# Surface of the globe

FMF
Science 19 Feb '10 06:05
1. FMF
Main Poster
19 Feb '10 06:05
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?
2. 19 Feb '10 06:102 edits
r(earth)*(sqrt(2)-1)
3. 19 Feb '10 06:121 edit
~2642 kilometers.
4. 19 Feb '10 06:52
what class is this for?
5. FMF
Main Poster
19 Feb '10 09:02
Originally posted by zeeblebot
what class is this for?
The realisation that I had essentially forgotten how to do such a calculation - having been good at maths as a kid - came to me, as such musings often do, as I was riding my motorbike this morning.
6. AThousandYoung
West Coast Rioter
19 Feb '10 09:14
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?
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.
7. FMF
Main Poster
19 Feb '10 12:11
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.
So what do you make the answer? (How many kilometres up?)
8. Palynka
Upward Spiral
19 Feb '10 13:131 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.
9. 19 Feb '10 19:44
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.
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.
10. AThousandYoung
West Coast Rioter
20 Feb '10 01:341 edit
Originally posted by FMF
So what do you make the answer? (How many kilometres up?)
The Earth's mean radius is 6371 km, so...

~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.
11. AThousandYoung
West Coast Rioter
20 Feb '10 01:35
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.
Hmm. So you're taking into account the crinkliness of the surface in your calculation? How!?
12. Palynka
Upward Spiral
22 Feb '10 10:26
Yes, my post was completely wrong because I used the formula for a circle, not a sphere. ðŸ˜ž
13. AThousandYoung
West Coast Rioter
22 Feb '10 11:02
To take into account the crinkliness, I suppose I'd have to say ~0.4 times the REAL surface area whatever that is.
14. 23 Feb '10 21:01
Originally posted by AThousandYoung
To take into account the crinkliness, I suppose I'd have to say ~0.4 times the REAL surface area whatever that is.
not crinkliness, oblate sphericity.