damienpaul

2004-Jan-20, 04:37 PM

Okay, I have got to ask the question, at the risk of sounnding silly, what is a planet's albedo and how is it measured?

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damienpaul

2004-Jan-20, 04:37 PM

Okay, I have got to ask the question, at the risk of sounnding silly, what is a planet's albedo and how is it measured?

Bluewolf027

2004-Jan-20, 04:40 PM

A planets albedo is the ratio of the amount of solar radiation reflected from an object to the total amount incident upon it

Here try this

http://curriculum.calstatela.edu/courses/b...es3/albedo.html (http://curriculum.calstatela.edu/courses/builders/lessons/less/les3/albedo.html)

:D

Here try this

http://curriculum.calstatela.edu/courses/b...es3/albedo.html (http://curriculum.calstatela.edu/courses/builders/lessons/less/les3/albedo.html)

:D

Tiny

2004-Jan-20, 07:52 PM

Albedo means "White" <<< Physic class Lab ^^

Albedo is a Fraction of the sunlight falling upon a surface that is reflect back to the space. The albedo represents the average reflectivity over the entire visible surface. Hence it differs slightly from the reflectivity of different portions of it.

measure between 0 - 1...

Example: The moon has an albedo of 0.07 I think, Sunlight is 100% : as u calculate 7/100 = 0.07 this is the number that the sunlight reflect back to space from the moon, and 97% of the sunlight were absorb by the Moon...

Albedo is a Fraction of the sunlight falling upon a surface that is reflect back to the space. The albedo represents the average reflectivity over the entire visible surface. Hence it differs slightly from the reflectivity of different portions of it.

measure between 0 - 1...

Example: The moon has an albedo of 0.07 I think, Sunlight is 100% : as u calculate 7/100 = 0.07 this is the number that the sunlight reflect back to space from the moon, and 97% of the sunlight were absorb by the Moon...

jkmccrann

2005-Nov-07, 09:12 AM

If the albedo of the moon is 0.07, how does that compare to the albedo of the Earth? And for that matter, which is the most luminous of planets?

Kaptain K

2005-Nov-07, 10:48 AM

The albedo of the Moon is actually 0.12, not 0.07. The albedo of the Earth is 0.367, almost exactly 3 times that of the Moon. The most reflective of the planets is Venus, with an albedo of 0.65, nearly twice that of Earth.

Enzp

2005-Nov-08, 02:12 AM

Planet albedo is where you take a planet and cover it in a sauce made from parmesano cheese, butter, and....

TheThorn

2005-Nov-08, 03:09 AM

I thought a planet albedo was a mutant planet with no pigment - all white with pink eyes.

scorpionman_53

2005-Nov-08, 08:36 AM

thanks for your responses, but are there divices to measure planet albdo?, and how they work, or how scientist measure refluxes of planets.

Ricimer

2005-Nov-10, 11:15 PM

measuring albedo is actually pretty straightforward.

First, you figure out how much light, per square meter, is present at the distance of the planet (i.e. how much the light has spread out). you then figure out how much area the planet covers (so that it can reflect the light)...this is pretty close to the cross section of the planet.

Then, you know how much light is hitting the planet. If 100% is reflected, all the light that hits the planet is reflected, and you'll have a certain intensity.

You then measure how much light is actually reflected by the planet, divide this by the "ideal 100% figure" and viola! you've got an albedo.

First, you figure out how much light, per square meter, is present at the distance of the planet (i.e. how much the light has spread out). you then figure out how much area the planet covers (so that it can reflect the light)...this is pretty close to the cross section of the planet.

Then, you know how much light is hitting the planet. If 100% is reflected, all the light that hits the planet is reflected, and you'll have a certain intensity.

You then measure how much light is actually reflected by the planet, divide this by the "ideal 100% figure" and viola! you've got an albedo.

Tim Thompson

2005-Nov-14, 04:16 PM

... You then measure how much light is actually reflected by the planet, divide this by the "ideal 100% figure" and viola! you've got an albedo.

That will work at the short wavelengths of eyeball light, since a planetary atmosphere will not emit at those wavelengths. However, if you want a bolometric albedo (reflectivity over all wavelengths), or an albedo at longer wavelengths (like infrared), then you have to deal with the fact that the electromagnetic radiation you see coming from the top of the atmosphere is a combination of reflection & emission. So you will need a radiative transfer model for the atmosphere, to weed out the "emission" part (it will give you the "reflection" part too, so you can check against the total observed).

That will work at the short wavelengths of eyeball light, since a planetary atmosphere will not emit at those wavelengths. However, if you want a bolometric albedo (reflectivity over all wavelengths), or an albedo at longer wavelengths (like infrared), then you have to deal with the fact that the electromagnetic radiation you see coming from the top of the atmosphere is a combination of reflection & emission. So you will need a radiative transfer model for the atmosphere, to weed out the "emission" part (it will give you the "reflection" part too, so you can check against the total observed).

TriangleMan

2005-Nov-14, 04:58 PM

First, you figure out how much light, per square meter, is present at the distance of the planet (i.e. how much the light has spread out).

But what is the actual equation? The luminosity (per a given area) of the Sun divided by the square of the distance from the Sun to the planet? :think:

But what is the actual equation? The luminosity (per a given area) of the Sun divided by the square of the distance from the Sun to the planet? :think:

suntrack2

2005-Nov-14, 05:12 PM

can black hole comes in this theory?

Candy

2005-Nov-14, 05:26 PM

I didn't get very far with googling "albedo equation" (http://www.google.com/search?sourceid=navclient&ie=UTF-8&rls=RNWE,RNWE:2004-21,RNWE:en&q=%22albedo+equation%22), either. My brain hurts.

Eroica

2005-Nov-15, 04:12 PM

The albedo of the Moon is actually 0.12, not 0.07.

Has it been recently updated? When I first learnt about albedos, the Moon's was always given as about 0.07. Googling results suggest that there is still some disagreement.

Perhaps 0.07 is the visual albedo (fraction of visible light reflected), while 0.12 is the bolometric albedo (fraction of all electromagnietic radiation reflected)?

Has it been recently updated? When I first learnt about albedos, the Moon's was always given as about 0.07. Googling results suggest that there is still some disagreement.

Perhaps 0.07 is the visual albedo (fraction of visible light reflected), while 0.12 is the bolometric albedo (fraction of all electromagnietic radiation reflected)?

Eroica

2005-Nov-15, 04:22 PM

But what is the actual equation? The luminosity (per a given area) of the Sun divided by the square of the distance from the Sun to the planet? :think:

I presume you divide the Sun's radiant flux (measured in watts) by the area of a sphere whose radius = the Sun-planet distance.

That's a start, anyway. :)

I presume you divide the Sun's radiant flux (measured in watts) by the area of a sphere whose radius = the Sun-planet distance.

That's a start, anyway. :)

TriangleMan

2005-Nov-15, 06:45 PM

So as an example, using Venus: :think:

Sun luminosity (L) : 3.827 x 10^26 W

average distance from Sun to Venus (r): 1.082 x 10^11 m

luminosity of Sun per square meter at radius equal to Venus orbit:

L / (pi r^2) = 1.040 x 10^4 W/m^2

Then you would measure the amount of sunlight reflected off of Venus and compare it to the above number to get (a simple) albedo?

Sun luminosity (L) : 3.827 x 10^26 W

average distance from Sun to Venus (r): 1.082 x 10^11 m

luminosity of Sun per square meter at radius equal to Venus orbit:

L / (pi r^2) = 1.040 x 10^4 W/m^2

Then you would measure the amount of sunlight reflected off of Venus and compare it to the above number to get (a simple) albedo?

Eroica

2005-Nov-16, 10:12 AM

So as an example, using Venus: :think:

Sun luminosity (L) : 3.827 x 10^26 W

average distance from Sun to Venus (r): 1.082 x 10^11 m

luminosity of Sun per square meter at radius equal to Venus orbit:

L / (pi r^2) = 1.040 x 10^4 W/m^2 You have to divide the solar irradiance by the area of a sphere, not a circle! :wall:

3.827x1026 W / 4pi (1.082x1011)2 = 2.601x103 Wm-2

Wikipedia (http://en.wikipedia.org/wiki/Venus_%28planet%29#Atmosphere) gives 2.6139x103 Wm-2.

The Sun's visual magnitude is -26.8; Venus's visual magnitude when 1 AU away is -4.1. So the Sun is 2.511886422.7 times as luminous as Venus, so Venus's visual luminosity is approximately 3.827x1026 W / 2.511886422.7 = 3.183x1017 W.

The surface area of Venus is 4.6x1014 m2. So the visual luminosity per area is 692 Wm-2. Edit: divide by half the Venusian surface area, as this is the luminous area!]

Unfortunately 692 is only about 26% of 2613.9, while Venus's visual albedo is about 65%, so I'm going wrong somewhere. Perhaps my mixing of visual parameters with bolometric is the problem. :think:

Sun luminosity (L) : 3.827 x 10^26 W

average distance from Sun to Venus (r): 1.082 x 10^11 m

luminosity of Sun per square meter at radius equal to Venus orbit:

L / (pi r^2) = 1.040 x 10^4 W/m^2 You have to divide the solar irradiance by the area of a sphere, not a circle! :wall:

3.827x1026 W / 4pi (1.082x1011)2 = 2.601x103 Wm-2

Wikipedia (http://en.wikipedia.org/wiki/Venus_%28planet%29#Atmosphere) gives 2.6139x103 Wm-2.

The Sun's visual magnitude is -26.8; Venus's visual magnitude when 1 AU away is -4.1. So the Sun is 2.511886422.7 times as luminous as Venus, so Venus's visual luminosity is approximately 3.827x1026 W / 2.511886422.7 = 3.183x1017 W.

The surface area of Venus is 4.6x1014 m2. So the visual luminosity per area is 692 Wm-2. Edit: divide by half the Venusian surface area, as this is the luminous area!]

Unfortunately 692 is only about 26% of 2613.9, while Venus's visual albedo is about 65%, so I'm going wrong somewhere. Perhaps my mixing of visual parameters with bolometric is the problem. :think:

TriangleMan

2005-Nov-16, 12:58 PM

Thanks for the area correction. :o

Was the Venus albedo measurements taken from the perspective of someone on Earth? (In which case you'd need to compensate for the distance between Earth and Venus). It could be why the calculation is off.

Was the Venus albedo measurements taken from the perspective of someone on Earth? (In which case you'd need to compensate for the distance between Earth and Venus). It could be why the calculation is off.

eburacum45

2005-Nov-16, 02:48 PM

The albedo should be tha same no matter what the distance; it is a measure of the light reflected off a surface, independent of how large or small or distant that surface is. However albedo can change depending on the angle you look at a surface; if the surface is shiny, or particularly rough, or reflects particles back towards the light source, then the albedo can be very different when seen from the direction of the light source compared to when you see it from an angle.

The Moon, for instance, has a higher albedo when it is full than when it is half, or crescent shaped; the Moon's surface is rough, and this causes the lit region to have many shadows when seen obliquely, making the surface appear darker.

Conversely there is a small reflection effect from the moon dust, some of which consists of glass spheres like catseye reflectors. This effect combined with the lack of shadows when seen from the direction of the light source (i.e. when the Moon is full) is sometimes called heiligenshein.

The Moon, for instance, has a higher albedo when it is full than when it is half, or crescent shaped; the Moon's surface is rough, and this causes the lit region to have many shadows when seen obliquely, making the surface appear darker.

Conversely there is a small reflection effect from the moon dust, some of which consists of glass spheres like catseye reflectors. This effect combined with the lack of shadows when seen from the direction of the light source (i.e. when the Moon is full) is sometimes called heiligenshein.

artmso

2005-Nov-17, 02:39 AM

You have to divide the solar irradiance by the area of a sphere, not a circle! :wall:

Nope.

The Sun doesn't illuminate the whole planet - nor do we see the whole planet.

In either case, only half of the planet is relevant. Also, the relevant quantity is the projected area - the area you see when looking at the planet - which is that of a circle of the relevant radius.

To complicate things a bit, there is also an angle dependence of the albedo. When light hits a surface, most of it will be reflected according to

"outgoing angle" = 180o - "ingoing angle"

like billiard balls reflecting off the sides of a pool table. That is how a perfectly shiny surface works. Other surfaces will have some distribution of reflections in other directions. When you look at a smooth billiard ball, lit from behind you, it will be brightest in the middle because light hitting there is reflecting right back at you. As you look towards the edge, the light hits the ball at a more abtuse angle and more light gets scattered out of your line-of-sight - therefore the dimming towards the limb (called limb-darkening).

If you move your light-source, the bright spot will move and point in the direction between you and the light-source, according to the equation above.

The Moon hardly has any limb-darkening which tells you that the Lunar surface scatters light almost isotropically [same in all directions].

Cheers

Regner

Nope.

The Sun doesn't illuminate the whole planet - nor do we see the whole planet.

In either case, only half of the planet is relevant. Also, the relevant quantity is the projected area - the area you see when looking at the planet - which is that of a circle of the relevant radius.

To complicate things a bit, there is also an angle dependence of the albedo. When light hits a surface, most of it will be reflected according to

"outgoing angle" = 180o - "ingoing angle"

like billiard balls reflecting off the sides of a pool table. That is how a perfectly shiny surface works. Other surfaces will have some distribution of reflections in other directions. When you look at a smooth billiard ball, lit from behind you, it will be brightest in the middle because light hitting there is reflecting right back at you. As you look towards the edge, the light hits the ball at a more abtuse angle and more light gets scattered out of your line-of-sight - therefore the dimming towards the limb (called limb-darkening).

If you move your light-source, the bright spot will move and point in the direction between you and the light-source, according to the equation above.

The Moon hardly has any limb-darkening which tells you that the Lunar surface scatters light almost isotropically [same in all directions].

Cheers

Regner

Eroica

2005-Nov-17, 09:08 AM

Nope.

The Sun doesn't illuminate the whole planet - nor do we see the whole planet.

In either case, only half of the planet is relevant. Also, the relevant quantity is the projected area - the area you see when looking at the planet - which is that of a circle of the relevant radius.

To complicate things a bit, there is also an angle dependence of the albedo. When light hits a surface, most of it will be reflected according to

"outgoing angle" = 180o - "ingoing angle"

like billiard balls reflecting off the sides of a pool table. That is how a perfectly shiny surface works. Other surfaces will have some distribution of reflections in other directions. When you look at a smooth billiard ball, lit from behind you, it will be brightest in the middle because light hitting there is reflecting right back at you. As you look towards the edge, the light hits the ball at a more abtuse angle and more light gets scattered out of your line-of-sight - therefore the dimming towards the limb (called limb-darkening).

If you move your light-source, the bright spot will move and point in the direction between you and the light-source, according to the equation above.

The Moon hardly has any limb-darkening which tells you that the Lunar surface scatters light almost isotropically [same in all directions].

Cheers

Regner

Show me the figures! :)

The Sun doesn't illuminate the whole planet - nor do we see the whole planet.

In either case, only half of the planet is relevant. Also, the relevant quantity is the projected area - the area you see when looking at the planet - which is that of a circle of the relevant radius.

To complicate things a bit, there is also an angle dependence of the albedo. When light hits a surface, most of it will be reflected according to

"outgoing angle" = 180o - "ingoing angle"

like billiard balls reflecting off the sides of a pool table. That is how a perfectly shiny surface works. Other surfaces will have some distribution of reflections in other directions. When you look at a smooth billiard ball, lit from behind you, it will be brightest in the middle because light hitting there is reflecting right back at you. As you look towards the edge, the light hits the ball at a more abtuse angle and more light gets scattered out of your line-of-sight - therefore the dimming towards the limb (called limb-darkening).

If you move your light-source, the bright spot will move and point in the direction between you and the light-source, according to the equation above.

The Moon hardly has any limb-darkening which tells you that the Lunar surface scatters light almost isotropically [same in all directions].

Cheers

Regner

Show me the figures! :)

TriangleMan

2005-Nov-17, 12:46 PM

Nope.

The Sun doesn't illuminate the whole planet - nor do we see the whole planet.

In either case, only half of the planet is relevant. Also, the relevant quantity is the projected area - the area you see when looking at the planet - which is that of a circle of the relevant radius.

But the calcluations were in Watts per square meter, so I don't see how the area is a planet or half-planet has an impact.

The Sun doesn't illuminate the whole planet - nor do we see the whole planet.

In either case, only half of the planet is relevant. Also, the relevant quantity is the projected area - the area you see when looking at the planet - which is that of a circle of the relevant radius.

But the calcluations were in Watts per square meter, so I don't see how the area is a planet or half-planet has an impact.

Ricimer

2005-Nov-18, 01:35 AM

the reason the size of the planet is important is that's what you observe. Your telescope measures the light over the entire disk of the planet Venus, not just a square meter on venus. At best you can get a fraction of the surface, but nothing that small.

So you take the (Solar Flux at Venus) * (cross section of Venus) to get the total reflected light off of venus (assuming 100% reflectance). Compare to observations and you have a rough albedo.

Now, there are other "fudge factors" to throw in (i'm being a bit glib here, there are other factors used to flesh out the rough approximation). Such as the angle dependence, the fact that reflecting shape can have an impact (i.e. take into consideration the rounding edges due to it being a sphere), and the fact that the light will diverge from venus as well (further dropping the observed reflected light, not due to reflectance characteristics, but by the divergence of the beam itself). and you'll get more and more accurate answers.

So you take the (Solar Flux at Venus) * (cross section of Venus) to get the total reflected light off of venus (assuming 100% reflectance). Compare to observations and you have a rough albedo.

Now, there are other "fudge factors" to throw in (i'm being a bit glib here, there are other factors used to flesh out the rough approximation). Such as the angle dependence, the fact that reflecting shape can have an impact (i.e. take into consideration the rounding edges due to it being a sphere), and the fact that the light will diverge from venus as well (further dropping the observed reflected light, not due to reflectance characteristics, but by the divergence of the beam itself). and you'll get more and more accurate answers.

artmso

2005-Nov-18, 02:28 AM

Show me the figures! :)

Sorry Eroica. My objection to surface of sphere vs. surface of circle was placed

in the wrong part of your post of 16-November-2005 09:12 PM.

Finding the Solar flux at Venus, you obviously have to divide by the surface of a sphere of the radius of Venus' orbit.

Later on, however, you divide the visual luminosity of Venus, by its total surface area, whereas you only want to divide by the area presented to us (1.15x1014m2).

All in all we get

LSun pi RVen2/(4 pi dVen2)albedo(Venus) = ------------------------------

LSun/(2.511886426.8-4.4)

2.511886426.8-4.4

= ------------------------------- = 0.713 = 71.3%,

4 (dVen/RVen)2

where LSun is the Solar luminosity, RVen is the radius of Venus and dVen is the radius of Venus' orbit. I have also used a Venusian visual magnitude of -4.4 instead of -4.1 (found at http://nssdc.gsfc.nasa.gov/planetary/factsheet/)

It is very painful to enter math in these editors... Hope it came out okay.

Cheers

Regner

Sorry Eroica. My objection to surface of sphere vs. surface of circle was placed

in the wrong part of your post of 16-November-2005 09:12 PM.

Finding the Solar flux at Venus, you obviously have to divide by the surface of a sphere of the radius of Venus' orbit.

Later on, however, you divide the visual luminosity of Venus, by its total surface area, whereas you only want to divide by the area presented to us (1.15x1014m2).

All in all we get

LSun pi RVen2/(4 pi dVen2)albedo(Venus) = ------------------------------

LSun/(2.511886426.8-4.4)

2.511886426.8-4.4

= ------------------------------- = 0.713 = 71.3%,

4 (dVen/RVen)2

where LSun is the Solar luminosity, RVen is the radius of Venus and dVen is the radius of Venus' orbit. I have also used a Venusian visual magnitude of -4.4 instead of -4.1 (found at http://nssdc.gsfc.nasa.gov/planetary/factsheet/)

It is very painful to enter math in these editors... Hope it came out okay.

Cheers

Regner

Eroica

2005-Nov-18, 02:06 PM

Later on, however, you divide the visual luminosity of Venus, by its total surface area, whereas you only want to divide by the area presented to us (1.15x1014m2).

:wall: Of course! I was forgetting that only one side of Venus reflects the light!

:clap: Thanks for the math.

Redoing my calulations with this change (and 4.4 for Venus's magnitude) gives an albedo of 69.8%, which is pretty close to the actual value! :dance:

:wall: Of course! I was forgetting that only one side of Venus reflects the light!

:clap: Thanks for the math.

Redoing my calulations with this change (and 4.4 for Venus's magnitude) gives an albedo of 69.8%, which is pretty close to the actual value! :dance:

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