Maximum distance for earth eclipse First of all sorry if this question was already asked, but I made a small search here and on google and I couldn't find any answer, so.. Here I am.

I was reading NASA's webpage about the Webb telescope, and I noticed that they are saying that one of the sides will be very hot.

I thought that one of the reasons to put the telescope on the L2 point was to avoid "seeing" the Sun, but apparently the L2 point is too far to have Earth fully cover the Sun.

So.. What is the maximum distance from Earth you have to be in order to have our planet block the Sun? Of course neglecting the effect of the atmosphere, which creates a sort of "halo" even if the Sun is fully covered.

Moreover what will the telescope see towards the sun? Just the star with a very tiny dot in the middle, or a very big black dot (optionally with the moon moving around it), or... What?

Thank you
Since the Earth is approximately 109 times smaller than the Sun (regarding radius), and the distance between them is about 150 million km, the eclipse will be visible up to about 1.39 million km from the Earth. This distance will vary (1.36 - 1.41 mln km) because the distance Earth-Sun varies during the year. This is about 3.6 times more than the Moon-Earth distance.

At this position, the apparent size of the eclipsed Sun will be only 1.08% smaller than the one on the Earth. The Moon has the radius about 1/4 of Earth's, and it will be visible as such on average. However
when in front of the Earth, it will be apparently almost 2 times bigger (regarding radius) than when behind.

We could orbit Earth at such distance, because Lagrangian points L1 and L2 are a bit farther (1.5 mln km). This means also that at L2 we could not see full Solar eclipse, but only annular with magnitude 0.93.