1. SubscriberFMF
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    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?
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    19 Feb '10 06:102 edits
    r(earth)*(sqrt(2)-1)
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    19 Feb '10 06:121 edit
    ~2642 kilometers.
  4. silicon valley
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    19 Feb '10 06:52
    what class is this for?
  5. SubscriberFMF
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    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. SubscriberAThousandYoung
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    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. SubscriberFMF
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    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. Standard memberPalynka
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    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.
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    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. SubscriberAThousandYoung
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    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. SubscriberAThousandYoung
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    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. Standard memberPalynka
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    22 Feb '10 10:26
    Yes, my post was completely wrong because I used the formula for a circle, not a sphere. 😞
  13. SubscriberAThousandYoung
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    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. silicon valley
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    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.
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