finding the curvature of the Earth's surface over a five mile distance.?

Question

How do you finf the amount of curvature, in feet, over a five mile distance on the Earth's surface?

lithiumdeteurteride, there are 66 miles per degree, your deduction makes no sense.
jimschem, you are very close.

A sattelite orbiting the Earth at 200 miles altitude and a speed of 17240MPH travels 4.8 miles in one second and falls 14.5 feet in one second due to reduced g at that altitude. The curvature of the Earth's surface is not matched. What am I doning wrong?

Answer 1

I concur with the calculation immediately above.

One way of *measuring* this effect follows: A simple experiment would be to use a telescope to look at a buoy on the ocean 5 miles away from your location at sea level. You could stick a big ol measuring stick on the buoy and see what level just peaks over the horizon (since you and the buoy are bobbing up and down you would want to take several measurements and take the median).

Answer 2

Bernard b and jimschem are partially correct. We know the properties of water and how it always seeks level; that is why water levels are used. The Old Bedford level experiment has been reproduced 1000's of times on different bodies of water at varying distances with the same results; and jimschem's math is correct for a sphere that is roughly 960 miles radius; If these conditions are correct, then there is NO curvature. Water is always level.

Answer 3

The common answer for the rate of curve is 7.9 inches per mi. This describes a slope. So I looked further into it. Some say to square the distance. That doesn t work either.
Let 1000 mi equal 1inch. Draw a circle of 24" 8" diameter and 4" radius.
Quarter it. We know that 4000 mi out it is 4000mi down. Any calculation used must match this to be accurate. Once an equation is formed use it to calculate any distance outward from line of sight and it will tell how far below the horizon is.
No one has ever answered the question to my knowledge.

Answer 4

Curvature is not measured in feet. That doesn't make sense. However, we can measure curvature by a change in angle:

The circumference of the Earth is very close to 40,000 kilometers. So, if you travel 40,000 kilometers along the Earth, you will have gone all the way around, ending up where you started (a 360-degree change of angle). If you only travel 5 miles (8.04672 kilometers), your angular change will be

8.0467 / 40,000 = 0.0002012

Multiplying this fraction by 360 degrees, we see that you will have traveled

0.0002012 * 360 = 0.07242 degrees around the Earth.

Answer 5

The most famous way this was attempted was the 'Bedford levels experiment, repeated several times between 1838 and 'the early 20th century'. It was a version of Mr Quarks 'ocean buoy' experiment, only done on a 6 mile stretch of canal.

Unfortunately this experiment consistently demonstrated that even if the earth was not flat, it appeared to be flat in Bedford. Read this link:

Answer 6

If a person walked 5 miles along the earth's surface
in a straight line at a place where the earth's
surface went straight you would be 16 feet 8 inches
higher or away from the earth's center than if it did
curve.

The earth's radius is 3960 miles.

sqrt(3960^2 + 5^2) = 3960.00315656 miles from center.

3960.00315656 - 3960 = 0.00315656 miles

0.00315656 * 5280 = 16.66666 feet = 16 feet 8 inches.

The earth's surface CURVES 16 ft. 8 in. in 5 miles.

Answer 7

69 miles equals 3174 feet of curvature.
69 X 69 X 8 = 38,088 / 12 = 3174
8" X miles squared equals number of inches of curve. Dived by 12 for feet

Answer 8

Divide the number of feet around the equater into 360 degrees and then multiply by 5.

Answer 9

every degree is 69 miles