I could calculate it [

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If you think of an isosceles triangle,

3,474 km (diameter of the moon) on short side.

1,737 km (radius of the moon)

384,399 km (distance to the moon)

SOH CAH TOA

Ok, I need a "right triangle" for my trig functions, but I can use half an isosceles triangle, ant thus use the radius.

Tan(½θ) = opposite/adjacent = 1,737/384,399

½θ = arctan(1,737/384,399) = 0.0045 radians.

θ = 0.0090 radians

It should be about the same as using the sin function as the hypotenuse and adjacent side are similar in length.

Sin(½θ) = opposite/hypotenuse = 1,737/384,399

½θ = arcsin(1,737/384,399) = 0.0045 radians.

θ = 0.0090 radians

Now, work it out the opposite direction.

Say you want to calculate size at 100km (about the max for a balloon).

tan(½θ) = opposite/100km.

opposite = radius of balloon = 100*tan(0.0045)

And you get about a 0.45 km radius, or about 0.9 km diameter

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Oops.

I think I did this a bit complicated.

One should be able to do the same with merely fractions.

Consider [tex]\frac{diameter_1}{altitude_1}[/tex] = [tex]\frac{diameter_2}{altitude_2}[/tex]

[tex]\frac{3,474 }{384,399}[/tex] = [tex]\frac{diameter}{altitude}[/tex]

[tex]diameter[/tex] = [tex]altitude[/tex] x [tex]\frac{3,474 }{384,399}[/tex]

Ok, so for 100km, one gets, [tex]100[/tex] x [tex]\frac{3,474 }{384,399}[/tex]

And at 100km altitude, one gets 0.9km diameter

At 200km altitude, one gets 1.8km diameter

It is always great to do my calculations 3 different ways, and come up with the same answer [

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Your balloon would have to be quite large if you chose to put it into a geostationary orbit.

diameter = [tex]35,786km[/tex] x [tex]\frac{3,474 }{384,399}[/tex] = 323 km diameter. Fortunately, pressure is low enough that it wouldn't take much gas to fill the balloon, provided adequate elasticity of the material, and resilience to heat, cold, and micro-meteors.