PDA

View Full Version : Absolute Magnitudes of Asteroids



Eroica
2004-Nov-17, 08:56 AM
The Free Dictionary (http://encyclopedia.thefreedictionary.com/Absolute+magnitude) gives a formula for calculating the apparent magnitude of an asteroid from its absolute magnitude:

Absolute Magnitude for Planets (H)
For planets, comets and asteroids a different definition of absolute magnitude is used which is more meaningful for nonstellar objects.

In this case, the absolute magnitude is defined as the apparent magnitude that the object would have if it were one astronomical unit (au) from both the Sun and the Earth and at a phase angle of zero degrees [ie full, like the full-moon]. This is a physical impossibility, but it is convenient for purposes of calculation....

Apparent Magnitude
The absolute magnitude can be used to help calculate the apparent magnitude of a body under different conditions.

app mag = H - 2.5*log10[p(X)*(a*b)]

Where:
X = phase angle (0 for opposition)

p(X) = phase integral (integration of reflected light) (<1) = (⅔)*((1-χ/π)*cos(χ)+sin(χ)/π)

At opposition, an asteroid's phase integral is ⅔

a = distance in AUs betweeen Earth and asteroid
b = distance in AUs between Sun and asteroid

But when I tried using this formula on Vesta to confirm that it is a naked-eye object at favourable oppositions, I got an apparent magnitude of 2.59, instead of 6.1.

app mag = H - 2.5*log10[⅔*(a*b)]

H = 3.2
a = 1.152 (distance between Earth and Vesta when Vesta is at perihelion)
b = 2.152 (Vesta's perihelion)

app mag = 3.2 - 2.5*log[⅔(1.152*2.152)] = 2.59

What's wrong with this picture? ](*,) :-k

Vesta (http://encyclopedia.thefreedictionary.com/4%20Vesta)

Kullat Nunu
2004-Nov-17, 10:19 AM
Easier way:

m = V(1,α) + 5*lg(Δ*r)-2.5*lg(Φ(α))

where

m = apparent magnitude
V(1,α) = phase curve; at opposition it is H = absolute magnitude = 3.2
Δ = distance between Earth and Vesta at perihelion = 1.152 AU
r = perihelion distance of Vesta = 2.152 AU
Φ(α) = phase function = cos(α), when 0 <= α <= π/2; 0 otherwise; at opposition Φ(α) = cos(0) = 1

so

m = 3.2 + 5*lg(1.152*2.152)-2.5*lg(1) = 5.17

Actually Vesta's orbit is inclined, so it never has 0 phase angle at perihelion. Correct value is something like 5.5.

Heres equation for phase curve:

V(1,α) = -2.5*lg[a1*exp(-3.33(tan(α/2)^0.63) + a2*exp(-1.87*(tan(α/2))^1.22)]

where

a1 = (1-G)*10^(-0.4*H)
a2 = G*10^(-0.4*H)

G is slope parameter that can be found in asteroid catalogues (0.32 for Vesta).

A Thousand Pardons
2004-Nov-17, 10:23 AM
app mag = H - 2.5*log10[⅔*(a*b)]

H = 3.2
a = 1.152 (distance between Earth and Vesta when Vesta is at perihelion)
b = 2.152 (Vesta's perihelion)

app mag = 3.2 - 2.5*log[⅔(1.152*2.152)] = 2.59

What's wrong with this picture? ](*,) :-k
Bad formula. :)

Check it out--as the distance gets greater, the logarithm also gets greater--but it's multiplied by a negative 2.5. That would mean that H would get more and more negative--that is, brighter.

I tried to look at their equations quickly to see if there was some sort of sign change problem, and started with the last equations on the page, but I couldn't even get the arithmetic to work out the same way! They say:


0.21 -2.5*log10(2/3*(2.57x10^-3)^2) = -12.4

but that's not what I get, even changing the sign to a positive.

frogesque
2004-Nov-17, 10:29 AM
I am in awe that you guys understand all that. OK, I can plumb in the numbers to the formulae and probably get an answer but the concepts and derivation are way above me. BA never ceases to amaze and I'm not surprised the woos get such a hard time here. Magnitude to me is -ve and it's bright like Venus, Jupiter and sometimes Mars. +ve and it's dim, +5.5 and is it 6, 7 or 8 stars I can see in the Pliades!

Edit: Changed +6.5 to +5.5 ... my BAD :oops:

A Thousand Pardons
2004-Nov-17, 10:45 AM
app mag = 3.2 - 2.5*log[⅔(1.152*2.152)] = 2.59

What's wrong with this picture? ](*,) :-k
And, now that I look it over some more, I don't get that answer either! :)

As near as I can tell, you forgot the 2.5.

Eroica
2004-Nov-17, 05:13 PM
app mag = 3.2 - 2.5*log[⅔(1.152*2.152)] = 2.59

What's wrong with this picture? ](*,) :-k
And, now that I look it over some more, I don't get that answer either! :)

As near as I can tell, you forgot the 2.5.
:oops: Yes, I forgot to multiple the log by 2.5. But 1.67 is still the wrong answer!

Eroica
2004-Nov-17, 05:14 PM
Easier way ...
=D> Thank you!

Evan
2004-Nov-17, 05:35 PM
Forget the math problems for a moment.


In this case, the absolute magnitude is defined as the apparent magnitude that the object would have if it were one astronomical unit (au) from both the Sun and the Earth and at a phase angle of zero degrees [ie full, like the full-moon]. This is a physical impossibility, but it is convenient for purposes of calculation....

There are a few important things missing here. First, what is the albedo? But more important, what about the "shadow hiding" that makes the apparent magnitude increase non-linearly near zero phase angle (The Opposition Effect)?

More here (http://www.vapahtaja.com/parabolic/stuff/assignment.pdf)

Kullat Nunu
2004-Nov-17, 06:32 PM
Forget the math problems for a moment.


In this case, the absolute magnitude is defined as the apparent magnitude that the object would have if it were one astronomical unit (au) from both the Sun and the Earth and at a phase angle of zero degrees [ie full, like the full-moon]. This is a physical impossibility, but it is convenient for purposes of calculation....

There are a few important things missing here. First, what is the albedo? But more important, what about the "shadow hiding" that makes the apparent magnitude increase non-linearly near zero phase angle (The Opposition Effect)?

More here (http://www.vapahtaja.com/parabolic/stuff/assignment.pdf)

Geometric albedo (phase angle = 0)

p = (r*Δ/(a*R))^2 * 10^(-0.4*(m0-m_sun))

where

r = distance of the object from the Sun (AUs)
Δ = distance between Earth and the object (AUs)
a = observer's distance from the Sun (AUs)
R = object's radius (AUs)
m0 = visual brightness of the object at phase angle 0
m_sun = Sun's visual brightness

All of these are measured from observations.

Slope parameter takes account the change of brightness in other phase angles. If the parameter has a large value, the slope is very peaked (asteroid is much brighter in opposition than normally).

Evan
2004-Nov-17, 07:05 PM
How is the slope parameter approximated? It appears to be very sensitive to phase angle, packing density and material type and size. It also appears to have possible sudden discontinuities.[1] This would be important for detection of asteroids as they can be observed at a wide range of phase angles. It would also be important to take into account for any search program.

[1]Yes, I know that the major discontinuity in the experiment I referenced was due to procedure.

Kullat Nunu
2004-Nov-17, 10:28 PM
How is the slope parameter approximated? It appears to be very sensitive to phase angle, packing density and material type and size. It also appears to have possible sudden discontinuities.[1] This would be important for detection of asteroids as they can be observed at a wide range of phase angles. It would also be important to take into account for any search program.

[1]Yes, I know that the major discontinuity in the experiment I referenced was due to procedure.

The reference that I have (and quick googling) says only that the way to approximate slope parameter is to obtain enough observations, especially at phase angles under 10 when the curve becomes non-linear, and fit a proper curve to the chart. These observations need to have timespans large enough that the variability due to irregularities of the body etc. can be eliminated.

Evan
2004-Nov-17, 10:58 PM
Although I am familiar with the opposition effect I hadn't given it much thought until this topic came up. I have been thinking of doing some asteroid hunting with my new camera and it becomes obvious that my chances will be much enhanced if I take pictures in the direction of the anti-solar point on the ecliptic.

Eroica
2004-Nov-18, 09:13 AM
PDF: Light-Scattering Measurements of Planetary Regolith Analogs at Small Phase Angles (http://www.astro.helsinki.fi/~naranen/texts/gradu.pdf)

I found this a few days ago while googling this subject. It deals with the opposition effect.

Eroica
2004-Nov-18, 11:50 AM
Here are a few other formulae relevant to this discussion, which I found here (http://www.bitnik.com/mp/archive/Formula.html):

H = 15.618 - 5*log(D) - 2.5*log(pV),

D = 1329*[10^(-H/5)]/√[(pV)]

H = absolute magnitude
D = asteroid diameter in kilometres
pV = the asteroid's albedo
(H and pV are visual band.)

For Vesta (diameter = 468.3 km, albedo = 0.423):

H = 15.618 - 5*log(468.3) - 2.5*log(0.423) = 3.2

D = 1329*[10^(-3.2/5)]/√[0.423] = 468.12

badprof
2004-Nov-18, 02:47 PM
Although I am familiar with the opposition effect I hadn't given it much thought until this topic came up. I have been thinking of doing some asteroid hunting with my new camera and it becomes obvious that my chances will be much enhanced if I take pictures in the direction of the anti-solar point on the ecliptic.

Hi Evan,

While hunting around the opposition point will certainly make it easier to see asteroids, it will not necessarily be easier to make discoveries. The reason for this is the large surveys! Unless you have access to a large telescope, by the time an asteroid reaches the opposition point, it is likely to have been picked up already. :(

To try and make discoveries, you are best served by aiming more towards the east where you can try and pick up objects coming out of the Sun's glare before they reach where the surveys search.

Cheers,

Maurice

Evan
2004-Nov-18, 04:26 PM
Maurice,

It seems to me that in the course of a little over a year the opposition point seen at midnight will cover the entire belt. Also, through the course of a night it will cover approximately half the belt, albeit more distant than at midnight. The most usable time will be from 10 pm to 2 am each night which covers 60. There are plenty of undiscovered asteroids to find.

I have previously imaged objects down to as deep as 13-14th mag using only a 50mm lens and ASA800 film with my camera mounted on my double arm drive. (http://vts.bc.ca/astrophoto/drive.htm) My new camera goes to ISO 3200 and should do better, especially with stacking of images. I have been able to image 14th mag with a webcam on my 6" newt. (http://vts.bc.ca/astrophoto/scope.htm)

George
2004-Nov-19, 07:30 PM
Wow, Evan. Very nice. =D> It all looks excellent including the art work.