PDA

View Full Version : Magnitudes for Oort and Kuiper Belt objects



George
2006-Feb-10, 05:37 PM
Just thought someone might have a cool link that would explain how to calculate magnitudes for any new body found in our solar system. [So far, I haven't found much I like.] I am most interested in derivations which compare with other known bodies.

For those inclined on math today...
Since they reflect light, instead of generate it, the inverse square law gets, well, squared, as I see it.

I tried to derive the apparent magnitude value by comparing it with known planets. Any new body, I think, should have an apparent mag. based on....
[delta B] = (d1/d2)^4 * (r2/r1)^2 * (a2/a1)

[added: it really gives intensity change before log base 2.51 is used to convert to mag values]

d ~ distance to objects (d1 being a known planets distance, d2 the new object). Since the inverse square law applies for the light to reach the planet, and it applies for the light to return to us, it is a 4th power (squaring the inverse square), right?
r ~ The reflective surface increases, or decreases, as the square of the radius.
a ~ albedo

However, my few attempts seem to leave me about a mag. too bright. What am I missing? [Probably not dust with photons, but dust with neurons. ;)]

Saluki
2006-Feb-10, 05:53 PM
I think you are getting things a little backward. Magnitude is directly measurable. Things like radius, distance, and albedo are calculated from observations. Knowing apparent magnitude, and distance, we can calculate absolute magnitude from:

M = m + 5(log D - 1)

Where:

M is the absolute magnitude.
m is apparent magnitude.
D is the distance. For deep space objects, we need to use Luminosity distance, which is beyond the scope of this discussion.

tony873004
2006-Feb-10, 06:42 PM
d ~ distance to objects (d1 being a known planets distance, d2 the new object). Since the inverse square law applies for the light to reach the planet, and it applies for the light to return to us, it is a 4th power (squaring the inverse square), right?

You might not want to do (d1/d2)^4 all in one step. The d's are different. On one hand, you have to compare the ratio of planet's and KBO's solar distance. And on the other hand, you are comparing the ratio of their Earth distances. At the distance of the Kuiper Belt, the difference between their Earth distance and their Sun distance is not huge, but you do square the difference. And since you're inputting your (small error)^2 into a logarithmic expression, things can get crazy quickly.

You've already figured out that your method is computing brightnesses, or fluxes, not magnitudes. m1-m2=-2.5 log(flux1/flux2).


I think you are getting things a little backward.
The way he phrased the question, yes. But you could reword the question:

A CCD can record objects as faint as mag 16. Can it detect a 2000km diameter KBO at 100 AU?

George
2006-Feb-10, 07:59 PM
I think you are getting things a little backward. Magnitude is directly measurable. Things like radius, distance, and albedo are calculated from observations. Knowing apparent magnitude, and distance, we can calculate absolute magnitude from:

M = m + 5(log D - 1)
...
Yes, thanks. I did find that one for comparison. I still wonder why the approach I took is off a little. So, I am just curious as to why. [Also, my formula might even be easier for me to use than yours; and I like 4th power power vs. log power! :)]


You might not want to do (d1/d2)^4 all in one step. The d's are different. On one hand, you have to compare the ratio of planet's and KBO's solar distance. And on the other hand, you are comparing the ratio of their Earth distances. At the distance of the Kuiper Belt, the difference between their Earth distance and their Sun distance is not huge, but you do square the difference. And since you're inputting your (small error)^2 into a logarithmic expression, things can get crazy quickly.
Yes. I considered that, too. I varied it up to 1 au, and for each object (1 and 2). The variations are minor in net m.
For instance, in comparing EL61 (m=17.5, 51 AU) to Pluto (m=13.9, 39.5AU from sun)...
Changing the AU of EL61 from 51 to 52 produces only a .1 mag. difference. About the same if Pluto's distance is changed by 1 AU.

[Added: BTW, my computation method produces an apparent mag. of 15.9, in lieu of 17.5.)

Fr. Wayne
2006-Feb-10, 08:03 PM
maybe he could use (1) Ceres as a comparison and compute the maximum distance at 16 M. (at opposition)

tony873004
2006-Feb-10, 08:43 PM
I just tried it comparing your method and mine. You're right. There's only a small difference not worth considering. But I get 15.0, not 15.9. I'll check my numbers later to see if I keyed something in wrong.

The answer may lie in the fact that different references commonly give different magnitude values for the same object. I think a visual magnitude is supposed to either measure in the green band or an average of the whole visible spectrum. And changing the wavelength you observe in by even a little can give different results. I remember doing a problem like this for homework a few semesters ago, and the teacher accepted answers off by 1 magnitude from published values for the reasons I mention (assuming I explained it correctly).

Romanus
2006-Feb-10, 08:50 PM
For objects near the Earth, you need to take into account the distances from Earth *and* the Sun.

Here is a tidy formula I've found:

m = k' +2.5log10(r^2)(Δ^2)

where:

m = magnitude difference
k' = constant
r = distance from the Sun
Δ = distance from the Earth

Take for instance Comet Encke, which has a listed nuclear absolute magnitude of 14.5 (http://neo.jpl.nasa.gov/cgi-bin/db_shm?sstr=Encke&group=com&search=Search), at a unit distance of 1 AU from the Sun and the Earth. If we were to move Comet Encke out to 10,000 AUs, its new magnitude would be:

m = k'+2.5log10(10000^2)(9999^2)

m = 40

40+14.5 = 54.5

Needless to say, this is far beyond the capability of any telescope we'll have in the forseeable future.

(feel free to check my figures, anybody)

George
2006-Feb-10, 10:49 PM
BTW, thanks all for the help. I appreciate it. I'm reluctant to bring up math on Fridays. :)


I just tried it comparing your method and mine. You're right. There's only a small difference not worth considering. But I get 15.0, not 15.9. I'll check my numbers later to see if I keyed something in wrong. A 100% albedo with a diameter of 1600km would yield 15. I am using a 1500km diameter and .5 albedo.

I realize these distant objects are listed with a diameter based on an assumed albedo. My main interest is to demonstrate just how difficult it is for astronomers to see even very large planet-sized (even Jupiter-sized) objects deep in the theoritical Oort Cloud; which is why it has not been easy to verify its presence. My fumbling found faulty fruition in the Creationists takes on a comet thread. (http://www.bautforum.com/showthread.php?t=37852)

If the formula were right, Jupiter could only move out to about 13,000 AU before Hubble, or Keck, looses it (m=30). The Oort may extend as much as 200,000 AU. If Jupiter were to magically swell to 1 million km, it would be ostensibly invisible at about 34,000 AU (assuming its surface radiation stays the same). More trivia shows a billion km object would be visible to the naked eye up to about 130,000 AU.

Of course, these are in error, slightly, due to something wrong with my formula.

George
2006-Feb-10, 10:53 PM
For objects near the Earth, you need to take into account the distances from Earth *and* the Sun.

Here is a tidy formula I've found:

m = k' +2.5log10(r^2)(Δ^2)
Thanks. I like this one. What is the value of k'? [added: I guess 14.5 for k'.]


Take for instance Comet Encke, which has a listed nuclear absolute magnitude of 14.5 (http://neo.jpl.nasa.gov/cgi-bin/db_shm?sstr=Encke&group=com&search=Search), at a unit distance of 1 AU from the Sun and the Earth. If we were to move Comet Encke out to 10,000 AUs, its new magnitude would be:

m = k'+2.5log10(10000^2)(9999^2)

m = 40

40+14.5 = 54.5

Needless to say, this is far beyond the capability of any telescope we'll have in the forseeable future.

(feel free to check my figures, anybody)
That's cool. It isn't a comet formula which allows for the lack of tail lights (and turn signals, ;) hey, its Friday!)?

Romanus
2006-Feb-11, 12:54 AM
k' is just a constant...I suppose one could use it in the way you did. In fact, I'm guessing that's exactly what that k was intended for. I'm actually no math whiz, myself. ;)

George
2006-Feb-11, 11:46 PM
I have had a chance to cogitate on all this...

Essentially, it is the same, but I took into account what Romanus said...

For objects near the Earth, you need to take into account the distances from Earth *and* the Sun.

Therefore, the formulae is now...

delta B = (s1/s2)^2 * (d1/d2)^2 * (r2/r1)^2 * (a2/a1)


Object 1 further than reference (object 2). [Reverse all ratios if opposite is true.]

s - distance from sun
d - distance from Earth
r - radius
a - albedo

Fortunately, Astronomy shows each month s, d, & apparent mag.

Results with Neptune as reference:
(My resulting mag., Astronomy pub. mag.)
Uranus: (6.0,5.9)
Jupiter: (-2.1, -2.1)
UB313: (19,19) [not in Astronomy]
Pluto: (15.2,14.0)

When I used Pluto as a reference, I was consistently off by about .3 mag, so my Pluto data is off, apparently. I used Pluto values of 2306 km dia. and 60% albedo. Hopefully, someone will correct one, or both, of these values.

Hopefully, this formula is now acceptable. Can anyone confirm its validity?

[Edit: radius (r) and albedo (a) values inverted to match the original OP formula, which was correct.]

George
2006-Feb-13, 04:05 AM
FWIW, I applied the simple equation on inner planets. [Neptune is my reference object]

Compilation: (Mine,Published)
Mercury: (-1.33, -1.5)
Venus: (-3.81, -3.8)
Mars: (-2.3,-2.3)
Jupiter: (-2.11,-2.1)
Uranus: (5.97,5.9)
Pluto: (15.14,14) [w/ albedo of .3]
Pluto: (14.4,14) [w/ albedo of .6]
UB313: (18.72,19)

Hopefully, this will be a simple way to play with magnitudes for objects out there; for hypothetical objects, or for actual bodies.

Notice Pluto's variation. Pluto is still a mystery to me. The diameter, I think, was based on a .3 albedo, yet some references place it between 0.4 and 0.6. However, this would change the diameter and still throw me off, I think.

Do we need a Pluto oddity thread?

I am also off slightly with Mercury, but I don't know why.

tony873004
2006-Feb-13, 07:10 AM
Phase angle plays a huge role with Mercury (as well as Venus)

George
2006-Feb-13, 04:32 PM
I presume that is also known as ellipticity. IIRC, it was the July of 2000 Astronomy issue where I found Mercury at 95%, I think. I did not reduce the magnitude accordingly since it was only 5%, and I am already less than their value. :( If ellipticity (phase angle) is incorporated, I can find numerous other values for Mercury and see how they compare. Maybe the data I used was in error slightly.

The bigger issue is with Pluto. Does anyone have the "best" set of parameters for Pluto I can use? Surely, they should work with the formula, else we could get a game going here (albeit, another thread). :)

tony873004
2006-Feb-13, 09:16 PM
I'm looking at the current issue of Astronomy. I don't see the work ellipticity. Their pull-out centerfold chart has a row labeled Illumination. Mercury and Venus have low numbers, while Mars is at 90% and Jupiter - Pluto have 99%+.

Mercury will only be close to 95% when it is near superior conjunction, and not visible from Earth due to solar glare. But that doesn't mean you can't compute its magnitude.

George
2006-Feb-13, 11:27 PM
You're right. Illumination was what we are thinking, but I confussed it with another parameter which I was thinking of incorporating in the formula, too. Ellipticity gives you the differnece in the ratios of the equatorial diameter with the polar diameter. Another term is oblateness, which is synonymous with ellipticity, I think.

Illumination is given in Astronomy and would be a factor if it were much less than 95%. I had to dig back to 2000 to get a high illumination for Mercury (where the data also includes the needed distances).

For objects beyond Pluto, with full illumination, only one distant factor should suffice, since Earth can only alter its distance by 1 AU.

If you see any illuminations for Mercury around 95% or better, along with accurate distances from it to Earth and the sun, please let me know and I will see if they fit. Thanks.

tony873004
2006-Feb-14, 06:15 AM
You can get all sorts of accurate info for any time and date on the JPL Horizons web page. horizons.ssd.jpl.nasa.gov
If you use their telnet service, it will give you magnitude, illumination, distance, RA, Dec, x,y,z positions and velocities relative to anything you want. It's tricky to learn how to use it. If that sounds helpful to you, try it and you can ask me questions. I've already pulled out my hair figuring it out, so that you can keep your hair. :)

George
2006-Feb-14, 02:25 PM
Wow! If I were Candy, I'd probably hug ya! ;) The perfect place I needed. Thanks.

I think I understand it, but it's hard to absorb the degree of ostensible accuracy.

The "range", r, (distance, obviously) for Pluto is shown as 31.5661882128807. This is about 1/10th of a meter in accuracy, I think. Amzaing!

They also note Charon is combined into the geometric (visible) magnitude. I wondered about that last nite and I found it helped toward reconciling my values with Astronomy's values. The biggest variation may simply be the change in albedo with Pluto, as well as, Mercury. Both, apparently, have abledos which vary. My computed magnitudes matched nicely when I played with the albedo value.

This table, apparently, takes these albedo variations into account when computing apparent magnitude. I assume this because they show variations in surface brightness, "S-brt". Is this your understanding, too?

Is there a way to get this output in a spreadsheet file format I can grab?

Tonite I will use these values in my formula, and see if NASA is correct!! :)

Thanks again. BTW, you are too late about the hair. :( :)

[Added: This is your reference link. (http://ssd.jpl.nasa.gov/cgi-bin/eph) ]

tony873004
2006-Feb-14, 05:07 PM
Actually, I've never used the one you linked to. It's probably easier than the telnet one that I use, but telnet lets you grab x,y,z and x,y,z dot which is what I'm usually after. I've placed telnet data in a spreadsheet.

But there does seem to be a spreadsheet check box on the form you link to.

I doubt that they know Pluto to 1/10 meter of precision. Its just that after computing, they don't bother to round their answers to the appropriate significant figures. I've had teachers mark me off for that.

George
2006-Feb-14, 10:36 PM
Actually, I've never used the one you linked to. It's probably easier than the telnet one that I use, but telnet lets you grab x,y,z and x,y,z dot which is what I'm usually after. I've placed telnet data in a spreadsheet.
Is x,y,z what I think it is - Euclidian coordinates? What is the centerpoint - the sun or the barycenter (or do you choose)? I'm not able to connect the "dot", is it a time parameter?

After choosing the spreadsheet format, I tried to cut and paste into Quatro Pro and Word, but no luck. It might work in Excel, but that's on my other computer. I'll just cut and paste the few items I need.


I doubt that they know Pluto to 1/10 meter of precision. Its just that after computing, they don't bother to round their answers to the appropriate significant figures. I've had teachers mark me off for that. Me too. They also deducted for graph truncatation.

tony873004
2006-Feb-15, 12:13 AM
x, y, and z are coordinates, in kilometers from what ever reference you specify. x,y,and z dot are velocities in km/s from what ever reference you choose.

George
2006-Feb-15, 07:11 PM
Thanks, that's simple enough.

Looks like something is wrong in the formula, but I can't guess what; the formula's aparrent mag. values do not match JPL's. The variation in surface brightness for each day in the ephemeris may be a factor. Possibly the geometric albedo is messing me up in this regard. Do you understand how to incorporate surface brightness values? The Mercury ephemeris shows increasing surf. brightness, but decreasing Ap. Mag. However, the illumination is also decreasing; which would explain the decrease in apparent mag.

I'm at a loss as to what could be wrong with the formula. I'll play with Astronomy's values later and see how their values compare.

George
2006-Feb-16, 02:24 AM
Odd. I just did a quick comparison of JPL's apparent mag. with Astronomy's. Most match, some don't. For instance...
Jan. 06 Astronomy planetary data shows Mercury to have an apparent mag. of -0.7, whereas JPL shows -0.47; a surprising difference.

Venus and Jupiter were off by only 0.1 mag. The others seemed to match. All other parameters were the same.

Going back Jan. 03 Astronomy, a BA favorite and one of mine, the apparent magnitudes match.

Is there something I'm missing in how JPL defines apparent magnitude? Does anyone have another ephemeris we can use for comparison????

tony873004
2006-Feb-16, 06:06 AM
Astronomy magazine gives Mercury's value only for a certain date, or maybe for Mercury they give you the 1st, 15, and 31st. So you must be comparing to the exact date, as (1) Mercury moves very fast, only a few months to complete an orbit. (2) Mercury goes through all the phases, so its magnitude should be quite different for different parts of its orbit. Venus too, but to a smaller degree.

Magnitudes are part of a logrithmic scale and are very sensitive to the values used in the computations. It's not completely unexpected that two different computations would not yield exactly the same answer.

You could download and use Celestia and Stellarium to compare. (They're fun programs you should have anyway! Just Google for them)

George
2006-Feb-16, 04:23 PM
Astronomy magazine gives Mercury's value only for a certain date, or maybe for Mercury they give you the 1st, 15, and 31st. So you must be comparing to the exact date, as (1) Mercury moves very fast, only a few months to complete an orbit. (2) Mercury goes through all the phases, so its magnitude should be quite different for different parts of its orbit. Venus too, but to a smaller degree.
Yes. The comparisons were based on the same dates, since your cool jpl ephemeris allows any date.


Magnitudes are part of a logrithmic scale and are very sensitive to the values used in the computations. It's not completely unexpected that two different computations would not yield exactly the same answer.I would think the log scale would reduce the variation. The biggest factor by far is the distance, which is a 4th power variable. I would assume the distance calculations would be quite accurate. Maybe the albedo variations common to Mercury and Pluto is the simple explanation. Yet, I would have to drop below 10% albedo for Mercury to match JPL's app. mag. :think:


You could download and use Celestia and Stellarium to compare. (They're fun programs you should have anyway! Just Google for them) I forgot I have Celestia. How do you get app. mag. out of it?

George
2006-Feb-16, 10:23 PM
I emailed JPL, and have recieved a helpful response.

Their "apparent magnitude" equation for Mercury is....

P1= 5.D0 * LOG10( R*DELTA )

VMAG= -0.42D0 + P1 + 3.8D0 *(I/100.D0) - 2.73D0*(I/100.D0)^2.D0 +
+ 2.00D0*(I/100.D0)^3.D0

for phase angles (I, deg) between 3 and 123 degrees. R is target-sun heliocentric distance, Delta is target-observer distance, both in AU.

He found an updated expression which differs in the first coefficient, -0.42, which had been -0.36. [He is updating Horizon.] All other coefficients are the same. This .06 difference is not close to the variation I am calculating using my super duper formula. Also, the variation between JPL's values and mine are not phase angle related since I am using 95% to 100% illumination values.

For Pluto...
VMAG= -1.01D0 + P1 + 0.041D0*I

Here, he also found the first coefficient to be off, but only by .01.

What does the "D0" designate, a numeric terminator? Also, I don't understand dividing the phase angle by 100 when it is given in degrees???

It is interesting the log of the product of the distances is incorporated. I guess my inverse square law thinking is.... too square. ;)

I now realize my illumination ratio in the formula is erroneous, since illumination is not a linear relation with brightness. Your statement "Phase angle plays a huge role" was not taken serious enough by me; partly because I am thinking beyond Pluto, where phase angle shouldn't be an issue, right?

However, since Mercury's apparent magnitude is an issue now raised, what is the formula for brightness as a function of phase angle? I know 0 deg. is 100%, so I would guess cos(p). However, 90 deg. is half illumination but is something like 10% [edit: brightness, 50% illumination], so is there a simple equation?

tony873004
2006-Feb-17, 01:32 AM
Was it Jon Giorgini who replied to you? I imagine it is since you said "He is updating Horizons". He is awesome. Everytime I've e-mailed him a question, he's replied within a few hours, and his responses are always informative. He is also one of the lead scientists trying to get a fix on Apophis' exact orbit.

George
2006-Feb-17, 01:50 AM
Yes. :) I sent it to Don Yeomans (Group Leader) and he delegated it to Jon.

Fortunately, he did not spend time on my simpleton equation. After I sent my request, I realized the illumination ratio was wrong; otherwise, I still think I'm correct, at least for KBO's and Oort objects.

I have downloaded numerous Horizon's ephemeris data sets to play with and compare with mine, and Astronomy's. If you would like, I'll post the results.

I've avoided phase angles by choosing near 100% illumination positions. However, is there a simple equation to determine brightness change due to phase angle?

[Added: BTW, do you have any answers to the questions in my prior post?]

tony873004
2006-Feb-17, 02:29 AM
Sorry, I've never tried to compute that.

George
2006-Feb-17, 03:42 PM
Here are my results, using my simple formula and comparing with Horizons (SSD)...

(My result, SSD)

Venus as reference:
Jupiter (-2.06, -2.12)
Uranus (5.92, 5.92)
Neptune (7.96, 7.98)
Pluto (14.7, 14.0)

Mercury (-1.31, -.77) [ Ill = 98%] [Added: SSD is limited to values > 3 deg. phase angle for accuracy.]
Mars (1.13, 1.49)

Uranus as reference:
Jupiter ( -2.1, -2.06)
Neptune (7.97, 7.98)
Pluto (14.7, 14.0)

The results were essentially the same when using Jupiter and Neptune for references.

Note the big difference in Mercury. I really think there is a problem here with their equation. They do state the range for phase angles between 3 and 123 degrees, but I would rather a real astronomer address it, if any would be interested. Any takers?

My Mars value is a little too bright because I am using 100% illumination vs. the 95.5% actual amount used by SSD.

What this says, IMO, is the equation I am using is correct and should work fine for [I]KBO’s and Oort objects. That is satisfying.

For these distant objects, the simple form is…

B = (r1/r2)^4 * (d2/d1)^2 * (a2/a1)
Ap Mag = Ref. Mag. + log(B, 2.51) [using Excel function for log base 2.51]

> B - Change in brightness
>r1 - reference object heliocentric range
>r2 - target object heliocentric range
>delta1 - reference object range from Earth
>delta2 - target object range from Earth
>d1 - ref. obj. diameter
>d2 - target obj. diameter
>a1 - ref. obj. albedo
>a2 - target obj. albedo

The slightly more accurate one is…
B = (r1/r2)^2 * (delta1/delta2)^2 * (d2/d1)^2 * (a2/a1)

George
2006-Feb-24, 01:07 AM
I compared very small phase angles with apparent magnitudes for an outer planet, Uranus, and found they did not have much effect. From near 0 phase angle to 1 degree changed apparent mag. from 5.71 to 5.72 (.01 change).

http://img126.imageshack.us/img126/3569/uranusphaseangleresult9xb.th.jpg (http://img126.imageshack.us/my.php?image=uranusphaseangleresult9xb.jpg)

It appears, for gas planets at least, that at near 0 phase angles, the apparent magnitude does not change hardly at all due to the angle itself. The gegenschein, as Grant has pointed out to me in another thread, could still be a factor but does not seem to be incorporated in the JPL ephemeris.

I tried the little formulae on Pluto's new moons, P1 and P2. It produced P1 at 23.2 vs. observed value of 23.38, P2 at 22.5 vs. observed value of 22.93 (based on .35 albedo, like Charon's, and 61km dia. and 46km dia. respectively). That's not too bad.