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I wanted to start astrophotography.

Let's assume that I want to take pictures of planets when they are the closest to the planet earth, which planet would look the biggest thought a telephoto lens? Some planets are smaller but closer to the planet earth (Mars), some are more far away but way bigger (Like Jupiter), so I don't know which planet is the easiest to take a picture. I know that a 800mm lens with an APS-C camera is enough to see some small details of Jupiter, but what about the other planets?

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    \$\begingroup\$ Note: although I don't think this is off-topic here, my first stack of call would have been astronomy. If you decide that you might get better answers there, you can use the "flag" link to ask the moderators to migrate it. \$\endgroup\$ Commented Mar 31, 2018 at 20:11
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    \$\begingroup\$ The pedantic answer would, of course, be "Earth." \$\endgroup\$ Commented Mar 31, 2018 at 23:09
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    \$\begingroup\$ What is your budget? Do you already have a telescope? What are your expectations? \$\endgroup\$ Commented Apr 1, 2018 at 14:03
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    \$\begingroup\$ I think it's worth mentioning that the planet that looks biggest through a camera is the same as the planet that looks biggest with the naked eye (all other things being equal). Adding a camera lens to the view makes no difference to the relative sizes. \$\endgroup\$
    – osullic
    Commented Apr 1, 2018 at 22:29
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    \$\begingroup\$ @osullic That's not necessarily true. Because planets are so small, how large they look to our brains is based more on their brightness than on actual size. \$\endgroup\$
    – Michael C
    Commented Apr 2, 2018 at 3:57

6 Answers 6

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Because the distance from Earth to each of the other planets varies due to orbital mechanics, the size of each planet as seen from Earth can vary significantly. Which planet is the largest and the order of relative sizes changes frequently.

For example, right now as of April 1, 2018 the following are the angular sizes of the planets as viewed from Earth:

  • Jupiter - 42.69" (arcseconds)
  • Saturn - 16.68"
  • Mercury - 11.27"
  • Venus - 10.59"
  • Mars - 8.49
  • Uranus - 3.38"
  • Neptune - 3.21

Venus will pass Mercury in size on April 12, 2018.
Mars will pass Mercury in size on April 19, 2018.
Mars will grow larger than Venus on May 7, 2018.
Mars will grow larger than Saturn on June 18, 2018.
Venus will overtake Saturn in size on July 20, 2018.
Venus will once again be larger than Mars on August 15, 2018.
Venus will grow larger than Jupiter on September 12, 2018.
Mars will shrink to smaller than Saturn on September 26, 2018.
Venus will peak in angular size at 1'1.33" (one arcminute and 1.33 arcseconds) on October 27, 2018.

By October 27, 2018 (less than seven months from now), the list will look like this:

  • Venus - 1'1.33"
  • Jupiter - 31.44"
  • Saturn - 15.79"
  • Mars - 12.28"
  • Mercury - 5.70"
  • Uranus - 3.73"
  • Neptune - 2.33"

By mid-December of 2018 Venus will once again be smaller than Jupiter.

At the end of July, 2019 the pecking order will look like this:

  • Jupiter - 42.68"
  • Saturn - 18.25"
  • Mercury - 9.68"
  • Venus - 9.66"
  • Uranus - 3.56"
  • Mars - 3.53"
  • Neptune - 2.34"

When it is closest to the Earth Venus has the largest angular size of any of the planets as seen from Earth. At its maximum, Venus is 0.01658 degrees wide. This is very close to exactly one arcminute, which is 1/60th of a degree. Venus is only larger than Jupiter for a few weeks (about 13-14 weeks from mid-September to mid-December in 2018) once every year and a half or so. The rest of the time Jupiter is larger than the other planets.

Unfortunately, when Venus is closest to the Earth and at its largest angular size, this means Venus is also almost directly between the Earth and the Sun and most of the side of Venus facing the Earth is dark while the bright sun is almost directly behind it. On very rare occasions, Venus and Earth's orbit align just right and Venus passes directly in front of the Sun as seen from Earth. We call this event a transit. The last transit of Venus occured on June 5, 2012. The next one will not be until December in the year 2117 followed by another in December 2125. They occur in pairs about 8 years apart, then there is a gap that alternates between 121.5 years and 105.5 years before the next pair occurs.

enter image description here
The large dot near the upper right is Venus. The smaller dots in the middle are sunspots. There are some thin clouds at the bottom of the solar disc.

Since Venus and Earth are both interior planets, their relative distance varies greatly. During conjunction they are only 41.4 million kilometers apart. At opposition (when Venus is directly on the other side of the Sun from Earth), they are 257.757 million kilometers apart. At that distance, Venus is slightly less than 10 arcseconds (.16 arcminutes or 0.00278 degrees wide).

Jupiter varies from about 32 arcseconds at opposition to 49 arcseconds (0.817 arcminutes or 0.0136 degrees) at conjunction. Most of the time Jupiter is larger than 40 arcseconds. Since Jupiter is an outer planet and is five times further from the Sun than Earth, the distance between Earth and Jupiter is much less variable than is the case with the other inner planets. It also means that when Jupiter and Earth are closest, the sun is 180° on the other side of the Earth and almost all of the part of Jupiter seen from Earth is illuminated by sunlight and Jupiter is also at its brightest when at its largest.

enter image description here
Jupiter as observed on January 21, 2013. It was about 44 arcseconds wide at the time. Canon 7D + Kenko 2X Teleplus Pro 300 DGX + EF 70-200mm f/2.8 L IS II. Image is a 100% crop.

Mars varies from about 25 arcseconds (0.00694 degrees) at conjunction to 3.5 arcseconds (less than 0.001 degrees) at opposition. This sometimes means Mars is smaller than Uranus at opposition. Since Mars' orbit is outside of Earth's orbit, it is near fully illuminated as seen from Earth when it is largest and hidden behind or very near to the Sun when it is smallest.

Saturn averages about 16-20 arcseconds (not including the wider angular size of Saturn's ring system) as seen from Earth. Since its orbit is almost twice as large as Jupiter's, the variation in size between conjunction and opposition is even less than Jupiter's.

The other planets are much smaller than the average sizes of those listed above in terms of angular size as seen from Earth. Mercury (Maximum of about 10 arcseconds) and Uranus (Maximum of just over 3.5 arcseconds) can be larger than Mars at times when Mars is at its most distant (just under 3.5 arcseconds). Jupiter never drops below second place, while Venus can vary anywhere from largest to fifth largest (though it only falls past fourth largest on rare occasions when both Mercury and Mars are larger than Venus at the same time). Mars can be anywhere from second to seventh largest. Notice that the most variable planets are the ones whose orbits are closest to the Earth's orbit and the least variable planets are the ones with orbits that are much larger than earth's orbit.

In contrast, the Sun and Moon are both about 0.5 degrees, or 30 arcminutes or 1,800 arcseconds as seen from the surface of the Earth. That is 30 times the width of Venus at its closest (and least percentage illuminated) and 36 times wider than Jupiter at its closest and brightest.

enter image description here
Jupiter on the left and the moon on the right. Note the comparative sizes. Later in the evening of when this image was taken on January 21, 2013, they passed within less than one degree of one another. Jupiter was about 44 arcseconds in width at the time.

Of course if one is standing on a flat piece of the Earth, it has an angular size of 180 degrees (10,800 arcminutes or 648,000 arcseconds) which is 360X more than the Sun and Moon!

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    \$\begingroup\$ "The next one will not be until December in the year 2117 followed by another in December 2025." Is that last date supposed to be 2125, perhaps? \$\endgroup\$
    – Cornstalks
    Commented Mar 31, 2018 at 23:27
  • \$\begingroup\$ "They occur in pairs about 8 years apart, then there is a gap that alternates between 121.5 years and 105.5 years before the next pair occurs." I love orbital mechanics. So elegantly complex and yet so satisfyingly repeatable at the same time. There's no other way to say it: studying the relative motion of the planets rocks. \$\endgroup\$ Commented Apr 1, 2018 at 12:36
  • \$\begingroup\$ @LightnessRacesinOrbit Well, there is a kind of precession to orbital mechanics as well. The pattern of 105.5, 8, 121.5 and 8 years is not the only pattern that is possible within the 243-year cycle, because of the slight mismatch between the times when the Earth and Venus arrive at the point of conjunction. Prior to 1518, the pattern of transits was 8, 113.5 and 121.5 years, and the eight inter-transit gaps before the AD 546 transit were 121.5 years apart. The current pattern will continue until 2846, when it will be replaced by a pattern of 105.5, 129.5 and 8 years. \$\endgroup\$
    – Michael C
    Commented Apr 1, 2018 at 14:36
  • \$\begingroup\$ Note: When this answer was originally written it was based on data from a source that said it listed each planet's average angular size when some of the planets were listed at their maximum angular size. I used this data to extrapolate the maximum/minimum size of each planet based upon the mistaken assumption that the maximum size was the average size. The answer has been updated to reflect more accurate numbers. \$\endgroup\$
    – Michael C
    Commented Apr 1, 2018 at 16:25
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    \$\begingroup\$ Thank you very much, I really appreciate your very detailled answer. \$\endgroup\$ Commented Apr 6, 2018 at 9:24
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Normally Jupiter is easily the largest seen from Earth, but depending on orbits, it could sometimes be Venus (next time in September, and then next in 2020).

This site will answer about details relative to exact date: https://www.timeanddate.com/astronomy/planets/distance

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    \$\begingroup\$ Venus, like Mercury, is an inferior planet (with respect to Earth). This means when it is closest to Earth, it presents its dark side (night side), so like a New Moon it is almost invisible. In addition, its angular separation from the Sun, as seen from us, is small when it is closest. So for inferior planets, the time when they are closest to Earth is not the best time to take pictures of them. In contrast, for superior planets, like Mars and Jupiter, it is perfect to observe when they are closest. \$\endgroup\$ Commented Apr 2, 2018 at 9:48
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Though the angular size of Venus in the Earth sky is larger than any other planet, because Venus is an inferior planet that largest angular size happens only when Venus in the the direction of the Sun. Jupiter has the next-largest angular size and it occurs when Jupiter is in opposition, thus is is also in it's most well-lit state (for an observer on Earth). Also, Venus' angular size varies by an order of magnitude as it and the Earth orbit the Sun, whereas much further Jupiter has a more subtle variation from largest- to smallest-diameter. This is very obvious in telescopes and cameras.

Note that Jupiter has very large features (bands, Great Red Spot) that Venus lacks, so if you are interested in seeing detail as opposed to a blank circle then Jupiter can provide that detail. Venus will, however, show a crescent similar to the Moon's phases, whereas Jupiter will not.

Note also that Jupiter has four very large moons, and these are very easy to photograph. So though you may or may not be able to resolve the bands or Great Red Spot on Jupiter, you will most likely be able to photograph the moons and see how their position changes from night to night. You don't even need Jupiter to be at opposition to photograph them, they are clearly visible throughout Jupiter's orbit.

For example, here is a photo of Jupiter made with stacked images taken through a Logitech webcam attached to a telescope:

enter image description here Image Source Includes other pictures of Jupiter shot through common DSLR cameras from Nikon and Canon.

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  • \$\begingroup\$ Your answer almost makes it appear the image of Jupiter was taken with a webcam pointed at the sky. Was this the case? Or was it taken by pointing the webcam at the eyepiece of a telescope? \$\endgroup\$
    – Michael C
    Commented Apr 1, 2018 at 15:03
  • \$\begingroup\$ The image source implies that image stacking was used. That technique takes several hundred images and selects the best pixels from each of them, as atmospheric distortion varies it will sharpen and soften different parts of the image. EDIT: Further down the page is in fact a photo of the webcam attached to a dobsonian telescope. \$\endgroup\$
    – dotancohen
    Commented Apr 1, 2018 at 15:15
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    \$\begingroup\$ @MichaelClark telescope+good webcam is a solid starting point for planetary imaging. You record a few minutes of video (longer and the planets rotation will begin to blur the result of stacking), break it into individual frames, and pick out the several dozen sharpest ones (atmospheric conditions will cause this to vary from moment to moment) and then combine them into a single whole one. \$\endgroup\$ Commented Apr 2, 2018 at 21:12
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    \$\begingroup\$ There're now dedicated planetary imaging cameras, but ~15 years ago before that happened people stacking images from a webcam could get better results than those with cameras 10x as expensive because the higher total number of images meant they were more likely to get ones where the atmosphere was momentarily still and the focus was at its sharpest maximum. \$\endgroup\$ Commented Apr 2, 2018 at 21:12
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    \$\begingroup\$ sucess with stacking for planetary images puts a heavy focus on recording at a high FPS, the more total images you could take before the planet rotated enough to cause blur stacking the 1st and last images the more you could be selective about which images had snapshotted the best seeing to combine into the final result. In the early 2000's top of the line webcams ruled that segment of astro imaging. General use astrocams and DSLRs could take better single images but couldn't compete with stacking by webcam imagers recording dozens of times more total frames. \$\endgroup\$ Commented Apr 3, 2018 at 18:50
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Short answer: Venus subtends the largest angle, followed by Jupiter.

Intermediate-length answer: Randall Munroe provides the following helpful visualisation (extracted from a larger visualisation at https://xkcd.com/1276/):

Angles subtended by major solar system bodies

Long answer: there is some variation due to relative positions in orbits. See Wayne's answer for an animation which shows how the relative sizes change over time.

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  • \$\begingroup\$ Thank you very much, that's what I wanted to know, have a nice day \$\endgroup\$ Commented Mar 31, 2018 at 20:13
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    \$\begingroup\$ Good luck viewing Venus when it is close enough the Earth to be that size, since the Sun is almost directly behind it at that point. \$\endgroup\$
    – Michael C
    Commented Mar 31, 2018 at 20:20
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    \$\begingroup\$ There is significant variation in the angular sizes of Venus and Mars and the order changes quite a bit. \$\endgroup\$
    – Michael C
    Commented Apr 1, 2018 at 4:33
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    \$\begingroup\$ @MichaelClark Luck isn't even needed. In march 2017, Venus had a magnitude of -4, had an apparent diameter of around 55'' (larger than Jupiter at opposition) and was in the sky for more than an hour after sunset. It was a beautiful crescent. \$\endgroup\$ Commented Apr 1, 2018 at 14:00
  • \$\begingroup\$ @EricDuminil Yes it was, but most of Venus as visible from Earth was in total darkness. The total area of reflected light visible from Earth was only a fraction of the size of a 55" disc. \$\endgroup\$
    – Michael C
    Commented Apr 1, 2018 at 15:59
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Don't buy that 800mm f/5.6 yet

Astrophotography with a DSLR is typically either done :

  • with a fast, wide angle lens in order to avoid star trails.
  • or mounted on a telescope with an adapter.

The first method is great to capture large structures in the sky (e.g. the milky way, Andromeda Galaxy, clusters or nebulas...)

The second one can be used for planets.

A 800mm is actually not that long for a telescope, and the corresponding aperture at f/5.6 is around 145mm, which isn't very large either. The 800mm f/5.6 is huge, expensive and would be hard to use for astrophotography.

Enjoy some visual astronomy first

From your question, I gather that you don't have much experience looking at planets. Visual astronomy could give you the experience needed in order to get good pictures.

Astrophotography is hard and requires a lot of money, experience and patience. You need to know where to point, at what time and under which sky conditions.

There are excellent, affordable amateur telescopes for $250 (e.g. this small dobsonian, a 900mm f/8). Many astrophotography adapters cost much more. You can see every planet with it, the Cassini division on Saturn rings, the great red spot on Jupiter as well as the Jovian moons or the ISS. With decent skies, you can see wonderful deep sky objects (e.g. Andromeda Galaxy, the Orion Nebula, the double cluster ...).

To change the magnification, you simply need another eyepiece, which are much more affordable than DSLR lenses.

Switch to astrophotography.

You can even use a webcam or a DSLR to take pictures through the telescope. Here's an example of Jupiter with the great red spots, 2 moon transits and Io:

enter image description here

It was taken as a single exposure with a Fuji X100s through a $600 dobsonian (1250mm f/5). 1/50s, f/4, ISO 1600. I had to :

  • manually track the telescope
  • manually focus the eyepiece (6.7mm)
  • hold the camera to point through the eyepiece
  • focus the camera
  • release the shutter.

Some amateur astrophotographers manage to take incredible pictures of the planets. Here are some examples.

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Just as there is no "best" camera or "best" lens ... there is no "best" telescope -- there are merely telescopes better suited to certain tasks than others.

While you can certainly attach a camera, point a telescope toward a planet, and capture an image, the quality of that image will depend on quite a few other factors (some of which are beyond your control).

Atmospheric Seeing Conditions

Due to the very tiny apparent size of another planet as seen from Earth, image quality is very sensitive to atmospheric stability here on Earth. Astronomers refer to this as "seeing conditions". The analogy I prefer to use is to imagine a coin resting on the bottom of a pool of clear water. If the water is still you can see the coin. If someone starts creating waves (either small ripples or large waves) the view of the coin will begin to distort and wobble. This same issue happens with our atmosphere when viewing the planets.

To get a stable atmosphere you want to make sure you are not within a couple hundred miles of either the jet-stream, a warm-front, or a cold-front. You also want to be located in some place where the geography is flat (and preferably water) to allow for smooth laminar airflow. Hot land will create thermals ... so cool land (high up in mountains) or looking over cool water will be helpful. Also the optical surfaces of the telescope should have time to adapt to ambient temperatures. Otherwise the image won't be steady ... it will wobble and distort image quality.

Sampling Theorem

There is also a question of magnification and there's a bit of science to this ... based on Nyquist-Shannon sampling theorem.

A telescope will be limited in it's resolving power based on aperture size. The camera sensor has pixels and these also have a size. The short-version of the sampling theorem is that the sensor needs to have double the resolution of the maximum resolving power that the telescope can offer. Another way to think of it is that based on the wave nature of light, a "point" of light actually focuses to something called an Airy Disk. The camera sensor pixel size should be 1/2 of the diameter of the Airy Disk. You would use some form of image magnification (such as eyepiece projection or barlow lens (preferably a tele-centric barlow) to reach that desired image scale.

This sampling theorem helps you make the best of the data your scope is able to capture without under-sampling (losing information) or over-sampling (wasting pixels that aren't actually able to resolve any more detail.)

Example

I'll pick on a camera & telescope combination as an example.

The ZWO ASI290MC is a popular planetary imaging camera. It has 2.9µm pixels.

The formula is:

f/D ≥ 3.44 x p

Where:

f = focal length of the instrument (in mm)

D = Diameter of the instrument (also in mm to keep the units the same)

p = pixel pitch in µm.

Basically f/D is the focal ratio of the telescope -- if that's an easier way to think about it. This formula says the focal ratio of your instrument needs to be greater-than or equal-to the pixel pitch of your camera sensor (as measured in microns) multiplied by the constant 3.44.

If you plug in the numbers for the 14" f/10 telescope using the camera with 2.9µm pixels, you get:

3556/356 ≥ 3.44 x 2.9

Which reduces to:

10 ≥ 9.976

Ok, so this works because 10 is greater than or equal to 9.976. So this would probably be an ok combination.

It turns out my actual imaging camera doesn't have 2.9µm pixels... it has 5.86µm pixels. When I plug in those numbers

3556/356 ≥ 3.44 x 5.86 we get 10 ≥ 20.158

That's no good... this means I need to magnify the image scale on the telescope. If I used a 2x barlow here, that doubles the focal length and focal ratio ... bringing it up to 20 ≥ 20.158. If I don't worry too much about the ".158" then I this works. But remember the symbol between the left and right sides is ≥ ... which means I could go higher. If I were to use a 2.5x barlow then it increase the focal ratio to f/25 and since 25 ≥ 20.158 this is still a valid combination.

If you use an APS-C camera (suppose you use one of the many Canon models with the 18MP sensor ... such as T2i, T3i, 60D 7D, etc. etc.) the pixel size is 4.3µm.

Suppose you use a smaller scope such as a 6" SCT. That's 150mm aperture and 1500mm focal length (f/10)

1500 / 150 ≥ 3.44 x 4.3

That works out to

10 ≥ 14.792

That's not quite enough ... you would get better results by using a 1.5x or stronger barlow.

Lucky Imaging (Using Video Frames)

BUT... before you run out and buy barlow lenses (and ideally... tele-centric barlows such as TeleVue PowerMate) it's probably better to consider a different camera and avoid using an traditional camera with APS-C sensor.

The planet is tiny. It will occupy only a very small spot on the center of the camera. So most of the sensor size is wasted.

But what's more ... getting ideal atmospheric conditions is a bit like winning the lottery. It isn't that it never happens... but it sure doesn't happen very often. Depending on where you live, it may be extremely rare. Of course if you happen to be high in the Atacama Desert ... this may be your every-day weather.

Most planetary imagers don't grab single images. Instead they grab about 30 seconds worth of video frames. They don't actually use all the frames ... they just grab a small percentage of the best frames and these are used for stacking. The technique is sometimes referred to as "lucky imaging" because you end up rejecting most of the bad data ... but for fractional moments of time you get a couple of clear frames.

DSLRs that can record video typically use a compressed video technique that is lossy. That's no good when you just want a few good frames. You need full non-lossy frames (preferably RAW video data ... such as .SER format). For this to work, you'd want a camera with a fairly fast video frame-rate. Cameras that can do video via a global electronic shutter are ideal ... but also a bit more expensive.

Before I continue... an important note: I will use specific camera models as examples. The ZWO ASI290MC is a very popular camera for planetary image at the time of this writing. It is likely that next year or the following year ... it'll be something else. Please don't take-away the message that you need to buy camera make/model _____. Instead take-away the ideas of how to work out the important features that make a camera better suited for planetary imaging.

The ASI120MC-S is a budget camera and able to capture frames at 60fps. It has a pixel size of 3.75µm. 3.44 x 3.75 = 12.9 ... so you'd want a scope with a focal ratio at or better than f/13.

This is what makes the ASI290MC such a good choice ... it has a capture rate of 170fps (assuming your USB bus and storage on the computer can keep up) and a small pixel pitch of just 2.9µm (3.44 x 2.9 = 9.976 so it works well at f/10)

Processing

Having captured the frames (and for Jupiter you want to keep it down to around 30 seconds worth of frames) you need to process the frames. The frames are typically "stacked" using software such as AutoStakkert. The output of that is typically brought into software that can enhance the image via wavelets such as Registax (btw, AutoStakkert and Registax are both free applications. There are also commercial apps that can do this as well.)

This is beyond the scope of the answer. There are numerous tutorials in how to process the data (and this becomes a bit subjective -- which isn't really the purpose of Stack Exchange.)

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