IOAA 2019, Теория — Задача 14. Photographing a nanosatellite

a)

As it was suggested in the hint the key is a comparison of the sunlight scattered from a 10 cm × 10 cm aluminium plate and the Moon to the observer (both are nearly at the local zenith).

Let the distance of the Moon and the satellite from the observing site be $d_{\mathrm{Moon}}$, and $d_{\mathrm{MASAT}}$, respectively. Under the mentioned "ideal" circumstances, when both the Moon and the satellite is near the local zenith: $d_{\mathrm{MASAT}} = 900\ \mathrm{km}$, while

(2 p)

where the average distance between the centre of Earth and Moon $s_{\mathrm{Moon}}$ is included in the Table of constants.

Let the intensity coming from the Sun and scattered back to the direction of the observer be $I$, this is considered practically the same at the solar distance of the Moon and the satellite, but necessarily multiplied by the albedo of the given surface quality (the lunar albedo $a_{\mathrm{Moon}}$ is given in the Constant table, while $a_{\mathrm{MASAT}}$ of the satellite is in the text).

Finally, for using the magnitude formula, to find the observable magnitude of the satellite, we have to consider the resulted flux of the scattered light at the observer, which is directly proportional to the intensity, the albedo and the scattering area, but is inversely proportional to the distance on the square:

(3 p)

Thus, the magnitude formula gives:

(3 p)

With numerical values:

(2 p)

b)

A moving point-like object will generally cause a line on the CCD image, resulting less photoelectrons in a given pixel, than a similarly bright unmoving object. Since the MASAT-1 satellite is so small, it is not resolved by the telescope – it will cause a seeing-sized spot at the focus of a telescope, like any other stars on the sky, but since the text of the problem described: we should omit the seeing. Thus, the image of any point source will be formed by the diffraction – so we shall use the approximate formula of the Airy disk.

By the diffraction-limited imaging, the point-like sources illuminate a spot in the focal plane having the angular diameter of the so-called Airy disc determined by the formula:

(2 p)

while its linear size at the focus of $f$ focal length telescope is:

(2 p)

For the telescope of Baja Observatory, described in the text, at a typical visual wavelength of $0.55\ \mu\mathrm{m}$:

(2 p)

This is very close to the pixel size of the applied CCD. However, this number is not necessary for our calculation, since both the stars and the MASAT-1 are point-like sources, illuminating the same number of pixels, thus the number of pixels will drop out from all comparative calculations. So does the quantum efficiency value of the applied CCD.

For deriving the effect of the moving object, let us first calculate the scale of the image at the focus of the 50 cm RC telescope of Baja ($N$ is the f-number of the telescope, $N = f/D$):

(3 p)

This means, that one pixel of the attached CCD describes $P = 0.049''\,\mu\mathrm{m}^{-1} \times 9\,\mu\mathrm{m} = 0.44''$ side length area on the sky. (3 p)

Now let us estimate the relative flux, resulted by a star having the limiting magnitude of the telescope-CCD system ($19.5^{\mathrm m}$), by the magnitude formula:

(2 p)

And similarly, for the MASAT-1 brightness – observed with the same system, under the same conditions, if it were an unmoving object:

(2 p)

Next, we have to calculate the angular speed of MASAT-1 at 900 km altitude above the observing site, in order to derive the fluxes falling onto one pixel of the CCD on the telescope.

The velocity of the motion on a circular orbit with orbital radius $r$:

(2 p)

where $r = h + R_{\mathrm{Earth}}$. (2 p)

With numerical values:

(1 p)

Hence, the visible angular speeds (aside from the Earth rotation):

(2 p)

With numerical values:

(1 p)

when the satellite is near the local zenith.

For deriving the relative speed, which will be really exposed onto the CCD, we have to calculate the angular speed of the observing site caused by the terrestrial rotation. At the latitude of $\varphi$ it is given as:

(2 p)

For Baja its numerical value:

(1 p)

Its component parallel to the direction of the satellite's orbital motion:

(2 p)

For Baja its numerical value:

(1 p)

Finally, the relative angular speed of the nanosatellite above the telescope:

(2 p)

With numerical values:

(1 p)

Since the satellite image is moving with a given $\Delta\omega$ angular speed, this means that the flux coming from the satellite is falling onto a given pixel only for a very limited time (independently of the exposure time of the image, but acting similarly, than that). Considering our very simple approximation we get an approximate "illumination time" for one pixel falling onto the line of the orbit:

(2 p)

With numerical values:

(1 p)

For finishing our estimation – to get the magnitude level of any pixel on the observable trace of this nanosatellite, let us use the magnitude formula for them and one non-moving light source at the detection limit. For doing this we have to multiply the relative fluxes $\Phi_{\mathrm{MASAT}}$ and $\Phi_{19.5}$ with the respective illumination times. (For a $19.5^{\mathrm m}$ stars we should use the 2 min-long exposure time mentioned in the text, which is needed to collect enough photoelectrons for the safe detection of such a faint star.)

(2 p)

With numerical values:

(2 p)

So, the trace of the moving $23.9^{\mathrm m}$ magnitude nanosatellite looks so faint on a (2 min-long exposure) CCD image, as a $\sim 24^{\mathrm m}$ stars would be seen, but they fall surely below the detection limit for the described telescope. Thus, the answer is definitely

NO, WE COULD NOT RECORD THE MASAT–1 (2 p)

using that telescope.

c)

If we consider a real atmosphere, due to the effect of the atmospheric turbulences (the so-called seeing) during the 2 minutes-long exposure the images of the stars will never be described by the ideal Airy disk, but expanded to a much larger "seeing spot", which can be approximated by a Gaussian profile. As mentioned in the text: in Hungary, its diameter at half-maxima is typically $3.5''$. This means, that the flux coming from a $19.5^{\mathrm m}$ star will be distributed over about a

(1 p)

pixels area. Thus, now the $F_{19.5}$ flux will be distributed amongst $4 \cdot 4\pi \approx 50$ pixels (for the sake of simplicity: homogeneously), i.e. one pixel gets:

(1 p)

As we have seen above, the satellite image is moving so fast along the focal plane ($0.26\ \mathrm{ms}$ per a pixel), that during this short time, the atmosphere can be considered to be still. Thus, the image of MASAT–1 can further be approximated with the Airy-disc size also in a real atmosphere.

Let us correct our last magnitude formula for this fact:

(2 p)

With numerical values:

(2 p)

Under real conditions the situation is much better: the trace of the small nanosatellite is near the detectability! Thus, owing to further secondary effects (pixel sensitivity inhomogenities, temporarily changing transparency of the atmosphere, etc.), the trace of the cubesat can easily be seen on the CCD of the 50 cm reflector. So in real cases, the answer can be YES.

YES, WE COULD HAVE RECORDED THE MASAT–1. (2 p)