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

Using the III Kepler law ($T^2/a^3=\mathrm{const}$) one can find that for the circular speed the following relation is valid:

and so:

It means that the circular speed ($V_2$) for Mars orbit is $1.52^{1/2}$ smaller than one ($V_1$) for Earth orbit.

During the Great oppositions Mars is near the perihelion of its orbit.

Using the II Kepler law one may also write (for the orbits with small eccentricities – see note*):

where $R_{\mathrm{circ}}$ and $V_{\mathrm{circ}}$ – the mean distance from the Sun to Mars and its mean orbital speed. So

(*) More accurate using of the II Kepler law here is

where $a$ and $b$ are the semiaxes of the orbit, $b=a(1-e^2)^{1/2}$.

In this case the answer is