[Space] Trojan Objects - Page 3

AlexanderHowl · 03-15-2018, 12:38 PM

I made a mistake with the luminosity of the brown dwarf, it should be 0.002 rather than 0.0119 at 5 billion years. The distance from the brown dwarf for a habitable planet in that system would be 0.2 AU rather than 0.4 AU. Everything else is correct though.

The warmth of the planet would be Terran due to the cumulative effect of the two sources of heat on the blackbody radiation of the planet. The G2V star would supply 139K while the brown dwarf would supply 132K (the two sources of heat are cumulative because the planet does not orbit both stars), meaning that the total backbody radiation would average 271K (having two sources of energy from two different directions means that the planet cannot efficiently release excess heat through radiative cooling, meaning that the temperature reaches much higher levels than if the star and brown dwarf acted as one source). If the G2V and the brown dwarf were close companions, their luminosity would be added together because they would be approximately the same distance from planet, but the way that the orbits work, they are separate sources of heat.

If you put the brown dwarf at 1 AU and the planet at 0.05 AU, the planet would have two sources of heat that would each cause a temperature of 278K, meaning that the total blackbody temperature would be 556K by my calculations. You could technically place the planet at 0.2 AU from the brown dwarf, but the blackbody temperature would still be 410K by my calculations. Due to the heat contributed by the brown dwarf, the planet must be further away from the G2V star than the Earth.

The illumination of the surface would be much less than that of the Earth, but that is not necessarily a bad thing. While photosynthesis would be less active, it would not be impossible, though I am sure that the local equivalents of plants would probably evolve to use heat as well as light for photosynthesis. If not, it would probably be an Ocean standard planet.

Anthony · 03-15-2018, 12:45 PM

Quote:

Originally Posted by AlexanderHowl

The warmth of the planet would be Terran due to the cumulative effect of the two sources of heat on the blackbody radiation of the planet. The G2V star would supply 139K while the brown dwarf would supply 132K

It doesn't work that way. You add up luminosity first, then compute blackbody temperature from the total. Alternately, add up the fourth powers of the temperatures, and take the fourth power of the result.

AlexanderHowl · 03-15-2018, 01:17 PM

No, not really, since distance (D) is a more important factor than luminosity (L) for calculating temperature (T). Since T goes up by the fourth power of L and down by the square of D, D is always going to be more important than L for the purpose of calculating T. For example, a star needs to have a L of 16 to double the T of a planet at 1 AU. Alternatively, a planet can just double T by orbiting a star with a L of 1 at 0.25 AU.

The blackbody temperature calculation in GURPS Space only calculates the blackbody temperature based on the L of the primary star of the system. In the case of adding L, it would only work if both sources of L are equal distances and an equivalent direction from the planet (the planet possesses a night side that allows them to cool by radiating heat into space). Without a consistent night side, the assumptions of blackbody temperature change because the night side receives heat (rather than light and heat) from the brown dwarf.

You would have the same effect if you had two G2V stars with a separation of 10 AU and each had a planet at 1 AU. When the planets are directly between the two stars (with a night side illuminated by the more distance star), their blackbody temperature would increase by 88 K. As they continued their orbit, however, they would become more efficient at radiating access thermal energy from the two stars until, at opposition, they would add the effective luminosity of the stars together.

The same thing would apply to the planet around the brown dwarf as well, though optimal cooling on the nightside would have a shorter duration because of its shorter orbit of 58 days. The summers would be temperate, but the winters would be brutal since the effective temperature would drop due to the efficient cooling of the night side. Of course, this is assuming that a planet possesses an atmosphere to provide insulation. If not, the temperature would drop even further.

Anthony · 03-15-2018, 02:02 PM

Quote:

Originally Posted by AlexanderHowl

No, not really, since distance (D) is a more important factor than luminosity (L) for calculating temperature (T). Since T goes up by the fourth power of L and down by the square of D, D is always going to be more important than L for the purpose of calculating T.

You're missing the point. You're assuming you can add together the blackbody temperature from two sources, and you can't. You have to add up the adjusted luminosity (L/D^2) from all sources, then take the fourth result of the sum.

RyanW · 03-15-2018, 02:43 PM

The proper blackbody temperature formula would be:

T=278*(L1/R1²+L2/R2²+L3/R3²...)^(1/4)

Where L# is the luminosity of each star and R# is the planet's mean distance from that star. You can't directly add the temperatures. You've got to add the luminous fluxes.

(Fluxes sounds so wrong)

AlexanderHowl · 03-15-2018, 03:17 PM

Where does it say in Space that you add the luminosity of both stars together when a planet is not orbiting a close binary? Even in the case of a close binary, there should be differences in temperature depending on the relative position of the stars (though the above formula should adequately address the differences in temperature caused by changes in star position). It would still not address the effects on the planet of losing the ability of the night side to radiate excess thermal energy though.

Anthony · 03-15-2018, 03:25 PM

Quote:

Originally Posted by AlexanderHowl

Where does it say in Space that you add the luminosity of both stars together when a planet is not orbiting a close binary?

Space doesn't address that situation. However, physics does.

Quote:

Originally Posted by AlexanderHowl

It would still not address the effects on the planet of losing the ability of the night side to radiate excess thermal energy though.

Remember, the day side radiates away thermal energy too; it just simultaneously absorbs it. It gets colder at night and warming during the day because of the difference in amount absorbed, not amount radiated away, and with a non-contact binary the net effect is that the total amount absorbed is the same, it just occurs over more than half of a planetary rotation (the expected situation for a planet orbiting a brown dwarf that is orbiting a larger star is being tide locked, so the side that's away from the dwarf will have a 'day' equal to one orbital period that is half dark, and the side that's towards the dwarf will have a day equal to one orbital period but never have a dark sky).

AlexanderHowl · 03-15-2018, 04:34 PM

Luminosity is the product of the square of the radius (relative to Sol radius) times the fourth power of the temperature (relative to Sol temperature). When plugged into the planetary blackbody temperature formula in Space, the result is that the fourth power of (planetary temperature x the square root of the orbital distance) is equal to the fourth power of star temperature × the square of star radius.

To simplify, the temperature of the planet (relative to Earth) is proportional to the temperature of the star (relative to Sol). While I am doing the conversions in my head because I am at the gym, the resulting formula should be T=(278 x [star temperature/Sol temperature] × [square root of (star radius/Sol radius)])/(square root of orbit in AU). I will double check it on paper when I get out of the gym. If anyone wants to check the accuracy of the conversions though, please feel free.

Assuming the conversions are correct, in order to gain the effect of a binary system, you need to use the temperatures of the stars (modified by the square root of the stellar radius relative to Sol), not the luminosity of the stars. Adding the luminosity of the stars together and then taking the fourth root creates an incorrect sum of temperatures, so the subsequent results will underestimate the increase in temperature for the orbiting planet caused by a second star. In essence, a binary star matters a lot for determining the temperature of a planet, even more so than I initially proposed.

Anthony · 03-15-2018, 04:52 PM

Quote:

Originally Posted by AlexanderHowl

To simplify, the temperature of the planet (relative to Earth) is proportional to the temperature of the star (relative to Sol).

Assuming the ratio of orbital distance to star radius is constant, sure.

Quote:

Originally Posted by AlexanderHowl

Assuming the conversions are correct, in order to gain the effect of a binary system, you need to use the temperatures of the stars (modified by the square root of the stellar radius relative to Sol), not the luminosity of the stars.

And here's where you go off the rails.

The way blackbody temperature works is simple: it's the temperature at which the amount of heat radiated away is equal to the amount of stellar radiation absorbed. For a blackbody, the amount radiated away is 4pi*r^2*T^4*(blackbody constant), the amount absorbed is pi*r^2*stellar flux. If there are multiple sources, add together absorbed amounts. Stellar flux is (constant)*luminosity/r^2.

03-15-2018, 12:38 PM	#21
AlexanderHowl Join Date: Feb 2016	Re: [Space] Trojan Objects I made a mistake with the luminosity of the brown dwarf, it should be 0.002 rather than 0.0119 at 5 billion years. The distance from the brown dwarf for a habitable planet in that system would be 0.2 AU rather than 0.4 AU. Everything else is correct though. The warmth of the planet would be Terran due to the cumulative effect of the two sources of heat on the blackbody radiation of the planet. The G2V star would supply 139K while the brown dwarf would supply 132K (the two sources of heat are cumulative because the planet does not orbit both stars), meaning that the total backbody radiation would average 271K (having two sources of energy from two different directions means that the planet cannot efficiently release excess heat through radiative cooling, meaning that the temperature reaches much higher levels than if the star and brown dwarf acted as one source). If the G2V and the brown dwarf were close companions, their luminosity would be added together because they would be approximately the same distance from planet, but the way that the orbits work, they are separate sources of heat. If you put the brown dwarf at 1 AU and the planet at 0.05 AU, the planet would have two sources of heat that would each cause a temperature of 278K, meaning that the total blackbody temperature would be 556K by my calculations. You could technically place the planet at 0.2 AU from the brown dwarf, but the blackbody temperature would still be 410K by my calculations. Due to the heat contributed by the brown dwarf, the planet must be further away from the G2V star than the Earth. The illumination of the surface would be much less than that of the Earth, but that is not necessarily a bad thing. While photosynthesis would be less active, it would not be impossible, though I am sure that the local equivalents of plants would probably evolve to use heat as well as light for photosynthesis. If not, it would probably be an Ocean standard planet.

03-15-2018, 01:17 PM	#23
AlexanderHowl Join Date: Feb 2016	Re: [Space] Trojan Objects No, not really, since distance (D) is a more important factor than luminosity (L) for calculating temperature (T). Since T goes up by the fourth power of L and down by the square of D, D is always going to be more important than L for the purpose of calculating T. For example, a star needs to have a L of 16 to double the T of a planet at 1 AU. Alternatively, a planet can just double T by orbiting a star with a L of 1 at 0.25 AU. The blackbody temperature calculation in GURPS Space only calculates the blackbody temperature based on the L of the primary star of the system. In the case of adding L, it would only work if both sources of L are equal distances and an equivalent direction from the planet (the planet possesses a night side that allows them to cool by radiating heat into space). Without a consistent night side, the assumptions of blackbody temperature change because the night side receives heat (rather than light and heat) from the brown dwarf. You would have the same effect if you had two G2V stars with a separation of 10 AU and each had a planet at 1 AU. When the planets are directly between the two stars (with a night side illuminated by the more distance star), their blackbody temperature would increase by 88 K. As they continued their orbit, however, they would become more efficient at radiating access thermal energy from the two stars until, at opposition, they would add the effective luminosity of the stars together. The same thing would apply to the planet around the brown dwarf as well, though optimal cooling on the nightside would have a shorter duration because of its shorter orbit of 58 days. The summers would be temperate, but the winters would be brutal since the effective temperature would drop due to the efficient cooling of the night side. Of course, this is assuming that a planet possesses an atmosphere to provide insulation. If not, the temperature would drop even further.

03-15-2018, 02:43 PM	#25
RyanW ☣ Join Date: Sep 2004 Location: Southeast NC	Re: [Space] Trojan Objects The proper blackbody temperature formula would be: T=278*(L1/R1²+L2/R2²+L3/R3²...)^(1/4) Where L# is the luminosity of each star and R# is the planet's mean distance from that star. You can't directly add the temperatures. You've got to add the luminous fluxes. (Fluxes sounds so wrong) __________________ RyanW - Actually one normal sized guy in three tiny trenchcoats.

03-15-2018, 03:17 PM	#26
AlexanderHowl Join Date: Feb 2016	Re: [Space] Trojan Objects Where does it say in Space that you add the luminosity of both stars together when a planet is not orbiting a close binary? Even in the case of a close binary, there should be differences in temperature depending on the relative position of the stars (though the above formula should adequately address the differences in temperature caused by changes in star position). It would still not address the effects on the planet of losing the ability of the night side to radiate excess thermal energy though.

03-15-2018, 04:34 PM	#28
AlexanderHowl Join Date: Feb 2016	Re: [Space] Trojan Objects Luminosity is the product of the square of the radius (relative to Sol radius) times the fourth power of the temperature (relative to Sol temperature). When plugged into the planetary blackbody temperature formula in Space, the result is that the fourth power of (planetary temperature x the square root of the orbital distance) is equal to the fourth power of star temperature × the square of star radius. To simplify, the temperature of the planet (relative to Earth) is proportional to the temperature of the star (relative to Sol). While I am doing the conversions in my head because I am at the gym, the resulting formula should be T=(278 x [star temperature/Sol temperature] × [square root of (star radius/Sol radius)])/(square root of orbit in AU). I will double check it on paper when I get out of the gym. If anyone wants to check the accuracy of the conversions though, please feel free. Assuming the conversions are correct, in order to gain the effect of a binary system, you need to use the temperatures of the stars (modified by the square root of the stellar radius relative to Sol), not the luminosity of the stars. Adding the luminosity of the stars together and then taking the fourth root creates an incorrect sum of temperatures, so the subsequent results will underestimate the increase in temperature for the orbiting planet caused by a second star. In essence, a binary star matters a lot for determining the temperature of a planet, even more so than I initially proposed.