Tall building sighting distance

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From what distance can a person on the ground see the Burj Khalifa?

Or: How tall does the building have to be to be seen from 250 kilometers away?
 
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What you said: The building needs to be 4.9 km high to be visible from 250 km away?
I would like an explanation of the equation that produced this result.
 
russ_watters said:
I scribbled it on a tiny kid's toy so I may have made a mistake, but I don't think so.

I used the polar Earth diameter from Wikipedia, ##12713\,\text{km}## and the distance ##250\,\text{km}## as line of sight. I reviewed it, and it is a little more than ##4.9\,\text{km}.##
Solve: ##0=(r+h)^2-r^2-(250)^2##
 
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You are a champion.
This is my equation:
h=63712+2502−6371h = \sqrt{6371^2 + 250^2} - 6371
 
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Echoing Stars said:
You are a champion.
This is my equation:
h=63712+2502−6371h = \sqrt{6371^2 + 250^2} - 6371
I think you have been a bit sloppy in the first equation. The same formula (Pythagoras) also allows you to solve for the distance ##x## if you use the radius 6374.344 km and height (peak) 0.8298 km of the actual building. Then ##x=102.857\,\text{km}.##
 
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fresh_42 said:
אני חושב שהיית קצת מרושל במשוואה הראשונה. אותה נוסחה (Pythagoras) מאפשרת לך גם לפתור את המרחק ##x## אם אתה משתמש ברדיוס 6374.344 ק"מ ובגובה (שיא) 0.8298 ק"מ של הבניין בפועל. ואז ##x=102.857\,\text{km}.##
And it's just like that..
Look here
 
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fresh_42 said:
I scribbled it on a tiny kid's toy so I may have made a mistake, but I don't think so.

I used the polar Earth diameter from Wikipedia, ##12713\,\text{km}## and the distance ##250\,\text{km}## as line of sight. I reviewed it, and it is a little more than ##4.9\,\text{km}.##
Solve: ##0=(r+h)^2-r^2-(250)^2##
Sorry, there were two questions asked: I thought you were answering the first and I skimmed past the second.
 
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