Ideas on how to solve Curvature of the earth problem?

Question

Being on top of a hill, you are watching someone swim through a telescope (which is assumed to be height 0), because of the curvature of the earth they will eventually dissapear across the horizon. How far away can they get before dissapearing from view? the radius of the earth is said to be 2.09x10^7...how can i solve this problem??

the hill is 66 feet above

Answer 1

Plane geometry will answer your question. Let the hill be h meter high. Now continue the altitude of the hill to the center of the earth, and draw a radius from the center of the earth to the point where the swimmer is when he "disappears". Call that rs. Let the chord from the base of the hill to the swimmer be d, what we want to know, and draw a line from the swimmer to the top of the hill we call T. rs and T are sides of a right triangle whose hypotenuse is r+h. From that, we can find T. Now, h and d are sides of a right triangle whose hypotenuse is T. Since we know T and h, we can find d.

Now,

Answer 2

The distance is about 9.95 miles, or 10 miles in round terms.

(The round figure of 10 miles is doubtless why you were told to consider a height of 66 feet!)

How was this found? :

To a first approximation, the classical geometry of a circle tells you that :

d^2 = 2 R h,

where ' h ' is the height, ' R ' is Earth's radius and ' d ' is the distance to the horizon.

ALL these lengths need to be expressed in COMPATIBLE UNITS, of course!

Which brings up the question : What BIZARRE kind of UNITS are YOU using ? !! In round terms, the radius of the Earth is about 6400 km = 6.4 x 10^8 cm, or about 4000 miles. Oh, there are 5280 feet in a mile, so that 4000 miles is about 2.1 x 10^7 FEET. O.K., so I guess you were given the Earth's radius expressed in feet (!) --- I've just never seen that quantity so expressed before!

So we have:

d^2 = 2 (66) 2.09x10^7, or d = 52500 ft (approx.) = 9.95 miles.

Live long and prosper.

P.S. Incidentally, this reminds me of something I worked out in my youth, some 55 years ago. To a rather CRUDE approximation, the distance to the horizon is given by :

d (in miles) is approximately the number h^(1/2), where ' h ' is measured in feet!

In the case of ' h ' = 66 (feet), this crude estimate would give you (66)^(1/2) or about 8.12 miles, a bit less than, but not too much less than, the 9.95 miles I worked out above.

The accurate implication of the actual data for the Earth is that:

d (in miles) = 1.2245... h^(1/2) where ' h ' is in feet.

So, if you just work out h^(1/2) and add 20% to it, you're going to be fairly close. For h = 66 feet, that would give you 1.2 x 8.12 = 9.75 miles, not really much different from the 9.95 miles I found above.

So here's a quick and easy way to impress your friends : Go with them to the top of a 95 ft building and ask them how far it is to the horizon. (I'm allowing 5 feet for the height of your feet above the surface you're standing on.) After they've all guessed, you can pull out your handy "1.2 h^(1/2)" formula, and tell them it's 12 miles.

Answer 3

If u r sitting on a hill watching the swimmer, he could be seen for quite a distance, depending where the sun is. You could probably see him get tired and turn back! You didn't say how high the hill was.



( And......nobody can swim through a telescope..Nyuk!)

Answer 4

Hmm, well you need the height of the hill to solve the problem. Once you got that I think you can form a triangle to calculate the distance. Is this physics? Took it last year, didn't really like the class.

Answer 5

You solve:

dist_horz = sqrt[(Re + hill)^2 - Re^2]

for example, a 100-m hill results in horizon distance of about 20 km.

Answer 6

Think of this as a big circle.
Using the "radian" measure you can find out haw far you can see.
However you are missing one part. It will also depend on the height of the mountain which is not mentioned in your problem...??!!