Link, can you figure out a way to use trig
that will confirm that the sun-jupiter distance is 5.2 times the
sun-earth distance?
If not then (this is maybe cheating but) you can use Kepler's 3rd Law.
It seems to me that using a telescope you could, by yourself, determine that the orbital period of Jupiter is 11.86 years
the 2/3 power of 11.86 is 5.2
so you could tell that the sun-jupiter distance is 5.2 times the sun-earth.
Now comes the hard part. It is essentially "measuring the AU in kilometers".
You know there is a distance X, and that sun-earth is X and that
sun-jupiter is 5.2X
this means that when Jupiter is overhead at midnight the earth-jupiter distance is 4.2X
and when we are farthest from jupiter, and it is overhead at noon, then our distance is 6.2X
Indeed when Jupiter is overhead at sunrise or sunset, so that Jupiter and the sun make a right angle, then 5.2X is the hypoteneuse and
our distance to Jupiter is 5.1X
Now you can determine what X is, in kilometers, by timing the orbit of Io.
By simply timing the orbit you know how often occultation should occur.
And you can predict the day and hour when it should happen. But you will find, when we are nearest Jupiter, that you observe it about 8 minutes earlier than you expect.
because you are closer----at 4.2X rather than, for example, at 5.1X----
and therefore the news reaches you that much sooner.
from this 8 minutes, you can estimate X, because it is the distance light travels in that time.
By occultation I mean eclipse, especially the moment when Io goes into the shadow of the planet, or emerges. This should be a well defined instant that one can record, to see whether it is happening 8 minutes early or 7 minutes late or whatever.
If you think of a better way to determine the sun-earth distance from scratch, just using your telescope, then tell me.