English to Math

[MacroQuantum]

At this site:

http://www.ess.sunysb.edu/fwalter/AST101/habzone.html

the author makes some fairly innocent statements that should arrive at a quick formula. However, my english to math translations have never been good.

What I am looking for should be two formulas, one to determine the inner and outer diameter. Would anybody care to take a shot at making the translations?

[Sisyphus]

The author says T^4 = L/(D^2), where T is planetary temperature, L is solar luminosity, and D is orbital radius. This is more or less what I think you want, already. Rearrange to get D = sqrt[L/(T^4)], or D = sqrt(L)/T^2, if it helps. Thus you have radius as a function of temperature. Plug in the upper and lower extremes for temperature to get the inner and outer radii for the limits of the habitable zone. (Diameters, obviously, are just twice the radii.)

[MacroQuantum]

The author says T^4 = L/(D^2), where T is planetary temperature, L is solar luminosity, and D is orbital radius. This is more or less what I think you want, already. Rearrange to get D = sqrt[L/(T^4)], or D = sqrt(L)/T^2[/b'], if it helps. Thus you have radius as a function of temperature. Plug in the upper and lower extremes for temperature to get the inner and outer radii for the limits of the habitable zone. (Diameters, obviously, are just twice the radii.)

 

Does this article basically imply that solar luminosity, like planetary luminosity is equal to [surface] Temperature^4? (A site that details the approximate surface temperatures of the various types of stars I already have, but I don't beleive the site specs the luminosity.)

And we start talking about diameters, or radii, are we talking the planet should be X number of planetary raddii, or solar radii, from the sun? [i'm getting lost again in possibilities!] Or do I still not have a clear picture?

I think given those two items, I should be able to plot the placement of a planet that should be within the describeded habitable zone. Since most scientist agree that liquid water is required for life. Thus the upper and lower temperature range of the planet would be the temperature at which water would remain a liquid (at the equator).

And obviously, thanks for your help.

[Martin]

At this site:

http://www.ess.sunysb.edu/fwalter/AST101/habzone.html

the author makes some fairly innocent statements that should arrive at a quick formula. However' date=' my english to math translations have never been good.

What I am looking for should be two formulas, one to determine the inner and outer diameter. Would anybody care to take a shot at making the translations?[/quote']

 

I looked at the page and I think the passage you want to "translate" is:

 

"You determine the temperature by equating the planetary luminosity (proportional to its temperature raised to the fourth power, T4) to the solar irradiance (L/D2, where L is the solar luminosity and D is the distance to the Sun). The distance at which a planet is at temperature T is proportional to 1/T2. Merely plug in the values of the upper and lower temperature to get the radii of the inner and outer radii of the habitable zone."

 

he has left out a constant that physicists typically write with a "sigma" called Stefan-Boltzmann constant.

 

there is a question of what units you want to use. to get any good from a formula like this you need to decide on units.

also I am curious what are the two temperatures you have in mind?

 

I suggest you use METRIC units and the absolute temp scale KELVIN

 

Ice melts at 273 K, so you might pick your two temps to be 260 K and 320 K, for instance.

 

IF YOU CHOOSE TO WORK IN METRIC THEN THE LUMINOSITY OF THE SUN WILL BE RATED IN WATTS. just like a big lightbulb.

 

IIRC the luminosity of our sun is about 4 x 10²⁶ watts.

any sun-like star you are studying will probably have some comparable raw power output, like roughly speaking somewhere 2 to 8 times 10²⁶ watts.

 

IIRC our distance from our sun is such that the power PER SQUARE METER is about 1300 watts per square meter. this power per unit area is sometimes called "the solar constant" and you can look it up on the web.

 

[math]\frac{L}{4\pi D^2} = 1300 W/m^2[/math]

 

I am not going to crank out the final formula for you. Somebody may, but I think it is better to advance one step at a time. Is what I just wrote understandable?

 

What about this?

 

[math]4\pi D^2 = \frac{L}{1300 W/m^2}[/math]

 

Can you plug in the sun's power (luminosity) and solve this for the distance?

 

[math]4\pi D^2 = \frac{4 \times 10^{26} W}{1300 W/m^2}[/math]

 

[math]12 D^2 = 3 \times 10^{23} m^2[/math]

 

[math] D^2 = \frac{1}{4} \times 10^{23} m^2[/math]

 

[math] D^2 = \frac{10}{4} \times 10^{22} m^2[/math]

 

[math] D = 1.5 \times 10^{11} m[/math]

 

this is just an approximate calculation, the real luminosity is more like 3.7 instead of 4 IIRC, but saying 4 keeps the numbers simple. the real "solar constant" might be more like 1350 W/sq. meter. but this is the no-frills treatment.

 

====================

 

This much gives you a handle on the habitable zone if you can specify rough limits on the "solar constant"-----like suppose you decide the earth would be habitable with the power per unit area being anywhere from 1000 to 1500 watts per square meter.

 

then you can just plug in for L (the luminosity of the star) and solve for D

with these simple assumptions

the closest "habitable" distance will be given by

 

[math]4\pi D^2 = \frac{L}{1500 W/m^2}[/math]

 

the farthest "habitable" distance will be given by

 

[math]4\pi D^2 = \frac{L}{1000 W/m^2}[/math]

 

any of several people could jump in here and give you the final formula handsomely gift wrapped, with temperature limits in Kelvin and the Stefan-Boltzmann "sigma" constant and all. But this is as far as I go on day one. I suggest you play around with different stellar luminosities (watts) and different values of the "solar constant" (watts per sq. meter) and see what distances you get

[Martin]

under reasonable assumptions the equilibrium temp of a ball

 

 

(which has surface area to radiate heat away which is 4 times its cross section area)

 

is given by the formula that

 

4 sigma T⁴ = solar constant or level of illumination

 

this is a form of the stefan-boltzmann "fourth power" law.

 

So you CAN relate average temperature to the power per sq. meter of sunlight. In fact it is simple!

 

But I advise you to take it gradually

[Martin]

MacQ seems to have gone, so I will just finish up

 

we could make some of the numbers and formulas LOOK simpler by using special astronomy units but the most straightforward thing is to use familiar metric

 

and then sigma = 5.7 x 10^-8 watts per sq. meter per Kelvin⁴

 

and the fourthpower law says, for a generic (unreflective, blackbody type) surface of TEMPERATURE T that the power radiated per unit area---the radiant heat it gives off in watts per sq. meter---is

 

sigma T⁴

 

example so if the temp is 300 Kelvin, you raise that to the fourth and get 81 x 10⁸ Kelvin⁴

and you multiply by sigma, or

5.7 x 10^-8 watts per sq. meter per Kelvin⁴

 

and the powers of ten cancel, and Kelvin⁴ terms cancel

and you are left with a clean

 

5.7 x 81 x 10^-8 watts per sq. meter

or about 460 watts per sq. meter

 

the rough idea is that if the surface of a planet GETS on average 460 watts per sq. meter, then (neglecting cloud reflectance and loss like that, and neglecting greenhouse) it will tend to have a surface temperature of 300 K because that is just warm enough so that it RADIATES OFF 460 watts per sq. meter-----so it stays in balance

 

But there is a catch. the radiating surface of a planet is 4 times more area than the crosssection intercepting the sunlight

 

so to keep the planet at 300 K equilibrium temp, the incoming sunlight has to be 4 times the wattage per sq. meter of cross section

 

460 x 4 = 1840, so the "solar constant" for that planet would have to be 1800 some watts per sq. meter. (because the incoming SPREAD OUT over the whole surface including the nightside, just equals the 460 watts outgoing)

 

the equilibrium equation says if L is the wattage of the star, and D is the distance, then the temperature is determined by

 

incoming = outgoing

 

L/(4 pi D²) = 4 sigma T⁴

[MacroQuantum]

So let me see if I got this right. Given a G5 type star which has an average temperature of 5770K (according to http://curriculum.calstatela.edu/courses/builders/lessons/less/les1/StarTables.html) and has a radius of .893 earth suns or 69,600,000,000m or 69,600,000km then

solar lumonisity = 4 * sigma * 5770^4

(Although I'm not exactly sure what sigma represents here? If I'm not mistaken though sigma should be representing the solar constant which is 1.37 kW/m^2, the results being:

solar lumonisity = 4 * 1.37 kW/m^2 * 5770^4

But I'm not sure how to simply this into something that I can work.)

Given Sisyphus's formula then it should be possible to say that our planet (at the equator since that is the closet the planet comes to the sun) should have a temperature of 273K to 360K (since we don't actually want it turning into steam) which means that:

Inner = (SQRT(4*1.37kW/m^2*5770^4)/273^2)*Albedo (or 0.37 for earth-like)

Outer = SQRT(4*1.37kW/m^2*5770^4)/360^2

 

Okay how far off am I?

[Martin]

you are way off, in certain details

 

and this may be partly due to a mismatch of our styles----I am saying things in the wrong words for you----or I am going too slow with some things and too fast with others.

 

but on the other hand it is great that you want to calculate, and can calculate. so I am saying great! great! even tho some things are wrong.

 

Swansont (tom swanson) may jump in and help out. he is very good.

 

We could make you a special, easy to use, formula with ASTRONOMY UNITS. Like, solar luminosity is ONE SOLAR LUMINOSITY unit. Instead of 3.7 x 10²⁶ watts

 

Would you like this? Or would you prefer to stick to familiar metric units.

If you work in metric it is messier, but the physical learning can carry over to other non-astronomy situations! Other things glow. Life is not just planets.

 

I am going to continue talking to you as if you have decided to continue working in metric. Tell me if you change your mind.

[Martin]

WHOA! that website you are using has everything WAY SIMPLIFIED so you do not need anything I was telling you!

I think they do a great job of setting everything up with special Astronomy Units-----like distance in AU and star radius in Solar Radii.

 

 

Let's work with that. they have set everything up to be a no-brainer.

 

HERE IS THE BASIC FORMULA.

 

in the case of the sun, our distance is 215 times the sun's radius.

 

that means the SIDEWALK MAX, the hottest an absorbant surface can get in direct uninterupted sunlight at our distance from the sun is EQUAL TO THE TEMPERATURE ON THE SURFACE OF THE SUN DIVIDED BY THE SQUARE ROOT OF 215.

 

So get a calculator and figure it out, macroQ. Sun temp is 5770 K, so divide that by sqrt(215). This is one of these miracle shortcut tricks they give you when they dont want to teach you a couple of pages out of a college physics text.

 

Now you have a very powerful tool because THIS WORKS FOR ANY STAR.

 

say the star radius is twice Sol, and the planet distance is 3 AU (3 times earth's) then the planet distance is 3/2 x 215 times the radius of the star.

 

do you get this? go in small steps. do you see that?

 

OK that is 320, now suppose the star surface temp is 6500 K

then SIDEWALK MAX (the "equilib temp" of a flat surface facing direct light) is going to be 6500 K divided by the sqrt of 320.

 

Suddenly you are very powerful. You can calculate the equilibrium temp at any distance from any star.

 

You never had to use metric units or learn about the Stefan-Boltzmann constant (what you are doing uses the fourth-power law implicitly, but you skip over it)

 

Of course the average temp on the planet is going to be less than the sidewalk max. But that is all right, it is proportional, like maybe it is always the sidewalk max divided by 1.414. That is a secondary detail.

In the real world it depends on other things like cloud reflectivity, albedo, greenhouse. But dont worry it is already good to be able to calculate the equilibrium temperature of an ideal flat surface in uninterrupted light.

 

Do you have questions?

[Martin]

...http://curriculum.calstatela.edu/courses/builders/lessons/less/les1/StarTables.html[/url'])

 

you have picked a good table. it streamlines the whole calculation.

FORGET what I told you at first, about metric units and StefanBoltzmann sigma.

 

use their method.

 

Look at an example-----F0 star with temp 7200 K and radius 1.64 solar.

 

much bigger and hotter than Sol.

 

because it is hotter by a factor of 72/57 = 1.25

to find a comfortable distance from it multiply 1.64 x 1.25²

which is about 2.5-----so that planet has to be 2.5 AU from the star!

 

this has really been simplified

 

WOW, just take the F0 star data (7200 K and 1.64 solar radius) and guess a distance, like 2.5 AU (2.5 times earth-sun distance) and

DIVIDE 7200/5770 BY THE SQUARE ROOT OF 2.5/1.64

and you get around 1.01 which means about 1 percent hotter than earth.

 

I am rushed and cant neaten this up. But I think we now have a streamlined formula. for the F0 star data, if you want a planet that is like 0.9 of earth temp, then you solve this

 

7200/5770 x sqrt( 1.64/D) = 0.9

 

that will give you distance D in the astronomer's AU units.

 

D = (7200/5770)² x 1.64/ 0.9²

[MacroQuantum]

That being my general point, and why I was looking at that other page (http://www.ess.sunysb.edu/fwalter/AST101/habzone.html). My goal is to find the habitable range. This way I can set two planets into that range (if possible). Then I can determine the temperature of both of those planets (and hopefully some other minor details) and then continue on about my way.

The next step would be to see if I could get some sort of concept of the visible here.

Perhaps I have not stated my goal correctly, or perhaps I don't follow what you are saying.

[MacroQuantum]

Sisyphus has given me a straight forward formula that will determine the diameter that I am looking for. However, in that formula, what I am lacking is Luminosity. So the question (that I understand) becomes how do I determine the Solar Luminosity given the facts that I have (Star temperature, mass, radius)?

But every time I try to figure out what Martin is trying to tell me I get lost. Anyone want to clarify?

[Martin]

But every time I try to figure out what Martin is trying to tell me I get lost. Anyone want to clarify?

 

I'd be glad to clarify

look at the "bolometric luminosity" column in this table

http://curriculum.calstatela.edu/courses/builders/lessons/less/les1/StarTables_Z.html

 

for example for an F8 star it is 2.10

call that L' date=' it means that the star is putting out 2.1 times as much power as Sol

 

Now pick a temperature, referred to average equilibrium Earth temp. Say you are willing for it to be 10 percent warmer. So T = 1.1

 

the orbit radius R =sqrt L/T[sup']2[/sup] = sqrt 2.1 divided by 1.1²

 

have to go, back later.

[MacroQuantum]

What? Radius = SQRT(L)/T^2

Radius = SQRT(2.1)/1.1^2

Radius = 1.45/1.21

Radius = 1.2 (Whats?)

 

Okay, let's see if I can do this right now. According to a reference I found the bolometric luminosity of earth's sun = 3.83 * 1026W or 3929.58. Also the temperature range that water is liquid at 273K to 360K (actually 373). Last, but not least, the albedo of an earth type planet would be equal to 0.37. Assuming that I am working with a G5 type star, this should work:

Inner = (SQRT(Luminosity * Percentage)/Temperature^2)*Albedo

Inner = (SQRT(3929.58*0.79)/273^2)*0.37

Inner = (SQRT(3104.37)/74529)*0.37

Inner = (55.72/74529)*0.37

 

Am I making this harder than it should be? Because obviously this doesn't work, either.

This is starting to get real frustrating.

[Martin]

What? Radius = SQRT(L)/T^2

Radius = SQRT(2.1)/1.1^2

Radius = 1.45/1.21

Radius = 1.2 (Whats?)

...

 

AU

 

I think you did it right and got the correct answer.

 

I mentioned in my first post that there is a choice of what system of units.

If you choose to use metric, then you will get answers in meters or kilometers. And the simple formula needs modifying. I thought that did not work very well for you.

 

So I'm suggesting now that you DONT use metric units. Use the easy units that astronomers have devised.

 

So the answer comes out in AU

One AU, or "astronomical unit of distance" is about 93 million miles or roughly 150 million kilometers.

 

Is that OK with you? CONGRATULATIONS, you have already worked a problem and gotten the right answer.

 

Be sure you use this table of luminosities, because the are referred to the sun as standard

http://curriculum.calstatela.edu/courses/builders/lessons/less/les1/StarTables_Z.html

 

If you are going to use the simple formula and work in astronomy units, then do not use any table of luminosities that says "watts" because that is a tip-off that it is in metric terms----watt is a metric unit.

 

I am basing my response on the star data table you originally gave the link to----that is in the Cal State site.

 

See if you can do this one: you just found the size of the orbit around an F8 star that makes the temp 10 percent warmer than earth----that is the temp is 1.1 relative to earth as standard.

 

Now pick a G8 star. Your table says the bolo lum. is 0.66 of Sol

(only 2/3 the power output of our sun)

What size orbit will give you the same 1.1 temperature?

 

Same formula as before:

I quote your post but put in L = 0.66 for the new luminosity.

 

Radius = SQRT(L)/T^2

Radius = SQRT(0.66)/1.1^2

Radius = ..../1.21

Radius = .... AU

[MacroQuantum]

Okay, I understand what we are doing here. But I'm not sure of the reference, because the original article (http://www.ess.sunysb.edu/fwalter/AST101/habzone.html) suggests that it should be possible to determine a range, or corridor, where life would be expected to survive. However, working from this percentage of earth temp, I don't see how to generate that. That is why I made the attempt to convert from 'percentage of earth' to a more generic range.

Shouldn't it be possible to convert that luminosity into something that is not a percentage of earth?

[Martin]

Okay' date=' I understand what we are doing here. But I'm not sure of the reference, because the original article (http://www.ess.sunysb.edu/fwalter/AST101/habzone.html) suggests that it should be possible to determine a range, or corridor, where life would be expected to survive. However, working from this percentage of earth temp, I don't see how to generate that. That is why I made the attempt to convert from 'percentage of earth' to a more generic range.

Shouldn't it be possible to convert that luminosity into something that is not a percentage of earth?[/quote']

 

I dont think it is possible to give EXACT temperature bounds to the habitable zone.

 

also the situation is complicated by albedo and greenhouse.

 

you could have two identical stars and two planets each the same distance and

 

A: one planet could be habitable but the other uninhabitable because greenhouse makes it too hot.

 

B: one planet could be habitable and the other could be so reflective that it doesnt get the benefit of all the sunlight (just reflects it back into space) and is in a permanent ice-age

 

So you have to realize that the brightness of the sun and the distance from it is not the whole story!

 

so when people talk about "habitable zone" it is an APPROXIMATE idea

 

=================

 

that said, the earth average surface temp, last time I looked it up, is about 283 K (roughly 10 Celsius)

 

this is the average for night AND day surface temp all over the globe from equator to arctic

 

compare this to freezing 273 K and boiling 373 K.

 

You should pick what you want to call the habitable range (there is no one right answer). suppose you pick from 260 K to 330 K------endpoints that are below freeze (but equator might be nice) and uncomfy hot (but arctic might be nice)

 

then you see what the endpoints are RELATIVE TO EARTH namely you calculate 260/283

and 330/283

 

that gives two numbers 0.92 and 1.17 as endpoints of the range----I would round them to 0.9 and 1.2

 

so either 10 percent colder or 20 percent warmer

 

and then you can plug them into the formula you already know how to use.

 

that is good because astronomy tables often give the luminosity of stars relative to Sol.

[MacroQuantum]

Mmmmm. According to the article that I quoted in my last post, an earth-like planet has an albedo of 0.37. What exactly that means, I don't know, but what would happen then, is the inner radius should be, as a last step, multiplied by said albedo (I think). So that what you would have would be:

Inner=(SQRT(Luminosity)/Temperature^2)*Albedo

Inner=(SQRT(0.79)/0.92^2)*0.37

Inner=(0.89/0.85)*0.37

Inner=1.05*0.37

Inner=0.39 AU

 

While the article does make mention of the greenhouse effect, it actually doesn't give any sort of parameters, so I guess I will have to live with just the straight formula.

Outer=SQRT(0.79)/1.27^2

Outer=0.89/1.61

Outer=0.55 AU

 

Thus, with a G5 type star, and a desired temperature range of 260K to 360K, we have a range of 0.39AU to 0.55AU. But wait a moment. Earth should fall somewhere in that range, having an average temperature of 283K. So, working the formula once more....

 

Radius=SQRT(0.79)/1^2

Radius=0.89/1

Radius=0.89

 

Huh?

[MacroQuantum]

Mmmmm. According to the article that I quoted in my last post, an earth-like planet has an albedo of 0.37. What exactly that means, I don't know, but what would happen then, is the inner radius should be, as a last step, multiplied by said albedo (I think). So that what you would have would be:

Inner=(SQRT(Luminosity)/Temperature^2)*Albedo

Inner=(SQRT(0.79)/0.92^2)*0.37

Inner=(0.89/0.85)*0.37

Inner=1.05*0.37

Inner=0.39 AU

 

While the article does make mention of the greenhouse effect, it actually doesn't give any sort of parameters, so I guess I will have to live with just the straight formula.

Outer=SQRT(0.79)/1.27^2

Outer=0.89/1.61

Outer=0.55 AU

 

Thus, with a G5 type star, and a desired temperature range of 260K to 360K, we have a range of 0.39AU to 0.55AU. But wait a moment. Earth should fall somewhere in that range, having an average temperature of 283K. So, working the formula once more....

 

Radius=SQRT(0.79)/1^2

Radius=0.89/1

Radius=0.89

 

Huh?

 

I'm not sure why this posted twice but you see my problem. A temperature that should fall into that range, didn't.

[Martin]

Thus' date=' with a G5 type star, and a desired temperature range of 260K to 360K, we have a range of 0.39AU to 0.55AU. But wait a moment. Earth should fall somewhere in that range, having an average temperature of 283K. So, working the formula once more....

 

Radius=SQRT(0.79)/1^2

Radius=0.89/1

Radius=0.89

 

Huh?[/quote']

 

I left the albedo business out of the formula, to make it simple for starters.

 

If we are going to include albedo and greenhouse effects then we have to use a more complicated formula.

 

In the earth case, conveniently enough, the albedo and greenhouse effects approximately CANCEL each other! so it is OK for a rough calculation to simply ignore both

 

albedo is the amount of light reflected (like off top of clouds) back into space.

 

so if 0.37 , it means 37 percent of light is reflected

so the effective luminosity of sun is only 63 percent what it would otherwise be. so it makes us COLDER

 

but if we include that fine point then we should also include the greenhouse blanket effect of the atmosphere that makes us WARMER.

 

and the formula should be tinkered with to include these two.

 

since we dont know anything about the alien planets we might as well assume that they are like earth and that the two effects cancel out and so let's just keep it simple and ignore both.

 

but you must do as you please. I've gone about as far as I want with this. Others may jump in.

[MacroQuantum]

I left the albedo business out of the formula' date=' to make it simple for starters.

 

If we are going to include albedo and greenhouse effects then we have to use a more complicated formula.

 

In the earth case, conveniently enough, the albedo and greenhouse effects approximately CANCEL each other! so it is OK for a rough calculation to simply ignore both

 

albedo is the amount of light reflected (like off top of clouds) back into space.

 

so if 0.37 , it means 37 percent of light is reflected

so the effective luminosity of sun is only 63 percent what it would otherwise be. so it makes us COLDER

 

but if we include that fine point then we should also include the greenhouse blanket effect of the atmosphere that makes us WARMER.

 

and the formula should be tinkered with to include these two.

 

since we dont know anything about the alien planets we might as well assume that they are like earth and that the two effects cancel out and so let's just keep it simple and ignore both.

 

but you must do as you please. I've gone about as far as I want with this. Others may jump in.[/quote']

 

Okay, I can live with simple. So leaving out the albedo and greenhouse effects, with a temperature range of 260K to 360K, and a G5 star we get

 

Outer=SQRT(0.79)/0.92^2

Outer=0.89/0.85

Outer=1.05

 

and

 

Inner=SQRT(0.79)/1.27^2

Inner=0.89/1.61

Inner=0.55

 

Then Earth, at 0.89 does fall within that range. To convert those ranges into kilometers should just be a case of multiplying by 14,959,787.691.

Thank you for your patience Martin!

[Martin]

Okay' date=' I can live with simple. So leaving out the albedo and greenhouse effects, with a temperature range of 260K to 360K, and a G5 star we get

 

Outer=SQRT(0.79)/0.92^2

Outer=0.89/0.85

Outer=1.05

 

and

 

Inner=SQRT(0.79)/1.27^2

Inner=0.89/1.61

Inner=0.55

 

Then Earth, at 0.89 does fall within that range. To convert those ranges into kilometers should just be a case of multiplying by 14,959,787.691.

Thank you for your patience Martin![/quote']

 

thank you for yours!

I got off on the wrong foot, thinking we should do everything in metric units (and use the Stefan-Boltzmann fourth power radiation law)

which turned out to be a very bad idea. better to use Astronomical units and then convert in the end (as you say, by saying one AU is 150 million km---but careful with the decimal point, dont call it 15 million)

 

also I think you (anyone) benefits from actually calculating----using some of the calculator's battery up just trying numbers in the formulas. so I am very glad when i run into people willing to do that. as I see you doing in this case. good luck