Measuring the rotation speed of Io

AI Thread Summary
It is possible to determine the distances from Earth to Jupiter, Jupiter to the Sun, and the diameter of Jupiter using a telescope, provided the speed of light is known. Historical observations of Io's orbit by Ole Roemer allowed for the calculation of these distances, and modern methods can confirm them using Kepler's laws. By timing Io's occultations, one can estimate the distance to Jupiter based on light travel time. The angular size of Jupiter can also be compared to the full moon to estimate its diameter. However, the gravitational effects of Jupiter on light are negligible and not significant enough to impact measurements.
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I'm wondering two things:

1. Is it possible to determine the distance from earth-jupiter, jupiter-sun and the diameter of Jupiter using only a telescope?

2. Is it possible to calculate the rotation speed of Io by comparing two CCD images taken a certain time apart?

Thanks.
 
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Link said:
I'm wondering two things:

1. Is it possible to determine the distance from earth-jupiter, jupiter-sun and the diameter of Jupiter using only a telescope?

2. Is it possible to calculate the rotation speed of Io by comparing two CCD images taken a certain time apart?

Thanks.

I am only replying to part 1.

Yes it is possible, if you know the speed of light independently.

In 1675 Ole Roemer determined the speed of light by observing occultations of Io (by timing Io's orbit) and in doing so he USED current estimates of those distances (the AU, the sun-jupiter distance, etc)

If you know the speed of light (from earth-based laboratory measurement) as we all do nowadays, then you should be able to work backwards and find the AU and jupiter's distance from the sun.

Do you want to perform the exercise of measuring these basic solar system distances by yourself?
 
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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.
 
If I had to estimate the diameter of jupiter, knowing the distance, by my own devices, then I would compare it in size to the full moon

the full moon makes an angle of about half a degree in the sky
or 1/100 radian
it varies because the moon orbit is not perfectly circular.

I believe that the angular size of jupiter, when we are closest, that is when the distance is 4.2 AU, would come out to be around 47 arc-seconds.
that is like 1/4400 radian. It is roughly 1/44 of the size of the full moon in the sky.

So if you determined that the AU was 150 million km.
and therefore that 4.2AU was 630 million km and you observed Jupiter (when we were closest to it) to have an angular size of 1/4400 radian.
then you could just divide

I have to go, back later
 
Thank you guys for helping!

When reading the first reply a thought popped up in my head. I was thinking, since we are using light as the measuring unit and actually analyzing pictures of Jupiter, is it possible that the gravity of Jupiter can disort the pictures in accordance of the principles of general relativity? I know Jupiter is large in mass, but i don't know if it is large enough to have a significant impact on measuring data
 
Not nearly enough. The effect is too small to measure. The sun is hugely massive compared to Jupiter and we can barely detect gravity effects on its spectral lines.
 
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