Three questions for astronomers regarding Elenin

Based on this image:

gustavomuler.fotografiaastronomica.com...

and this formula:

Tan (x) = Opposite/Adjacent

and using JPL to retrieve the "Adjacent" length for the 5th April 2011, between Earth and Elenin:

ssd.jpl.nasa.gov...

and using this converter to convert 24.4944 arc seconds to degrees (from the image):

www.convertunits.com...

The comet has a coma with a diameter of around 33824km.

The diamater of the yellow dot in the image is 3200km.

Can someone please:

1) Explain to me how a 4km object generates a 33824km coma?

2) Explain how an astronomer would mathematically determine the original 4km odd diameter of the nucleus?

3) Explain what the colours are in the image and what the yellow colour means?

Cheers

JS


Mods: can you please move this to the astronomy area.

reply to post by jumpspace


LoL... I gave you a star on the other thread because of your comment. Then you took it off, and I wondered what happened. But now I saw that you started your own thread and the mystery came to an end. In any regard, thanks for showing your information, I find it interesting and I have a hard time believing it is only 4km in size.

Thanks Gerizo

I thought I'd better ask the question in the astronomy thread instead but I stuffed up when I posted the questions...hopefully a mod will move it

Cheers

JS

what size are we talking about than OP?

reply to post by heineken

The nucleus heineken.

Cheers

JS

The comet has a coma with a diameter of around 33824km.


Where did this measurement come from exactly?

reply to post by jumpspace


At the distance that the comet is, 3,200 km is only resolveable as a tiny point of light.

The spread-out blur that is shown in the image is largely caused by the limitations of the telescope's resolution and atmospheric scattering.

If you look at the stars in the background, you will see that they also are blurred out (most stars are so far away that they only appear as points of light). You will notice that the brighter stars blur out further than the dull stars. This is an artefact of the resolution of the telescope system.

That being said, these astronomers have attempted to determine how big the comet and its tail (its coma) is by playing with various images and approximating size (see the inset pictures). Their science is called Astrometry or the measurement of astronomical objects.

Because of the distances usually involved and the relative tininess of the astronomical objects at those distances, astronomers have to use all sorts of "tricks" to estimate sizes and distances.

As a comet gets close to the Sun, it begins to melt and bits break off, forming the tail. A comet that is close to the Sun can have a tail of several hundred thousand km, but this one is still far away, so its tail is small.

The 4 km size estimate for the head of the comet was approximated by obseving how bright it is and then calculating what size of cometary material would reflect that magnitude of light.

Comets are a mix of ice and dust and all have quite similar "reflectivity" and so an estimate is usually quite close to the final size when the comets are within a distance making direct observational measurement possible.

reply to post by Fastercaddy

Fastercaddy

Look at the OP image as well as follow the steps I gave in the OP to come to this size.

It's calculated using trigonometry:

en.wikipedia.org...

Cheers

JS

reply to post by chr0naut

Thanks for that chr0naut, though I pretty well knew most of what you posted...but it is good for those who don't know much about comets.

As we have many images of Elenin:

gustavomuler.fotografiaastronomica.com...
www.aerith.net...

...and ignoring question 1) and 3) for now, do you have the credentials to show me mathematically how you would calculate the size of Elenin?

I'm very curious as to how this is done.

Cheers

JS

So your suggesting we have an incoming planetoid or something?

2nd..

Originally posted by jumpspace
reply to post by heineken

The nucleus heineken.

Cheers

JS

i meant what estimate do you give for the object? (or object's nucleus?)

reply to post by Redevilfan09

>So your suggesting we have an incoming planetoid or something?

LOL

reply to post by heineken

>i meant what estimate do you give for the object? (or object's nucleus?)

I'm giving no estimate heineken.

I thought the three question's were pretty self explanatory.

Cheers

JS

reply to post by jumpspace

What interests me about Elenin is that JPL predicts that it will get within 1/4 AU of earth around mid-October of this year. That's about 23 million miles. The moon, by comparison, is 250,000 miles away.

So the mass of Elenin is of some interest. So far, it seems the estimates are that it is much smaller than the moon.

So Elenin may not be a major player this time around.

reply to post by l_e_cox

I agree with you completely l_e_cox and I'm sure JPL/NASA have calculated the projected path correctly, based on the interactions with the planets etc

In order to calculate those trajectories though, they would have to know the mass of Elenin as well as its composition and to do this, they would have to know the size of the nucleus.

As such, I'm extremely frustrated as to why JPL/NASA haven't published these figures - they only seem to quote Leonid's original estimate.

Thus the reason for this thread; to find someone reputable who can show us figures/images/calculations/references or whatever is used to calculate the nucleus of a comet and put it all to rest once and for all.

Hello NASA...hello?

Cheers

JS

Originally posted by jumpspace
reply to post by l_e_cox
Thus the reason for this thread; to find someone reputable who can show us figures/images/calculations/references or whatever is used to calculate the nucleus of a comet and put it all to rest once and for all.

Well, I'm not reputable, but by looking around the net, the following seems to be one way of doing it...

The "absolute magnitude" of an object (such as a comet) is what it would look like at a standard distance.
For comets, we can use...
m = m0 + 5 log Δ + 2.5 n log r
where
m = apparrent magnitude
m0 = the absolute magnitude
Δ = distance from object to earth
n = constant, but seems to be that 4 is useful for comets
r = distance from object to sun

Now, in the middle of March, Elenin was at opposition, and 3 AU from the sun, so 2 AU from earth.
Amateur astronomer websites report it was about magnitude 15.

so...
15 = m0 + 5*log2 + 2.5 * 4 * log3
which gives m0 = 8.7

Thats the magnitude that Elenin would be if it was 1 AU away.

Is this a reasonable value for a comet, or abnormal???
Looking around the net, it seems to be so.
Hale Bopp was massive, unusually big at -0.8
Halley was reasonable 4.7
Hyakutake was fainter at 5.3
periodic comet Encke, a little thing, is 9.8
other small comets are about this range.

So my calculated value of 8.7 says Elenin is an ordinary standard small faint comet.

-----

Having said all that the Minor Planet Center says Elenin has an absolute magnitude of 8.0
the size (depending on the albedo) a few tens of km across...

Table

So there, its just a comet.

reply to post by alfa1

Thanks for that alfa1. Brilliant!

You've certainly opened a can of worms, haven't you

Based on an absolute magnitude of 8.7 and the Minor Planet Center's statement that "Most main-belt minor planet will have albedos in the range 0.05 to 0.25":

www.minorplanetcenter.org...

...I then used the following page to calculate the diameter:

www.physics.sfasu.edu...

At the lower end, using an Albedo of 0.25, Elenin has a diameter of 48.367 km

At the upper end, using an Albedo of 0.05, Elenin has a diameter of 108.153 km

Now this is where it gets interesting (as if it isn't already). From the wiki links below:

"A minor planet is an astronomical object in direct orbit around the Sun that is neither a dominant planet nor a comet."
en.wikipedia.org...

and

"Main-belt comets are bodies orbiting within the main asteroid belt..."
en.wikipedia.org...

So, we now know that Elenin isn't:

1) A minor planet
2) A main belt minor planet
3) A main belt comet

From further research, I found that a long period comet is actually called a "dirty snowball" and Elenin is a long period comet...apparently from the Oort cloud. See page 2 of the following document regarding fluffy snowballs:
www.planetarydefense.info...

Comets can "apparently" have an Albedo as little as 0.01 (though controversial):
www.lpi.usra.edu...

Let's hope Elenin's Albedo is larger than that

...and let's not even start on the "rate of brightening".

Cheers

JS

reply to post by jumpspace

In order to calculate those trajectories though, they would have to know the mass of Elenin as well as its composition and to do this, they would have to know the size of the nucleus.

The mass maybe, but not the composition. That does not need to be known.

Frankly, I do not believe that mass has to be known either as long as the mass of the object is small enough that the motions of the planets are not affected. This is a simplifying assumption that is likely to be true because any affects of the mass of a comet are so small that they cannot be measured. For all practical purposes then the comet does not affect the planet, but the planet affects the comet.

reply to post by stereologist

stereologist:

Well said!

"...but the planet affects the comet."

Very true. It will be interesting to see if it maintains the same JPL trajectory as it passes the Sun

Cheers

JS

reply to post by jumpspace


S&F for the data shared within this thread.