<h3>Direct viewing through an eyepiece</h3>
<p>When looking through an eyepiece, there is an <a href="https://en.wikipedia.org/wiki/Eyepiece#Field_of_view" rel="nofollow noreferrer">apparent field of view</a>. <a href="https://astronomy.stackexchange.com/a/40462/7982">This answer</a> says:</p>
<blockquote>
<p>Because I was lazy, I used the default telescopes and eyepieces.</p>
</blockquote>
<p>which means that the circles shown are what one would see if one had an eyepiece with a field of view equal to the default eyepiece used in the Stellarium simulation. If your eyepieces have a narrower apparent FOV then the circle should be drawn smaller.</p>
<p>The actual FOV is the apparent FOV divided by the magnification.</p>
<p>SO if you have f=1000 mm and a 10 mm eyepiece you have 100x magnification. If your actual eyepiece has an apparent FOV of 50 degrees then your actual FOV is 0.5 degrees or 30 arc minutes.</p>
<p><strong>Just recently</strong> they have become closer than that so from now until the end of December you would be able to see them with the eyepiece I've just described.</p>
<p><em>Your milage may vary</em></p>
<h3>Using your camera</h3>
<p><em><strong>The FOV of your camera's sensor is a totally different issue!</strong></em></p>
<p>It is rectangular not circular, and so you may have to rotate your camera body around the axis of the telescope to get the long direction or the diagonal direction parallel to the line between the two planets.</p>
<p>Your camera's documentation will give you the CCD sensor's dimensions in millimeters.</p>
<p>Let's say it's h x w = 4mm x 6mm.  Assuming there are no lenses in between and your Newtonian telescope's primary mirror's focus is directly on the sensor, the FOV of your sensor is h/f x w/f or 4/1000 x 6/1000 radians. Multiply by <span class="math-container">$180 / \pi$</span> to get degrees, so it will be 0.23 x 0.34 degrees, <strong>a lot smaller than for an eyepiece!</strong></p>
<p>Diagonally it's <span class="math-container">$\sqrt{(w/f)^2 + (h/f)^2}$</span> or 0.41 degrees.</p>
<p>Since your sensor will be a different size:</p>
<p><em>Your milage may vary</em></p>

CC BY-SA 4.0
2020-12-18T08:21:32.937Z
FWIW, I actually used a 10mm eyepiece in the simulation. By "using the default telescopes", I meant that I didn't add the exact parameters of the OP's telescope, but chose some of the presets that had the same focal length.
jng224
CC BY-SA 4.0
2020-12-18T10:58:12.330Z
@Jonas yep, the OP didn't state their FOV so there was no way to go further. I just added the eyepiece part here as supplemental; in case they do know their FOV they'll be able to calculate from scratch.
uhoh
CC BY-SA 4.0
2020-12-18T18:39:54.200Z
I followed @Jonas example and opened Stellarium and tried to enter the params of my telescope. For the scope it was easy, but for the eyepieces, I was prompted for the aFOV and I don't know what this is. I have 10mm and 20mm eyepieces that came with the scope. I also bought a 2x barlow adapter (you can remove the magnifying lens).
S. Imp
CC BY-SA 4.0
2020-12-18T23:46:09.700Z
@S.Imp "How to estimate my eyepiece FOV?" Is an *excellent* new question and it will receive some great answers! Include a photo and mention all writing you can find especially the type (e.g. Plössl, Ortho, Erfle, etc...) With no FOV printed on the eyepiece then I think you can guess or use a default setting in the software which is *their guess.* My guess is 50° to 55° for the ones that came with your telescope. There are ways you can measure it, so please post that new question to allow room for answers!
uhoh
CC BY-SA 4.0
2020-12-18T23:49:08.130Z
@S.Imp For the Barlow you use the same procedure as above but divide by $M$ the magnification of the Barlow. So with $f_T, f_E$ of 1000 and 10 mm and a 50 degree aFOV and $M$ of 2.2 you'll have an FOV about 0.23 degrees.
uhoh
CC BY-SA 4.0
2020-12-19T01:14:20.093Z
@uhoh i truly appreciate this extra info, and will post a new question when i get a moment. still trying to finish work. womp womp
S. Imp
CC BY-SA 4.0
2020-12-19T01:21:46.380Z
@S.Imp Okay great, it can be short, just link back to this page for reference. I've got an answer in the works already and there will likely be others.
uhoh
CC BY-SA 4.0
2020-12-19T02:14:48.933Z
@uhoh i posted the question, and tried to tag you in a comment.
S. Imp
CC BY-SA 4.0
2020-12-19T02:19:41.483Z
@S.Imp Great, thanks! I got a notification for your comment here because it's my post and because I've already commented here, which is good because I didn't get a notification for your other comment because it's not my post nor have I commented yet there. For more on how all of that works go to [Stack Exchange FAQ](https://meta.stackexchange.com/q/7931/303080) and then click [How do comment `@replies` work?](https://meta.stackexchange.com/q/43019/303080)
uhoh
CC BY-SA 4.0
2020-12-18T01:36:48.100Z
2020-12-18T01:36:48.100Z
uhoh
<p>This can easily be tested using software such as <a href="https://stellarium.org/de/" rel="noreferrer">Stellarium</a>, where you can visualize the field of view with given focal length.</p>
<p>If you have the software installed, click on &quot;ocular view&quot; (the most left button in the upper right corner). The following view is what you culd expect on December 21 with a 10mm eyepiece and a focal length of 1000mm:</p>
<p><a href="https://i.sstatic.net/wRfFl.png" rel="noreferrer"><img src="https://i.sstatic.net/wRfFl.png" alt="enter image description here" /></a></p>
<p>And this is the view when attaching a Crop-sensor (APS-C) Camera to a telescope of focal length 1000mm:</p>
<p><a href="https://i.sstatic.net/zBhNS.png" rel="noreferrer"><img src="https://i.sstatic.net/zBhNS.png" alt="enter image description here" /></a></p>
<p>So you should be able to see both planets and some moons in your telescope.</p>
<p><strong>Some notes:</strong></p>
<ul>
<li><p>These simulated views do not necessarily represent the actual view to 100% accuracy. I did leave the &quot;atmosphere&quot; setting turned on, but actual visibility and brightness will depend on your location (light pollution and other atmospheric effects such as haze might reduce visibility)</p>
</li>
<li><p>Because I was lazy, I used the default telescopes and eyepieces. This ignored aperture (or rather, it was different than on your telescope), but it won't have an effect on magnification. If you wanted, you could add your own telescope in the software's settings.</p>
</li>
</ul>

CC BY-SA 4.0
2020-12-17T21:36:25.970Z
Thanks, @jonas, for this great answer.
S. Imp
CC BY-SA 4.0
2020-12-17T19:49:05.483Z
2020-12-17T19:49:05.483Z
jng224
<p>I have a <a href="https://www.celestron.com/products/astromaster-114eq-telescope" rel="nofollow noreferrer">Celestron Astromaster 114</a> Newtonian reflector telescope. The specs:</p>
<ul>
<li>aperture: 114mm</li>
<li>focal length: 1000mm</li>
<li>focal ratio: f/8.77</li>
</ul>
<p>I also have a <a href="https://www.celestron.com/products/t-ring-for-nikon-camera" rel="nofollow noreferrer">t-ring</a> so I can connect a Nikon camera to it.</p>
<p>Will I be able to see both Jupiter and Saturn at the same time thru this telescope? I understand <a href="https://www.nationalgeographic.com/science/2020/12/see-rare-great-conjunction-of-jupiter-saturn-on-winter-solstice/" rel="nofollow noreferrer">from this article</a> that they will come within a tenth of a degree of each other. The article says this is &quot;fifth of the width of the full moon&quot; which, based on my experience, suggests that I will indeed be able to see both, but I'm hardly any expert.</p>
<p>I'm considering borrowing a friend's film camera to try and snap a photo of both in one image, but would like to avoid the trouble (thanks, COVID!) if this won't happen.</p>

CC BY-SA 4.0
2020-12-18T01:37:52.060Z
I've added another answer. It's usually a good idea to refrain from accepting an answer for a few days at least. There are several astrophotographers here but they don't necessarily check in every day.
uhoh
<telescope><photography><jupiter><saturn><solstice>
Will I be able to see both Jupiter and Saturn at the same time in my Astromaster 114?
S. Imp
# Will I be able to see both Jupiter and Saturn at the same time in my Astromaster 114?

(Asked by: username_0 on 2020-12-17)

I have a [Celestron Astromaster 114](https://www.celestron.com/products/astromaster-114eq-telescope) Newtonian reflector telescope. The specs:

* aperture: 114mm
* focal length: 1000mm
* focal ratio: f/8.77

I also have a [t-ring](https://www.celestron.com/products/t-ring-for-nikon-camera) so I can connect a Nikon camera to it.

Will I be able to see both Jupiter and Saturn at the same time thru this telescope? I understand [from this article](https://www.nationalgeographic.com/science/2020/12/see-rare-great-conjunction-of-jupiter-saturn-on-winter-solstice/) that they will come within a tenth of a degree of each other. The article says this is "fifth of the width of the full moon" which, based on my experience, suggests that I will indeed be able to see both, but I'm hardly any expert.

I'm considering borrowing a friend's film camera to try and snap a photo of both in one image, but would like to avoid the trouble (thanks, COVID!) if this won't happen.

**Tags:** telescope, photography, jupiter, saturn, solstice
> **Comments:**
> **username_1**: I've added another answer. It's usually a good idea to refrain from accepting an answer for a few days at least. There are several astrophotographers here but they don't necessarily check in every day.  

---

## 2 answers

### Answer by username_2 on 2020-12-17. Score: 5 | Accepted answer

This can easily be tested using software such as [Stellarium](https://stellarium.org/de/), where you can visualize the field of view with given focal length.

If you have the software installed, click on "ocular view" (the most left button in the upper right corner). The following view is what you culd expect on December 21 with a 10mm eyepiece and a focal length of 1000mm:

[![enter image description here](https://i.sstatic.net/wRfFl.png)](https://i.sstatic.net/wRfFl.png)

And this is the view when attaching a Crop-sensor (APS-C) Camera to a telescope of focal length 1000mm:

[![enter image description here](https://i.sstatic.net/zBhNS.png)](https://i.sstatic.net/zBhNS.png)

So you should be able to see both planets and some moons in your telescope.

**Some notes:**

* These simulated views do not necessarily represent the actual view to 100% accuracy. I did leave the "atmosphere" setting turned on, but actual visibility and brightness will depend on your location (light pollution and other atmospheric effects such as haze might reduce visibility)
* Because I was lazy, I used the default telescopes and eyepieces. This ignored aperture (or rather, it was different than on your telescope), but it won't have an effect on magnification. If you wanted, you could add your own telescope in the software's settings.

> **Comments:**
> **username_0**: Thanks, @jonas, for this great answer.  

---

### Answer by username_1 on 2020-12-18. Score: 4

### Direct viewing through an eyepiece

When looking through an eyepiece, there is an [apparent field of view](https://en.wikipedia.org/wiki/Eyepiece#Field_of_view). [This answer](https://astronomy.stackexchange.com/a/40462/7982) says:

> Because I was lazy, I used the default telescopes and eyepieces.

which means that the circles shown are what one would see if one had an eyepiece with a field of view equal to the default eyepiece used in the Stellarium simulation. If your eyepieces have a narrower apparent FOV then the circle should be drawn smaller.

The actual FOV is the apparent FOV divided by the magnification.

SO if you have f=1000 mm and a 10 mm eyepiece you have 100x magnification. If your actual eyepiece has an apparent FOV of 50 degrees then your actual FOV is 0.5 degrees or 30 arc minutes.

**Just recently** they have become closer than that so from now until the end of December you would be able to see them with the eyepiece I've just described.

*Your milage may vary*

### Using your camera

***The FOV of your camera's sensor is a totally different issue!***

It is rectangular not circular, and so you may have to rotate your camera body around the axis of the telescope to get the long direction or the diagonal direction parallel to the line between the two planets.

Your camera's documentation will give you the CCD sensor's dimensions in millimeters.

Let's say it's h x w = 4mm x 6mm. Assuming there are no lenses in between and your Newtonian telescope's primary mirror's focus is directly on the sensor, the FOV of your sensor is h/f x w/f or 4/1000 x 6/1000 radians. Multiply by $180 / \pi$ to get degrees, so it will be 0.23 x 0.34 degrees, **a lot smaller than for an eyepiece!**

Diagonally it's $\sqrt{(w/f)^2 + (h/f)^2}$ or 0.41 degrees.

Since your sensor will be a different size:

*Your milage may vary*

> **Comments:**
> **username_2**: FWIW, I actually used a 10mm eyepiece in the simulation. By "using the default telescopes", I meant that I didn't add the exact parameters of the OP's telescope, but chose some of the presets that had the same focal length.  
> **username_1**: @Jonas yep, the OP didn't state their FOV so there was no way to go further. I just added the eyepiece part here as supplemental; in case they do know their FOV they'll be able to calculate from scratch.  
> **username_0**: I followed @Jonas example and opened Stellarium and tried to enter the params of my telescope. For the scope it was easy, but for the eyepieces, I was prompted for the aFOV and I don't know what this is. I have 10mm and 20mm eyepieces that came with the scope. I also bought a 2x barlow adapter (you can remove the magnifying lens).  
> **username_1**: @S.Imp "How to estimate my eyepiece FOV?" Is an *excellent* new question and it will receive some great answers! Include a photo and mention all writing you can find especially the type (e.g. Plössl, Ortho, Erfle, etc...) With no FOV printed on the eyepiece then I think you can guess or use a default setting in the software which is *their guess.* My guess is 50° to 55° for the ones that came with your telescope. There are ways you can measure it, so please post that new question to allow room for answers!  
> **username_1**: @S.Imp For the Barlow you use the same procedure as above but divide by $M$ the magnification of the Barlow. So with $f_T, f_E$ of 1000 and 10 mm and a 50 degree aFOV and $M$ of 2.2 you'll have an FOV about 0.23 degrees.  
> **username_0**: @uhoh i truly appreciate this extra info, and will post a new question when i get a moment. still trying to finish work. womp womp  
> **username_1**: @S.Imp Okay great, it can be short, just link back to this page for reference. I've got an answer in the works already and there will likely be others.  
> **username_0**: @uhoh i posted the question, and tried to tag you in a comment.  
> **username_1**: @S.Imp Great, thanks! I got a notification for your comment here because it's my post and because I've already commented here, which is good because I didn't get a notification for your other comment because it's not my post nor have I commented yet there. For more on how all of that works go to [Stack Exchange FAQ](https://meta.stackexchange.com/q/7931/303080) and then click [How do comment `@replies` work?](https://meta.stackexchange.com/q/43019/303080)  

---

