nourishing obscurity

10 Responses to “Earth calling readers”

1. Andrew Scott, on November 6th, 2009 at 23:04 Said:

   Depends on the cloud cover. A different distance every day, every moment, you tease.

   
   
2. MTG, on November 7th, 2009 at 07:24 Said:

   Varies with degrees of latitude, unless earth is assumed to be spherical.

   
   
3. MTG, on November 7th, 2009 at 10:48 Said:

   Coincidentally a GEO satellite height of 22K miles?

   
   
4. MTG, on November 7th, 2009 at 17:30 Said:

   I was hoping somebody would venture the 3963Tan60 route by now.

   
   
5. James Higham, on November 7th, 2009 at 18:14 Said:

   The official answer is the earth’s diameter away but I feel your objections hold water, gentlemen.

   
   
6. Nigel Sedgwick, on November 7th, 2009 at 20:32 Said:

   I agree with James’ answer: an altitude of one Earth’s radius from the surface, or distance of one Earth’s diameter from the centre of the Earth.

   I get this by solving cos(theta) = r/(r+h) and 2*theta/360 = 1/3, where r is the Earth’s radius, h is the height of observation (above the surface) giving the 1/3 surface area view and theta is the angle between a line drawn from the Earth’s centre to the point of observation and a line drawn from the Earth’s centre to one of the places where the line-of-sight from the point of observation tangentially grazes the Earth’s surface. The preceding, with area variation from 1/3, also gives the height for any other fraction (<1/2) of the Earth's surface being visible.

   However, I am somewhat in dispute with those who raise real-world issues of cloud, uniformity of radius, etc. That is unless they whinge jestingly.

   The only way that a rationally deduced answer can be given to the question is if one assumes that the Earth is a perfect sphere, with no relevant atmospheric effects (clouds, refraction – hence need to assume a particular wavelength of light). Otherwise, there is insufficient information to answer the question.

   Of course, the best answer (Oxbridge entrance style) would be to give the perfect sphere solution, point out its lack of realism, and (lacking the detailed information) mention all the real-world effects and (briefly) how one might take them into account with more time and more detailed information (eg better approximation with ellipsoid rather than sphere and observation from specified direction WRT two axes – or just WRT the major axis assuming equal minor axes, and why that might be a reasonable approximation).

   Best regards

   
   
7. MTG, on November 8th, 2009 at 09:02 Said:

   Very smart Nigel – but the question was framed with far too many assumptions. Your equation only works with a snapshot in time, otherwise the reference point must be must be geosynchronous in space and that condition was unspecified. There should certainly be a correction for refraction and getting really pedantic, another for gravitational lensing.

   
   
8. Andrew Scott, on November 8th, 2009 at 10:05 Said:

   Nigel: Partly jesting, partly suspicions that James was sneakily trying to catch us out (as I think he likes to do), partly trying to think out of the box (as I like to do) and partly me just being a childish smart-ass. You’ve got to try to cover all bases with this Higham chap

   
   
9. James Higham, on November 8th, 2009 at 10:37 Said:

   

   
   
10. Lord T, on November 8th, 2009 at 11:09 Said:

    I’m with Andrew on this. The earth was specified and it is not a perfect sphere. I bet some one somewhere has done the maths on the earth and come up with the answer. Me. I’m to busy resting.