IAO 2002, Теория, Група α, английски — Задача 5

One can find from the first part of the table that the eccentricity of the orbit is $e=1.000812$. It is very near to a parabolic orbit. I.e. speed of the comet in perihelion is about $2^{1/2}$ of the circular speed with the same distance from Sun.

Than one can find from the ephemeris that the distance Sun–Comet in perihelion is $a_c=0.776$ a.u. (02.10.2002.). Using the III Kepler law ($T^2/a^3=\mathrm{const}$) one can find that for the circular speed

$V^2a=\mathrm{const}$

and so:

$V\sim a^{-1/2}.$

I.e.

where $V_E\approx30$ km/s is the circular speed of Earth.