umlaufzeit means period, though, which the game says is 50 min, so maybe I'm way off.
Edit Edit: And Grosse Halbachse would be semimajor axis, so 0.87 ls.
Edit Edit Edit: Organizing to be more coherent.
 
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Ozzie_J_Isaacs

Ozzie_J_Isaacs
MiloGreenRanger said:
Assuming that the difference between the star and planet distances (.345 ls) is the orbital radius, then if my math is correct it orbits in 749.5 s, or about 12.5 minutes.

M = 0.5859 solar masses = 1.16504x10^30 kg
a = 16771.675 - 16771.330 ls = 103,428,398 m
G = 6.67408 x 10^-11 m^3 / (kg s^2)

T (in seconds) = 2 pi sqrt(a^3 / (GM))

Edit: Umlaufzeit means period, though, which the game says is 50 min, so maybe I'm way off.
Click to expand...
Semimajor axis is 0,00174347AU, eccentricity 0.4205

That is what i get from Wolfram Alpha, maybe it helps you to calculate the speed
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MiloGreenRanger

M
Thanks. At first I was answering the wrong question, then I looked at the actual data, and then it sent me down this rabbit hole of how eccentricity is represented. 10/10 would math again.
 

Ozzie_J_Isaacs

Ozzie_J_Isaacs
CMDR SleepyDelegate said:
I've found a similar system last October...

I just didn't know how to calculate the distance...

View attachment 290260

I'm not able to pull the exact data at the moment, but this is what EDSM says:

View attachment 290261
Click to expand...


For this you need the semi major axis, but with the data that ED displays in the game, unfortunately, nothing works in your case. Only that the gas giant is closer than 5Ls to the star I can say, because 0.01 are 5Ls.
 

Orvidius

Orvidius
The actual semi-major axis value should be in the journal, and EDSM will have it too. Neither the game nor the EDSM website display enough digits to see what it actually is though. If you haven't sold the data yet, you might want to wait, but with the name I can look it up in my data too. Otherwise, you might want to dig through the journal, or EDDiscovery or something to see what the SMA value is.

The semi-major axis is basically half of the longer diameter of the orbit. It's a pretty decent approximation of the planet's average distance from the parent object, especially when the eccentricity is low, which it is here. (More accurately, the SMA is the average when averaging over the eccentric anomaly, rather than the time-based average, but for most purposes it's "good enough").
 

CMDR SleepyDelegate

CMDR SleepyDelegate
So, I looked in my journal, I did not sell the data yet and Im on an expedition that lasts till April, so the name will have to wait... :)

But here are the data:

Semi major axis: 133 554km
Orbital Excentricity: 0,024

So what is the distance in Ls?

BTW, I was not able to map it. I was too scared to get too close... I know how dangerous the exclusion zone of the WD can be...
 

CMDR SleepyDelegate

CMDR SleepyDelegate
Orvidius said:
Yeah, that's pretty close. If it's only 133554 km, that's about 0.445 light seconds.
Click to expand...
Is that a record perhaps? :)

EDIT: according to ED Astrometrics, it isn't... The lowest semi-major axis for Class V Gas Giant is 0.000100006129AU, which translates to roughly 15k km...
 
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