# Theoretical parameters Gliese1214b

## Theoretical parameters Gliese1214b

I decided to think about the possibility of study of this planet.

By formula depth of secondary eclipse in optics = Аg * (Rp / a) ^ 2 obtained the following values:

5% 3 ppm
10% 6 ppm
20% 12 ppm
30% 18 ppm
40% 24 ppm
50% 30 ppm
60% 35 ppm
70% 42 ppm
80% 48 ppm
90% 54 ppm
95% 56 ppm

If we take the planet for an absolutely black body, we can calculate the depth of secondary eclipse at infrared wavelengths, as (Tp/T*)^4 * (Rp / R*)^2

200 Kelvins 0,2 ppm
300 Kelvins 1,1 ppm
400 Kelvins 3,6 ppm

Ie turns out that near the Green Line, a small planet in the main light through the reflected light, rather than its own thermal radiation.

Borislav
Jovian

Number of posts : 561
Registration date : 2008-11-12

## Re: Theoretical parameters Gliese1214b

Borislav wrote:If we take the planet for an absolutely black body, we can calculate the depth of secondary eclipse at infrared wavelengths, as (Tp/T*)^4 * (Rp / R*)^2

200 Kelvins 0,2 ppm
300 Kelvins 1,1 ppm
400 Kelvins 3,6 ppm

Ie turns out that near the Green Line, a small planet in the main light through the reflected light, rather than its own thermal radiation.

Although apparently this formula is wrong. If we substitute the parameters are known hot Jupiter, then we are also very small value. What is the correct formula for calculating the depth of secondary eclipse at infrared wavelengths for a small, cold planet?

Borislav
Jovian

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Registration date : 2008-11-12

## Re: Theoretical parameters Gliese1214b

Although this slide, write that correctly use it without the fourth degree for the temperature ratio planet and star.
http://planetquest.jpl.nasa.gov/NavigatorForum/documents/May17-2007/8-JWST-Clampin.pdf

Then

200 Kelvin 789 ppm
300 1185 ppm
400 1579 ppm

This depth is probably available Spitzer even one observation. Although the question of the correctness of the formula remains (formula only work for hot Jupiter?).

Borislav
Jovian

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Registration date : 2008-11-12

## Re: Theoretical parameters Gliese1214b

I think the equation Fp / Fstar = Ag (Rp / a)2 works for the visible light, since Ag is a visible light reflectivity coefficient.

If the planet radiates as a blackbody then your second equation might be more accurate. I would not expect Gliese 1214 to do so because it does not receive as much irradiation from its star.

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Sirius_Alpha

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## Re: Theoretical parameters Gliese1214b

In addition, counted on these two formulas of depth of secondary eclipse to another well-known super Earth - Corot-7b.

for Ag=5-95% in optics 0.7-14 ppm, ie apparently available to Kepler, or Hubble

for the infrared range for Tp=1000-2000 Kelvin 70-130 ppm, ie Spitzer is available for a few observations.

For comparison, Spitzer observations of the primary transit of Corot-7b

http://online.kitp.ucsb.edu/online/exoplanets_c10/charbonneau/

Borislav
Jovian

Number of posts : 561
Registration date : 2008-11-12

## Re: Theoretical parameters Gliese1214b

If I did the math right, Gliese 1214 b will receive about half the insolation from its star that Gliese 876 d received, and about a third as much as Gliese 436 b.

Gliese 436 b --- 43.5 x Earth
Gliese 876 b --- 33.4 x Earth
Gliese 1214 b -- 17.0 x Earth

Complicating the issue further is that Gliese 1214 is a much dimmer star. Could be very challenging to detect with Spitzer.

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Sirius_Alpha

Number of posts : 3662
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