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max8166
2011-Nov-06, 11:12 PM
An Asteriod is going to buzz our planet in the next couple of days and NASA has prepared this Graphic to show how close it will be:

http://neo.jpl.nasa.gov/images/2005_YU55_approach_movie.gif

My question would be,
As the asteriod appears to be coming from the approx. Sun Location , can anyone say how close it was to the Sun, and considering it was on a parabolic orbit, how fast would it have been travelling at this closest approach?
Just 'cos it appears to be going pretty fast outwardly bound and must have been decelerating since passing the Sun 93 odd million miles ago.

Info Ref: http://neo.jpl.nasa.gov/news/news171.html (http://neo.jpl.nasa.gov/news/news171.html)

antoniseb
2011-Nov-06, 11:20 PM
The illusion is that it is coming from the Sun's direction, but really, it is just running in orbit parallel to the Earth's orbit, and crossing our path. If you could see the diagram, not relative to the Earth, but relative to a fixed line from the Sun to a star, it would be more clear.

Nowhere Man
2011-Nov-07, 02:31 AM
This animation may help. (http://www.curtrenz.com/asteroids09.html) From JREF.

Edit to add: Here is a static image. (http://www.curtrenz.com/asteroids06.html)

Fred

Peter B
2011-Nov-07, 02:46 AM
My question would be, As the asteriod appears to be coming from the approx. Sun Location , can anyone say how close it was to the Sun...

According to Wikipedia, its closest approach to the Sun is about 0.65AU, which is about 100 million kilometres.

...and considering it was on a parabolic orbit, how fast would it have been travelling at this closest approach?

I can'y help you with the speed, although I'll guess it'll be similar to Venus's orbital speed. It's worth clarifying thought that it's not on a parabolic orbit. It's in an elliptical orbit, with its aphelion at about 1.63AU.

Peter B
2011-Nov-07, 02:53 AM

YU55's Wikipedia page shows a radar image of the asteroid. But the image is of a crescent asteroid.

Why would a radar image show a crescent shape and not the entire asteroid?

StupendousMan
2011-Nov-07, 10:18 AM
When scientists bounce radio waves off an asteroid, the reflections will come back with slightly different frequencies if the asteroid is rotating. The waves which bounce off the limb of the asteroid which is rotating toward the Earth will have a slightly higher frequency, due to the Doppler effect; waves bouncing off the limb which is receding from Earth will have a slightly lower frequency.

If one chooses to make a picture using only the reflected radio waves which have, say, slightly higher frequencies than the original wave, then one will only see the limb which is rotating toward Earth. One _could_ choose to make a picture using all the waves, with all the frequencies, and that would give a picture which shows the entire asteroid.

So, my guess is that the creator of this particular image decided to choose only reflected radio waves with higher-than-average (or lower-than-average) frequencies.

IsaacKuo
2011-Nov-08, 03:30 PM
If one chooses to make a picture using only the reflected radio waves which have, say, slightly higher frequencies than the original wave, then one will only see the limb which is rotating toward Earth. One _could_ choose to make a picture using all the waves, with all the frequencies, and that would give a picture which shows the entire asteroid.

So, my guess is that the creator of this particular image decided to choose only reflected radio waves with higher-than-average (or lower-than-average) frequencies.

This guess is completely wrong. In fact, both sides of the asteroid are shown, superimposed on top of each other. This is an unavoidable artifact of the way these images are created. It's called a doppler-delay (http://planetary.org/blog/article/00002462/) image.

A doppler-delay image is not oriented the way you think. The point closest to Earth is actually the top of the image, while the point furthest away is actually the bottom of the image. The Y-axis represents the delay.

The X-axis is used for doppler shift. The entire range of doppler shift is used. This axis is roughly equivalent to the east/west dimension. However, there's a problem. For each pixel on the doppler-delay image, there are two points on the asteroid which provide returns. One is in the northerm hemisphere, while the other is in the southern hemisphere. So, the image we see actually squashes both the northern and southern hemispheres onto a single image. This complicates visual interpretation of the image.

Githyanki
2011-Nov-11, 02:27 AM
At work, I told my coworkers that the point of light that is Jupiter right next to the Moon was the asteroid and my, it's getting larger and larger.