# Astronomy problem

1. Oct 10, 2005

### big man

Hi guys,

I've been having trouble with this problem and would appreciate just some direction to assist me.

I have to estimate the motion of Pluto in "/hour from observations given that the telescope is a 14inch f/11 with an AP7 CCD camera with pixels 24microns on a side.

Now we have two pictures of Pluto at epochs separated by a little more than half an hour. I assume that since the resolution of the camera is given you'd have to measure something on the picture. The other though I had was that you would need to find the image scale of the telescope (which I've done).

Image scale = $$d\theta/ds$$ = $$206265/F$$ = 52.7" per mm.

Where F is the focal length determined by the formula f=F/D.

I swear I have to use the image scale of the telescope and the resolution of the camera together to get any further in this problem, but like I said I don't really have a clue.

2. Oct 10, 2005

### SpaceTiger

Staff Emeritus
I'm not sure why you think you would need the resolution to get the answer. If the separation of the object in the two images is greater than your resolution limit, then all you should need is the image scale and a ruler.

3. Oct 10, 2005

### big man

well haha like I said I didn't really have any idea. I just thought that since he gave us the information on the camera we would somehow need to incorporate it into the calculation. So then would I just find the distance moved by Pluto on the picture in maybe 40 minutes and then multiply that by the image scale???

4. Oct 10, 2005

### SpaceTiger

Staff Emeritus
Don't forget that you want the rate of motion, not the absolute motion (i.e. you have to divide by the time interval).

I suppose you could calculate the resolution and compare to the separation you measure to verify that it's beyond the limit. It should be obvious from direct comparison of the images, however.

5. Oct 10, 2005

### big man

yeah sorry my phrasing wasn't too good there, I meant the rate of motion when I said the distance travelled in 40 minutes.

Thanks again for your explaining this stuff to me.

It takes me a little longer than most people to understand this Astronomy stuff haha : )

6. Oct 10, 2005

### big man

Wait sorry for re-opening this Spacetiger, but I was just looking at the images and you can JUST see Pluto move a little. Sorry I thought I remembered it being more pronounced than that, but it is a very small movement. It's maybe a millimeter if you're lucky, but that estimation has a great deal of uncertainty...

7. Oct 10, 2005

### SpaceTiger

Staff Emeritus
Are you overlaying the images? If the apparent motion is noticably larger than the background stars, then it's probably resolved, but again, you can just do the calculation and see for yourself. Do you know the equation for calculating resolution?

8. Oct 11, 2005

### big man

I'm actually blinking the images using the CLEA software we have, but they could easily be overlayed.

I've attached one of the pictures with Pluto circled so you can see what it looks like.

For the resolution of the camera wouldn't it just be the dimension of the picture divided the dimension of a single pixel??

ie Res = Dimension/(24*24 microns)

#### Attached Files:

• ###### PlutoC1.jpg
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9. Oct 11, 2005

### SpaceTiger

Staff Emeritus
Pixel sizes are often chosen to roughly correspond with the resolution limit, but they don't determine it. Try this:

$$\theta=\frac{1.22\lambda}{D}$$

where $\lambda$ is the wavelength of the light, D is the diameter of the telescope's aperture, and $\theta$ is the minimum angle that can be resolved.

10. Oct 11, 2005

### big man

ahh k so you mean the smallest resolvable angle by the telescope.
Well I estimated it as 2.4*10^-6 radians (or 8.26*10^-3 minutes of arc) so yeah now I just compare that to the distance measured on the picture. Like you said it is obvious that it's resolved, but I guess it would be better to demonstrate it with the above calculation.

I was thinking that a better way to find the distance moved by Pluto would be to overlay the images and then use photoshop to find the x,y co-ordinates of pixels. Using that you can use the right-angle formula to determine the resultant change in position in terms of pixels and then multiple by 24 microns to get the distance moved. Can that be done or should I just use a ruler??

11. Oct 11, 2005

### big man

Just an additional question spacetiger and then this is it haha : )

I took a picture of a galaxy which is meant to be of size 10.2 x 9.5 arcmin (M74) and it fits perfectly in the middle of the picture taken by the ccd on the telescope. But the thing is the field of view of the telescope is 10 arcmin. So wouldn't the dimension with 10.2 arcmin be slightly truncated?

I've attached the pic.

#### Attached Files:

• ###### M74-R.jpg
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12. Oct 11, 2005

### SpaceTiger

Staff Emeritus
If that's the number, then your resolution limit is likely determined by atmospheric turbulence. Normal telescopes can't resolve below a few arcseconds.

Either way is fine.

13. Oct 11, 2005

### SpaceTiger

Staff Emeritus
It's hard to say. Either number could be an approximation or misnormalization. I wouldn't worry about it.

14. Oct 11, 2005

### big man

Thanks a lot for your help Spacetiger. I really appreciate it and now I know more about telescopes which is good :).

Thanks again