# A couple of questions

## A couple of questions

Here I have some questions for world building:

1 - Could eccentric and highly eccentric exoplanets (e>0.6) retain stable satellites? If yes, which mass upper limit?

2 - How to get planetary temperatures knowing planet's mass and radius and host star's luminosity?

Thank you for the attention

1 - Could eccentric and highly eccentric exoplanets (e>0.6) retain stable satellites? If yes, which mass upper limit?

2 - How to get planetary temperatures knowing planet's mass and radius and host star's luminosity?

Thank you for the attention

**Edasich**- dM star
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Location : Tau Ceti g - Mid Latitudes

Registration date : 2008-06-02

## Re: A couple of questions

1) Yes. The gravitational hill sphere of the planet will decrease as the planet gets closer to the star. The moon system's size will thus be determined by the planet's closest distance from the star.

HD 80606 b probably doesn't have any moons, because at its closet from the star, it's hill sphere is extremely small. If Earth were in an orbit identical to HD 80606, Earth would be unable to hold on to its moon as Earth got too close to the star. The only moons a planet will be able to have are those that it can hold on to when the planet is at its closest to the star.

As for the mass of the moons, that continues to remain a field of continuing study I'm sure. I wouldn't expect eccentric Jovians to have moon systems too different from non-eccentric Jovians.

However there is an issue with how the planet got into its eccentric orbit in the first place. If planet scattering led to the eccentric orbit, then this may have pumped eccentricity into the orbits of any moons (of both planets).

2) Planet temperature isn't a function of planet mass unless your planet is near the deuterium burning limit, or very young. I'm not sure that the planet's radius is a factor either.

I asked about this very thing earlier in the thread about planetary insolation, but no one replied, lol.

HD 80606 b probably doesn't have any moons, because at its closet from the star, it's hill sphere is extremely small. If Earth were in an orbit identical to HD 80606, Earth would be unable to hold on to its moon as Earth got too close to the star. The only moons a planet will be able to have are those that it can hold on to when the planet is at its closest to the star.

As for the mass of the moons, that continues to remain a field of continuing study I'm sure. I wouldn't expect eccentric Jovians to have moon systems too different from non-eccentric Jovians.

However there is an issue with how the planet got into its eccentric orbit in the first place. If planet scattering led to the eccentric orbit, then this may have pumped eccentricity into the orbits of any moons (of both planets).

2) Planet temperature isn't a function of planet mass unless your planet is near the deuterium burning limit, or very young. I'm not sure that the planet's radius is a factor either.

I asked about this very thing earlier in the thread about planetary insolation, but no one replied, lol.

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**Sirius_Alpha**- Admin
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## Re: A couple of questions

Thank you very much, SiriusAlpha

**Edasich**- dM star
- Number of posts : 1529

Location : Tau Ceti g - Mid Latitudes

Registration date : 2008-06-02

## Re: A couple of questions

Actually there are 4 formulas to get the temperature of a planet. First, you have to calculate to luminosity of the star with this formula: L= 4*pi*R^2sigmaT^4 (where sigma= 5.67*10^-8; T is the star's temp in Kelvin and R is the radius in meters). After you have to know the absolute magnitude of the star. Normally there's a formula but I never found it, so I use an alternative. Take the luminosity value of the star and divide it with the luminosity value of the Sun (which is 3.896*10^26 watts). Then you get a number and, making a comparison with the HR diagram, you get the absolute magnitude.

Now, with the absolute magnitude, you can calculate the albedo of the planet with this formula: A= ((1329*10^(-H/5))/D)^2; where A is the albedo, H is the absolute magnitude and D is the diameter of the planet in meters. But you get a number with a lot of zeros after the coma, from 7 to 15 in some cases (the albedo range is 0.01 - 1). To get a number in this range you just have to multiply it.

And the last formula, the temperature of the planet. With T= ((L(1-A)/(16*pi*sigma*D^2))^(1/4), where T is the temperature of the planet in Kelvin, L is the star's luminosity, A is the albedo of the planet, sigma = 5.67*10^-8 and D is the semi-major axis in meters.

Now you get everything to calculate the temperatures of your planets. If you have created tidally locked planets, their temperatures will, maybe, be above the temp of their stars (yes theorically it's possible !).

Bye

Sedna

Now, with the absolute magnitude, you can calculate the albedo of the planet with this formula: A= ((1329*10^(-H/5))/D)^2; where A is the albedo, H is the absolute magnitude and D is the diameter of the planet in meters. But you get a number with a lot of zeros after the coma, from 7 to 15 in some cases (the albedo range is 0.01 - 1). To get a number in this range you just have to multiply it.

And the last formula, the temperature of the planet. With T= ((L(1-A)/(16*pi*sigma*D^2))^(1/4), where T is the temperature of the planet in Kelvin, L is the star's luminosity, A is the albedo of the planet, sigma = 5.67*10^-8 and D is the semi-major axis in meters.

Now you get everything to calculate the temperatures of your planets. If you have created tidally locked planets, their temperatures will, maybe, be above the temp of their stars (yes theorically it's possible !).

Bye

Sedna

**Sedna**- Planetary Embryo
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