OWAO 2023, Теория — Задача 2. Straight Forward

• -81 km/s

  Official solution: $V_{r,2} = -81$ km/s.

First, we determine the current distance to the star from the known parallax $\varpi$:

Then, let us find the current tangential velocity of the star. It is quite convenient to use the ratio $1\ \mathrm{au} = 1\ \mathrm{pc} \times 1''$:

Next, we estimate the current radial velocity. According to the Doppler effect formula

The total velocity is

In 20 000 years, the star will cover a distance of

$S$ denotes the Sun, $M_1$ is the current position of the star and $M_2$ is the position of the star in future. Let us write the law of cosines for $\triangle M_1M_2S$:

where $|M_1S| = \dfrac{1}{\varpi\,['']} = 12.5$ pc, $\gamma = \arccos\dfrac{V_{r,1}}{V} = 19^{\circ}$.

Consequently,

Next, we write the law of cosines for the same triangle:

The radial velocity is

Marking Scheme:

• Current distance to the star $|SM_1| \equiv r$ — 1 pt.
• Current tangential velocity of the star $V_{\tau,1}$ — 1 pt.
• Current radial velocity of the star $V_{r,1}$ — 1 pt.
• Geometry of the problem (preferably, a drawing) — 1 pt.
• Distance covered in 20 000 years $|M_1M_2|$ — 1 pt.
• Direction of the star’s velocity vector, angle $\gamma$ — 1 pt.
• Distance to the star in 20 000 years $|SM_2|$ — 2 pt.
• Angle $\theta$ — 1 pt.
• Radial velocity of the star in 20 000 years $V_{r,2}$ — 1 pt.

$^{2}$ It is also remarkable that $30\ \mathrm{km/s} \approx V_{\oplus} = 2\pi\ \mathrm{au/yr} \approx 6.28\ \mathrm{au/yr} \approx V_{\tau,1}$.