In article <> of rec.puzzles, Thomas D Bending <> wrote:

I must admit that when I first came across this problem a while ago I just bashed out the integration, too. However, you've inspired me to think about it again and discuss it with some colleagues, and we found that it's actually quite easy:

I'm not sure Kepler's Laws apply to bodies which collide, even if they're point masses, so give the Earth a small push sideways as it starts to fall - it will follow a long narrow elliptical orbit passing very close to the Sun - and then let the size of the push -> 0. If the Earth's initial distance from the sun is 1 then the semi-major axis of the new orbit -> 0.5. By Kepler's ? Law the period of an orbit is proportional to the 3/2 power of the semi-major axis, so the period of the new orbit -> 0.5^(3/2) years, or 1/(2*sqrt(2)) years. The time taken for the fall is thus half this time, or 1/(4*sqrt(2)) years, or about 64.6 days.

Fortunately this is the answer I got by integration, too. Of course I never doubted it would be, you understand 8-)

Dr Thomas Bending


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