## Swedish winner for the Cassini-Huygens quiz.

**The winning answer.**

**Within this article:**

After a careful examination of all entries, Anders Nyholm from Sweden was selected as the winner. His reply was elegant and exhaustive, recalling all his scientific assumptions and describing step-by-step the procedure followed to solve the quiz.

## The winning answer

When answering this question, I make the following assumptions:

The "big telescope" to be used to observe Cassini is located in space, outside the disturbing atmosphere of the Earth

I assume that "mid-May" is 00:00, the 15^{th} of May, 2000 (GMT)

Since Cassini flies with its HGA pointed at the Earth since the HGA-manoeuver is in February, I assume that the visual appearance of Cassini is dominated by the antenna, and that the total geometric albedo of Cassini is 0.8

I find it suitable to begin with the second part of the question: "...at what angle respect to the Sun-Earth line would you see it?" this must be the elongation if I don't misunderstand you. The elongation of an object with respect to the Sun can be computed using

CosE=(s^{2}+R^{2}-r^{2})/2sR

Where **r** is the object heliocentric distance, **R** its geocentric distance and **s** is the Earth's heliocentric distance. Inserting values for Cassini on May 15^{th} (r=3.5590212 AU, R=4.5695704 and s=1.0108898), one obtains cosE=0.999737593, which gives an elongation of 1.312°.

This leads to the first part of the question "How big a telescope would be needed to see Cassini-Huygens in mid-may 2000?". To answer this, one has to account for how much of the spacecraft that is illuminated. That is, the phase of Cassini. The phase is given by the phase angle, which can be computed using

CosP=(r^{2}+R^{2}-s^{2})/2rR

With the same designations as before. Inserting the values, one obtains cosP=0.999978832, which gives a phase angle of 0.372°. Since the phase is calculated using: (1+cosP)/2, one finds more than 99% of Cassini is illuminated as seen from the Earth. Hence, I think that the spacecraft can be considered as fully illuminated for all practical purposes.

If one assumes that an object is fully illuminated by the sun (as I assume for Cassini, see above), its apparent magnitude ca be calculated using

V=V_{s}-2.5logp-5logs+5logrR

Where **V _{s} **is the absolute magnitude of the sun (-26.78),

**p**the object albedo,

**s**its radius,

**r**the object heliocentric distance and

**R**the object geocentric distance (s, r and R are expressed in AU). The values for Cassini in this case are:

p=0.8

s=1.3369174*10^{-11 }AU (the radius of the HGA)

r=3.5590212 AU

R=.5695704 AU.

Having inserted these values, one finds that the apparent visual magnitude of Cassini is 33.887.

Knowing the apparent magnitude, one can calculate how large the aperture of a telescope needed to observe Cassini has to be. This can be done using a rearranged version of the (quite well known) formula for the theoretical lighting magnitude of a telescope (M=7+5logA):

(M-7)/5=logA

where M is the limiting magnitude (which in this case equals the apparent visual magnitude of Cassini) and A is the aperture, given in cm. Since M=33.887, on obtains a value of logA=5.377556368, which in turn gives A=238537.3378 cm, or 2.38 km. The only reasonable way to achieve such an aperture is most probably to build some kind of interferometer

The answer to "*How big a telescope would be needed to see Cassini-Huygens in mid-May 2000?"* given above can only be considered as "half", since it also could be interesting to know how big a telescope one would need to resolve Cassini. To answer this, the apparent size of Cassini, as seen from the Earth has to be known. This value can be computed using the well-known trigonometric formula a/b=tanV, where a (in this case) is the diameter of Cassini's HGA, and b the spacecraft's geocentric distance, given in AU (for values see above). Using the appropriate values, one obtains tan V=5.851392*10^{-12}, which gives an apparent diameter of 3.3526*10^{-10} °, or expressed in arcsec: 0.000001206". Knowing the apparent size, ona can compute the aperture of the telescope needed to resolve it. This is done using the following formula

_ *206,000/_ =D

where **_** is the wavelength at which the object is observed, **_** the resolution in arcseconds and **D** the aperture of the telescope (in the same unit as the wavelength). I assume that Cassini is observed at a wavelength of 5000 Å. The resolution _ equals (in this case) the apparent diameter of Cassini HGA. With these values, one obtains D=8.534*10^{-14} Å, which equals 85340 m or 85.3 km. Quite a large telescope... It should be noted that this only is valid for a telescope located in space, outside the disturbing atmosphere of the Earth.

So the answer to your question should be something like: To see Cassini in "mid-May 2000)", one would need a telescope (most probably some kind of interferometer) with an aperture of at least 2.38 km. Since the elongation of Cassini is quite small at this time, certain practical difficulties associated with observing an object so close to the Sun should be expected! However to resolve Cassini, one would need an even larger telescope, with a minimum aperture of at least 85, 34 km. Cassini/Huygen's angle to the sun-Earth line (as you call it) at this

time is 1.71 °.