Ever since its Christmas present-like launch about two weeks ago, I have closely followed the progress of the James Webb Space Telescope (JWST, or Webb, for short). During one of the live streams, I was a bit surprised to learn that there are actually no cameras at all on JWST that can be used to visually monitor the progress of the deployments. There are a number of good reasons for this (including, e.g., the fact that they would require artificial light, add weight and complexity, and would have to work in pretty extreme conditions), but for me, that just made the whole thing even more impressive. Talking to a friend about this, we jokingly came up with the idea that, in case something goes wrong and JWST needs debugging, we could try to use the Hubble Space Telescope (Hubble, or HST) to take a closer look!

Now, I say “jokingly,” because intuitively, I already had a feeling that this almost surely was not going to work in practice. Yes, JWST is the size of a tennis court, but it is a tennis court that is already about a million kilometers away from us at the time that I am writing this article. I also recalled that a couple of days ago, somebody on Reddit claimed that Hubble can resolve “an orange on the moon as a single pixel”. Obviously, as several other users pointed out, that claim was nonsense. However, nobody had provided really solid numbers, so, to make a long story short, I decided the look up some values and try to work out the math properly.

First of all, Webb is about 22 meters along the longer side. Once it has reached its target position, it will be about 1.5 million kilometers away from us.1 (We can probably neglect the fact that Hubble itself is in an orbit around Earth at a height of about 550 km here.)

This allows us to compute Webb’s angular diameter (or apparent size), which is the quantity that astronomers usually use to describe the size of objects on the sky. The following schematic sketch should illustrate the notation:

I tried to keep JWST and Hubble to scale as much as possible, which was somewhat tricky, because it is actually surprisingly hard to find exact numbers for the size of the different parts.

With a bit of basic trigonometry, we find that the angular diameter $d$ is given by:

$$ \delta = 2 \arctan\left( \frac{d}{2D}\right) \,. $$

Plugging in the numbers for the size of Webb and its separation from Earth / Hubble, we get2:

$$ \delta = 2 \arctan\left( \frac{22\,\text{m}}{1.5 \cdot 10^9\,\text{m}} \right) = 8.4034 \cdot 10^{-7}\,\text{degree} = 0.003025’’ \,. $$

The unit of the latter quantity is arcseconds, which is, again, the convention that astronomers prefer to use. To give you some perspective (example shamelessly taken from the Wikipedia article): One milliarcsecond—which is still bigger than the $\delta$ we just computed—is the size of a dime (or a 2 Euro cent coin) on top of the Eiffel tower, as seen from New York City.

…I guess you already suspect where this is going?

Alas, I went to look up the actual resolution of the Hubble instruments, because, you know, there are actually several cameras on board of the HST. Perhaps the best for our (admittedly questionable) purposes would be the Advanced Camera for Surveys (ACS) . Since the high-resolution channel (HRC) has been defunct since 2007, we’d probably have to make do with the wide field channel (WFC). This is the camera that was used for a lot of the spectacular images produced by Hubble in the past 20 years.

According to the official ACS Data Handbook, the WFC has an angular resolution of about 0.049 arcseconds per pixel. Of course, this is bad news for us, because:

$$ \frac{0.003025’’}{0.049’’\,/\,\text{pix}} = 0.06173\,\text{pix} \,, $$

which means that, theoretically, the whole tennis court-sized star shade of JWST would make up less than one tenth of a pixel on the sensor of Hubble’s ACS/WFS camera. Considering that JWST is probably relatively bright2, and that there is also the instrument PSF that smears out everything, we might be able to see JWST as one or perhaps a few pixels. This should be enough to detect Webb (also because we know exactly where to look), but we will defintely not resolve it spatially, meaning that it will be pretty useless for monitoring the deployments.

Update (January 25, 2022): As it turns out, some people actually have taken an image of JWST! Not with Hubble, but with the Virtual Telescope Project, an array of remotely controlled robotic telescopes here on Earth. They indeed manage to resolve JWST as a blob of a few pixels, which I think is kind of impressive considering that their angular resolution is only 1.2 arcsec per pixel and that, unlike a space telescope, they also have to look through the Earth’s atmosphere.

  1. Actually, the distance between Webb and the Earth will not be constant: JWST is heading towards L2, but unfortunately, you can’t just “park” it there. Instead, Webb will be circling around L2 on a somewhat complicated halo orbit at a distance between approximately 1.2 and 1.7 million kilometers from Earth. ↩︎

  2. That’s the other thing we need to keep in mind here: We can’t really see things if they are too dark. However, since Webb on its course to L2 is flying “outwards” into the solar system (i.e., away from the Sun), the side of JWST that is facing us is the star shade, which is designed to reflect as much light as possible, so one might hope that this would not be the biggest issue here. ↩︎ ↩︎