Question about if something would be visible from the surface of a sphere. [closed]

So, I am trying to see if something would be visible to someone standing on the surface of a planet or the top of mountain on it.

So, imagine a perfect sphere for the planet, then imagine a moon which orbits perpendicular to the line from the observer to the centre of the sphere.

Let's give a orbital distance of A for the moon, a observer height of $$h$$ for the observer, and a planetary radius of $$R$$.

How would you calculate the necessary height H required to see the moon, given a specific A and R? It would be helpful if a formula could be given for solving this.

I tried doing it myself, to find out the necessary height to see it despite the curve of the planet, but I just can't work out how to get it. So I decided to come ask here.

This isn't about the physics of it, this is just about the geometry and such. This also is not taking into account angular size of the moon.

I tried doing something said in the comments, but my result was just:

$$h = \sqrt{2R^2 + 2Rh + h^2 + a^2 + 2aR} - R$$

From $$R + h = \sqrt{R^2 + (R+a)^2}$$

Not sure what other work to show. And it is hard to use mathjax on mobile. This however, seems wrong as it makes it so $$a < h$$. Which seems wrong.

I also tried:

$$R + h = \sqrt{R^2 + R^2}$$ which got $$h = \sqrt{2} R - R$$, which seems more reasonable, as $$h < a$$. But not sure if the math works out, especially as it doesn't take $$a$$ into account.

I could be missing something obvious. But I just want help for finding if something would be visible given the variables. I don't think what I am doing now helps with that.

• What have you tried to do in solving your question? Have you some work you can show where you attempted to find the line of sight from observer to moon as a function of the variables given? Apr 15 at 23:58
• I tried going off of the Horizon distance formula, but the results didn't seem accurate. As I know objects high in the sky can be seen at a further distance than just using the Horizon Distance formula, afaik. I also tried working out lines that I measured. To keep everything proportioned right, but that didn't work out. Apr 16 at 0:00
• Basically I couldn't turn anything into a functional formula. Apr 16 at 0:15
• Draw a picture with the smallest possible $A$. Then the right triangle with angle at the observer with opposite side $R$ and hypotenuse $R+H$ is similar to the right triangle with legs $R$ and $A+R$. Then you can finish with the Pythagorean theorem and algebra. Apr 16 at 0:31
• @EthanBolker I did that and got an answer that seems wrong, H= Sqrt(2R^2 + 2Ra + a^2) - R, which is greater than a. Which seems wrong. If I understood you correctly. Apr 16 at 0:41

If I understand the question correctly, you're asking about the height $$H$$ above the surface of the planet where you can just barely see the moon on the horizon, as in this diagram where I've greatly exaggerated the radius $$R$$ of the planet compared to the distance $$a$$ to the moon (of negligible size).

In this diagram all three triangles are similar right triangles. Using the inverse Pythagorean theorem, $$\frac{1}{(H + R)^2} + \frac{1}{a^2} = \frac{1}{R^2},$$ which we can rearrange to give (try it yourself before revealing)

$$H = \frac{aR}{\sqrt{a^2 - R^2}} - R = \Biggl( \frac{a}{\sqrt{a^2 - R^2}} - 1 \Biggr) \, R$$

If the center of Earth is at $$(0,0)$$, and the moon is at $$(A, 0)$$, and the observer is at $$(0, R + h)$$, then the line connecting the moon and the observer is

$$y = R + h - \dfrac{ R+ h}{A} x$$

The distance between the origin (which is the center of Earth) and this line is

$$d = \dfrac{ R + h }{ \sqrt{ 1 + \dfrac{(R+h)^2}{A^2} } } = \dfrac{ A (R + h) }{\sqrt{A^2 + (R + h)^2} }$$

And we want this distance to be greater than $$R$$. Hence we want to solve for $$h$$ the equation

$$A (R+h) \gt R \sqrt{ A^2 + (R + h)^2 }$$

Square both sides

$$A^2 (R+h)^2 \gt R^2 (A^2 + (R + h)^2 )$$

From which,

$$(A^2 - R^2) (R + h)^2 \gt R^2 A^2$$

i.e.

$$(R + h)^2 \gt \dfrac{ R^2 A^2 }{A^2 - R^2}$$

And finally

$$R + h \gt \dfrac{ R A }{\sqrt{A^2 - R^2} }$$

which gives

$$h \gt -R + \dfrac{ R A }{\sqrt{A^2 - R^2}}$$