Hello, matgrl!

Penthouse apartments rent for the highest prices because of terrific views.

Assuming the atmosphere is clear and there are no obstructions,

how far could you see to the horizon from the top of a 100-meter fire tower?

From the top of 165-meter skyscraper?

From the top of the tallest building in the world?

. . (the Petronis Tower in Kuala Lumpur, Malaysia, 1,467 ft. high)

From an airplane flying at 35,000 feet?

From the Space Shuttle flying at 300 km?

The radius of the earth if approximately 6378 km or 3963 miles. Code:

A
o
|\
| \
h| \
| \ d
C| \
* o * \
* | *\
* | o B
* | * *
r| * r
* | * *
* o *
* O *
* *
* *
* *
* * *

The center of the Earth is $\displaystyle \,O.$

The radius is: $\displaystyle \,r \:=\:OB = OC.$

The observer is at $\displaystyle \,A.$

His altitude is: $\displaystyle h \,=\,AC.$

He can see a distance of: $\displaystyle d \:=\:AB.$

. . Note that $\displaystyle \angle B = 90^o$

Pythagorus: .$\displaystyle d^2 + r^2 \:=\:(h+r)^2 \quad\Rightarrow\quad d^2+r^2 \:=\:r^2 + 2rh + h^2$

. . . . . . . . . $\displaystyle d^2 \:=\:2rh + h^2 \quad\Rightarrow\quad d \:=\:\sqrt{h(2r+h)} $

We can simplify further.

Consider the factor $\displaystyle (2r + h)$

Compared to the diameter of the Earth $\displaystyle (2r)$,

. . the height $\displaystyle \,h$ is negligible.

We can let: .$\displaystyle 2r \:\approx\;2r+h$

The distance formula becomes: .$\displaystyle \boxed{d \:\approx\:\sqrt{2rh}}$

Be sure to use the same units (feet/miles, meters/kilometers).