To put some numbers.
Assuming you are on the ground then this formula is in effect:
(1)
\begin{align} d=\sqrt{D h} \end{align}
Where
h = The object height
D = diameter of the Earth
d = distance to the horizon
Put in some numbers
h = 2 meters
D = 12742 km
(2)
\begin{align} d=\sqrt{12742 \text{ x } 0.002} \text{ = 5.048 km } \end{align}
You can see 5 km far from a 2 meter height.
Now the reverse would be this, if you are d distance away how heigh must the object be to see it only just above the horizon?
(3)
\begin{align} h = \frac{d^2}{D_{\text{Earth}}} \end{align}
The South pole is
d = 5,815 km form Melbourn.
D = 12742 km (earth's diameter)
Fill in the numbers and we get:
(4)
\begin{align} h = \frac{5815^2}{12742}} = \text{ = 2,653 km heigh} \end{align}
2,653 km is the maximum height it could be to be just detectable from Melbourn when it is above the South pole.
But 2,653 km means low Earth orbit and that means that it has to move around the Earth because it cannot hover up there. And it also means that it is tiny having no influence whatsoever possible.
austyn (guest) 16 Mar 2011 05:53
