19 Feb 10
Originally posted by FMFSA = 4 pi r^2
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
Originally posted by AThousandYoungSo what do you make the answer? (How many kilometres up?)
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
1 edit
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/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
Originally posted by PalynkaThat 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.
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
1 edit
Originally posted by FMFThe Earth's mean radius is 6371 km, so...
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
Originally posted by PalynkaHmm. So you're taking into account the crinkliness of the surface in your calculation? How!?
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.