- 19 Feb '10 09:14

SA = 4 pi r^2*Originally posted by FMF***How many kilometres up from the Earth's surface would one have to go to be then on a notional sphere whose surface area would be double the actual surface area of the Earth?**

2SA = 4 pi R^2

SA = 2 pi R^2

4 pi r^2 = 2 pi R^2

2r^2 = R^2

2^(1/2) r = R (Square root of 2 times radius of earth = larger radius)

[2^(1/2) - 1]r = R - r = distance above the ground = (square root of 2 minus 1) times Earth radius ~ 0.4 times the Earth's radius above the surface. - 19 Feb '10 12:11

So what do you make the answer? (How many kilometres up?)*Originally posted by AThousandYoung***SA = 4 pi r^2**

2SA = 4 pi R^2

SA = 2 pi R^2

4 pi r^2 = 2 pi R^2

2r^2 = R^2

2^(1/2) r = R (Square root of 2 times radius of earth = larger radius)

[2^(1/2) - 1]r = R - r = distance above the ground = (square root of 2 minus 1) times Earth radius ~ 0.4 times the Earth's radius above the surface. - 19 Feb '10 13:13 / 1 editThe surface is not a sphere, though.

The estimated real surface area is about 510,072,000 km2, while calculating the surface by using the mean radius would get you 2*Pi*6371^2 = 255,032,236. A huge difference.

So interpreting the question literally, I would say that the radius of the notional sphere R is:

2*510072000 = 2*Pi*R^2

R = sqrt(510072000/Pi) = 12742 km. Approximation errors aside, that's exactly double the mean radius 6371!

FMF: 6371 km above the mean radius, so slightly less for the equatorial radius and a bit more for the polar one. - 19 Feb '10 19:44

That should read 4*pi*6371^2. This is ~ 510,064,000 km^2, an error of .001%. So it's not a particularly bad approximation.*Originally posted by Palynka*

The estimated real surface area is about 510,072,000 km2, while calculating the surface by using the mean radius would get you 2*Pi*6371^2 = 255,032,236. A huge difference. - 20 Feb '10 01:34 / 1 edit

The Earth's mean radius is 6371 km, so...*Originally posted by FMF***So what do you make the answer? (How many kilometres up?)**

~2500 km above the surface using only my head and some minor estimations. A proper answer would take more work and I'm not in the mood but if you insist I'll do it.

EDIT - This assumes the Earth is a perfect sphere, which other posters have pointed out is not true. How you'd take into account the crinkliness is beyond me though. - 20 Feb '10 01:35

Hmm. So you're taking into account the crinkliness of the surface in your calculation? How!?*Originally posted by Palynka***The surface is not a sphere, though.**

The estimated real surface area is about 510,072,000 km2, while calculating the surface by using the mean radius would get you 2*Pi*6371^2 = 255,032,236. A huge difference.

So interpreting the question literally, I would say that the radius of the notional sphere R is:

2*510072000 = 2*Pi*R^2

R = sqrt(510072000/ ...[text shortened]... e the mean radius, so slightly less for the equatorial radius and a bit more for the polar one.