EPIC 201498078 - Eccentric super-Neptune

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EPIC 201498078 - Eccentric super-Neptune

Post by Sirius_Alpha on 11th June 2018, 8:55 pm

EPIC 201498078b: A low density Super Neptune on an eccentric orbit
https://arxiv.org/abs/1806.04073

We report the discovery of EPIC 201498078b, which was first identified as a planetary candidate from Kepler K2 photometry of Campaign 14, and whose planetary nature and orbital parameters were then confirmed with precision radial velocities. EPIC 201498078b is half as massive as Saturn (MP=0.179±0.021 MJ), and has a radius of RP=0.840±0.011 RJ, which translates into a bulk density of ρP=0.37±0.05 g cm−3. EPIC 201498078b transits its slightly evolved G-type host star (M⋆=1.105±0.019 M⊙, RP=1.669±0.022 R⊙) every 11.63364±0.00010 days and presents a significantly eccentric orbit (e=0.420±0.034). We estimate a relatively short circularization timescale of 1.8 Gyr for the planet, but given the advanced age of the system we expect the planet to be engulfed by its evolving host star in ∼1 Gyr before the orbit circularizes. The low density of the planet coupled to the brightness of the host star (J=9.4) makes this system one of the best candidates known to date in the super-Neptune regime for atmospheric characterization via transmission spectroscopy, and to further study the transition region between ice and gas giant planets.

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Re: EPIC 201498078 - Eccentric super-Neptune

Post by Drix on 12th June 2018, 12:44 pm

Sirius_Alpha wrote:EPIC 201498078b: A low density Super Neptune on an eccentric orbit


We report the discovery of EPIC 201498078b, which was first identified as a planetary candidate from Kepler K2 photometry of Campaign 14, and whose planetary nature and orbital parameters were then confirmed with precision radial velocities. EPIC 201498078b is half as massive as Saturn (MP=0.179±0.021 MJ), and has a radius of RP=0.840±0.011 RJ, which translates into a bulk density of ρP=0.37±0.05 g cm−3. EPIC 201498078b transits its slightly evolved G-type host star (M⋆=1.105±0.019 M⊙, RP=1.669±0.022 R⊙) every 11.63364±0.00010 days and presents a significantly eccentric orbit (e=0.420±0.034). We estimate a relatively short circularization timescale of 1.8 Gyr for the planet, but given the advanced age of the system we expect the planet to be engulfed by its evolving host star in ∼1 Gyr before the orbit circularizes. The low density of the planet coupled to the brightness of the host star (J=9.4) makes this system one of the best candidates known to date in the super-Neptune regime for atmospheric characterization via transmission spectroscopy, and to further study the transition region between ice and gas giant planets.

Hello,

I'm new and sorry for my bad english.

I don't know if you know this website : exoplanet .eu

But when I look an exoplanet profil for example the profil of epic 201498078 b, there are some things that I do not understand.
The first thing is : what represents Mass * sin (i) (I don't understand why we can't calculate this because we have the Mass and the Inclination)
Thank you in advance  Very Happy .
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Re: EPIC 201498078 - Eccentric super-Neptune

Post by Sirius_Alpha on 12th June 2018, 8:53 pm

Welcome Smile

Yeah we're all pretty familiar with exoplanet.eu. It was among the first websites that kept track of exoplanets.

Mass * sin (i) is much more important when you have a planet detected by Doppler spectroscopy -- where we don't know the inclination of the orbit, and thus the m sin i represents a lower limit to the true mass of the planet. For transiting planets, the papers that announce their discoveries rarely report m sin i, since the inclination is known. If that's the case, presumably exoplanet.eu does not list the m sin i.

That being said, exoplanet.eu has occasionally been rather sloppy with this sort of stuff.

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Re: EPIC 201498078 - Eccentric super-Neptune

Post by Drix on 13th June 2018, 5:54 am

Sirius_Alpha wrote:Welcome Smile

Where we don't know the inclination of the orbit, and thus the m sin i represents a lower limit to the true mass of the planet. For transiting planets, the papers that announce their discoveries rarely report m sin i, since the inclination is known. If that's the case, presumably exoplanet.eu does not list the m sin i.


Thank you for your answer. Sorry, but I still don't understand why we can calculate this with radial velocity method because we don't have the inclination of the orbit.
Indeed, in Mass * sin (i), i represents the inclination, no? So how can we calculate Mass * sin (i) without i ?

For example, for HD 147379 b we have Mass*sin(i) but we don't have the inclination. How it is possible? Surprised

Thank you very much Very Happy
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Re: EPIC 201498078 - Eccentric super-Neptune

Post by Sirius_Alpha on 13th June 2018, 7:54 pm

For a transiting planet, we know the inclination of the orbit from the transit. A planet with an inclination of 90° will always transit, for example.

For HD 147379 b, since it does not transit the star, we don't know the inclination. Exoplanet.eu records a mass value for the planet, but this is not actually known.

As a counterexample: The outer two planets at Upsilon Andromeda. We know their inclinatons (from a different method), but they don't transit. So their inclination is much less than 90°, and their true mass is very different from their m sin i.

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Re: EPIC 201498078 - Eccentric super-Neptune

Post by Drix on 13th June 2018, 8:20 pm

Ok ok I understand all of what you said here but yesterday you said :
"Mass*sin (i) is much important when you have a planet detected by Doppler spectroscopy where we don't know the inclination of the orbit"

So,
For Doppler spectroscopy they don't know i (the inclination),  so, I still don't understand how Mass*sin (i) can be calculated. Indeed, they need i for calculate Mass*sin (i) no ?

I'm so sorry :/ Thank you so much again Wink


Last edited by Drix on 14th June 2018, 3:38 pm; edited 1 time in total
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Re: EPIC 201498078 - Eccentric super-Neptune

Post by Sirius_Alpha on 13th June 2018, 8:54 pm

The value of m sin i is determined by the amplitude of the radial velocity signal in the star. It represents the true mass of the planet multiplied by the sine of the inclination. Neither of which are known in the case of a planet that has only been detected by radial velocity. Let's take the example of HD 147379 b. It has an m sin i = 22 Earth-masses. Here, m is the true mass of the planet, which is unknown If the planet's inclination is 60 degrees, then the planet's true mass is solved for algebraically by dividing both sides by sin(60°), giving m = 25 Earth-masses. Or if the inclination is 5 degrees, then the true mass is solved by dividing both sides by sin(5°) giving m = 253 Earth-masses.

So when you see m sin i, think of this as a lower limit to the true mass of the planet, with the planet's actual mass depending on whatever i is. We don't know anything about the mass of the planet or the inclination of the orbit, other than what you get when you divide whatever that unknown mass is by the sine of the (unknown) inclination. It's like if I told you the area of a rectangle was 12. You couldn't tell me what the length and width were, but you know that whatever L and W are, when you multiply them together you get 12.

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Re: EPIC 201498078 - Eccentric super-Neptune

Post by Drix on 14th June 2018, 4:51 pm

Thank you so much for your help I think that I understand.
Do you know the mathematical formula to calculate the value of Mass*sin(i) ?
😊
You said that I ntheeed the amplitude of the radial velocity signal in the star (Is it the value of K (in m.s^-1) in RV parameters of the Cornell University PDF?
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Re: EPIC 201498078 - Eccentric super-Neptune

Post by Sirius_Alpha on 14th June 2018, 9:44 pm

The equation for the radial velocity amplitude, K, can be found on the NASA Exoplanet Archive.
https://exoplanetarchive.ipac.caltech.edu/docs/poet_calculations.html

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Re: EPIC 201498078 - Eccentric super-Neptune

Post by Drix on 16th June 2018, 7:12 pm

Sirius_Alpha wrote:The equation for the radial velocity amplitude, K, can be found on the NASA Exoplanet Archive.

Ok thank you, but I want to calculate the mass planet. So, we have already the radial velocity amplitude or not? If we have already K, can we resolve the equation to find Mp * sin(i) ?
Very Happy

For example, the Mass display on exoplanet. eu is calculate thanks to this equation or an other equation?
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Re: EPIC 201498078 - Eccentric super-Neptune

Post by Sirius_Alpha on 16th June 2018, 9:00 pm

Yes. You can determine Mp sin i from the equation on that site.
https://exoplanetarchive.ipac.caltech.edu/docs/poet_calculations.html


Exoplanet.eu seems to just take the Mp sin i values from the discovery paper.

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Re: EPIC 201498078 - Eccentric super-Neptune

Post by Drix on 19th June 2018, 1:39 pm

Omg I finally found the mass of exoplanets thank you Very Happy Very Happy

But, sometimes we don't have K, for instance for Imaging method they found the Mass but they don't have K, do you know how calculate the Mass without K? like here:
http://exoplanet.eu/catalog/11_oph_b/
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Re: EPIC 201498078 - Eccentric super-Neptune

Post by Stalker on 19th June 2018, 2:04 pm

Drix wrote:Omg I finally found the mass of exoplanets thank you Very Happy Very Happy

But, sometimes we don't have K, for instance for Imaging method they found the Mass but they don't have K, do you know how calculate the Mass without K? like here:
http://exoplanet.eu/catalog/11_oph_b/

With models of cooling. You know the temperature of the "planet", you know the age of the system (because you assume it's the age of the star) and you know that a more massive substellar object is cooling (and faint) slower than a light one.

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Re: EPIC 201498078 - Eccentric super-Neptune

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