What is the elevation factor value for Earth’s radius R = 20,906,000 ft and local elevation H = 6,320 ft?

What is the elevation factor value for Earth’s radius R = 20,906,000 ft and local elevation H = 6,320 ft?

• A. 0.999697785
• B. 0.9997
• C. 1.000000000
• D. 0.980000000

Answer: (A)

The elevation factor comes from how arc lengths on a sphere scale with its radius. For a height H above the reference radius, the local radius is R + H, so the ratio of the distance on the reference sphere to the distance at height H is R/(R+H). This is because arc length for a given central angle is proportional to the radius. Compute the numbers: R + H = 20,906,000 + 6,320 = 20,912,320 ft. Then the elevation factor = R/(R+H) = 20,906,000 / 20,912,320 ≈ 0.999697785. That value is very close to 1 since H is small compared to R. Rounding to four decimals gives 0.9997, but the exact value with the provided figures is 0.999697785. The other options are either a rounded form, exactly 1 (which would require H = 0), or far off.

The elevation factor comes from how arc lengths on a sphere scale with its radius. For a height H above the reference radius, the local radius is R + H, so the ratio of the distance on the reference sphere to the distance at height H is R/(R+H). This is because arc length for a given central angle is proportional to the radius.

Compute the numbers: R + H = 20,906,000 + 6,320 = 20,912,320 ft. Then the elevation factor = R/(R+H) = 20,906,000 / 20,912,320 ≈ 0.999697785.

That value is very close to 1 since H is small compared to R. Rounding to four decimals gives 0.9997, but the exact value with the provided figures is 0.999697785. The other options are either a rounded form, exactly 1 (which would require H = 0), or far off.