Can the transit of satellite in front of Sun be visible in 10-inch telescope?

Question

I have 10 inch Sky-Watcher telescope and I would like to know if a satellite passing in front of Sun can be visible. Suppose that satellite's angular diameter is 1˝. Yes, I know the formula $\theta=\frac{1.22\lambda}{D}$ for limiting angular size and yes, using this formula, the telescope could theoretically see it. But this applies to "stationary" double stars that you can observe for longer period of time. I don't know if this is true for rapidly moving satellite in front of bright Sun. You see, the main difference is that two stars are sources of light, but a satellite is actually dark and we can only see its shadow. That's why I think that actual angular resolution for satellites in front of Sun may be larger.

Certainly, you have more experiences on this field. Can such satellite be visible?

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As of answers, some additional information: I have proper solar filter so that full Moon is a bit brighter than Sun through filter. Therefore, damage on equipment won't be done. But please, ISS and HST are not 1˝ satellite; I am not interested in those.

Answer 1

Dark or bright features does not matter - the resolution stays the same. Resolution is achieved when the diffraction pattern is such that two maxima or two minima can be distinguished.

Many pictures exist of the ISS in front of the Sun or the Moon, and 10" is definitely a telescope with enough aperture. The question one has to ask is: do you mean to watch with your own eye or do you want to attach a camera? That defines the minimum feature you can detect.

Anyway, with the theoretical resolution of your telescope, one gets roughly 0.5" angular diameter while a 5m satellite in a low earth orbit at 400km height is about 2.6" angular diameter. So... clearly you can resolve that easily - and decent visibility only depends on the magnification and the pixel size you choose (or just magnification, if you really want to see it). There is even a tool by Sky&Telescope which shows you when ISS or HST will next pass near or through a bright object. Only other thing I managed to find were space shuttles and even space shuttle and HST simultaneous in front of the sun, though.

The calculation is different, of course, for satellites in geostationary orbits, where their angular diameter will be a factor of 10 smaller than the theoretical resolution of your instrument.

Answer 2

When you point your 25 cm aperture Newtonian reflector at the Sun you're concentrating sunlight to about 50 watts per square centimeter. About half of that is in IR/UV and will be absorbed in many kinds of optical glass and the rest will be available for imaging, and way too much for it!

If you want to use your full aperture, then you must put a special solar filter over the entrance to your telescope that is designed for this purpose, or find a special filter that can survive at focus.

We generally say that an aperture larger than six inches is not helpful for resolution because of astronomical seeing but this is at night. During the day the turbulence in the atmosphere can be worse and problems can arise even in the air close to the surface heated by the ground. This is why solar telescopes are often located near bodies of water.

• Why are solar telescopes built on lakes?

So normally it is both dangerous to use a full 10 inch aperture (unless you really really have the correct, safe filters) and unhelpful due to atmospheric turbulence.

However, with plenty of light you have the opportunity for a short exposure time and high frame rate and so even if the transit is fast you may be able to use lucky imaging where one or some of the frames in your rapid photo sequence can be used to make a less-distorted image.

The fastest angular speed that the ISS could have is the orbital velocity divided by the minimum distance. Assuming the orbit is circular (which it nearly is) we can use the vis-viva equation to get

$$v = \sqrt{\frac{GM}{a}}$$

where $GM$ is the Earth's [standard gravitational parameter of 3.986E+14 m^3/2^2 and $a$ is the ISS' semimajor axis which is it's averge altitude (recently) of 400,000 meters plus (by definition) Earth's equatorial radius of 6378137 meters. This gives about 7669 m/s. At a minimum distance of 400,000 meters that's 0.019 radians/second or 1.1 degree per second, or 4 arcseconds in a 1/1000 second exposure.

So if you want to talk about 1 arcsecond resolution you first have to lay hands on a camera with a 1/4000 second shutter speed or faster, or wait for a transit that's closer to the horizon where the angular speed is lower (and the ISS' apparent size is also smaller).

How small is it?

Roughly 100 meters / 400,000 meters is about 50 arcseconds, way bigger than the 1 arcsecond resolution mentioned in the question.

So if you just want to resolve the ISS' shadow's shape against the Sun, you really may not need the full aperture of your telescope.

That means that maybe you can use an off-axis circular aperture covering most of your telescope with only say a 2 or 5 cm hole. This helps cut down on some of the thermal power that can destroy your equipment (or eyes, NEVER, EVER, Not Even A Little look through the telescope or through a camera pointed near the Sun).

Sill, as @planetmaker's answer points out, you will need something big and close like the ISS in order to resolve anything more than a dark dot.

Potentially helpful in Space SE: How does the ISS Transit Finder website get the position of the ISS so accurately?