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IOAA 2021, Теория — Задача 12. Hodograph (15 points).

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Theory — Q12 Hodograph (International Olympiad on Astronomy and Astrophysics, 14, 2021) — English (Official) · 15 т.

Внимание: Бележка към темата

Solutions document not provided in the archive for this paper; answers are derived, not transcribed from an official solutions file.

Условие

In curvilinear motion of a planet around a star, the direction of the velocity vector changes continuously. This can be represented by a so-called "trajectory in velocity space" and is obtained as follows: for each point on the spatial trajectory, the corresponding velocity vector is drawn so that its starting point is at the origin of the velocity space, and its magnitude and direction is the same as the velocity vector at that point. The tip of this variable velocity vector generates a curve in velocity space. (The name 'hodograph' was given to this curve by Hamilton in 1846.)

As an example, see figures 1 and 2 below. For a circular orbit (Figure 1), the magnitude of the velocity is constant and therefore, the hodograph (Figure 2) of the velocity vector for Keplerian circular motion is also a circle, the center of which is located at the origin of the velocity space. The radius of this circle is equal to the constant magnitude of the circular velocity.

A dotted circle of radius R around a star (yellow dot at centre); eight points numbered 1–8 around the circle, each with an outward tangent arrow labelled V showing the velocity direction at that point. — Fig. 1 Spatial trajectory of the Planet with Uniform Circular Motion around the star.

Velocity space diagram: a circle centred at the origin with eight velocity vectors of equal length labelled 1–8, drawn from the origin to evenly spaced points on the circle. — Fig. 2 Corresponding hodograph

12.1 Write an expression for the radius of the hodograph in Fig. 2, as a function of the mass $M$ of the star, and the radius $R$ of the circular orbit of the planet's motion. [1 т.]

12.2 For a planet in a Keplerian trajectory, write the expression for centripetal acceleration vector ( $\vec{a}$ ) and the magnitude of angular momentum ( $L$ ). For any Keplerian trajectory, it is true that

|\Delta v| = k \Delta \theta \qquad (1)

Where $k$ is a constant for each type of Keplerian trajectory. Find the expression for the constant $k$ as a function of the masses $M$ and $m$ of the star and the planet, respectively, and the angular momentum, $L$ .

(Eq.1) allows us to conclude that for any Keplerian trajectory, the hodograph ( $v$ as a function of $\theta$ ) is a circle, but except for circular motion, the centre of the hodograph does not coincide with the star. It is not necessary to prove this result, you may simply accept it as a given. For the hodograph of uniform circular motion, the compliance with (eq.1) is completely obvious, as evidenced in Fig. 3

Velocity space diagram: a circle centred at the origin with eight velocity vectors labelled 1–8; near vector 1 a small change Δv between two successive velocity vectors subtends a small angle Δθ at the centre.

[4 т.]

12.3 Determine the expression of the constant $k$ for the hodograph of circular planetary motion. [2 т.]

12.4 Given that the hodograph of the Keplerian elliptical motion is a circle, determine the radius of this hodograph and the distance between the center of the hodograph and the position of the star, as a function of the velocities at periastron and apoastron. Draw a rough sketch of the hodograph in the answer sheet as per the schematic shown in Fig. 4. The black circle is the star.

Left: an ellipse with the star inside; at the top the periastrom velocity V_P is a horizontal arrow pointing right (θ = 0) and at the bottom the apoastrom velocity V_A is a horizontal arrow pointing left (θ = π). Right: an empty answer-sheet schematic with a horizontal and a vertical axis crossing at a black dot (the star).

[4 т.]

12.5 Similarly, for the parabolic Keplerian trajectory, determine the radius of the corresponding hodograph and the distance from the center of that hodograph circle to the star. Express the radius as a function of the velocity at periastron. Draw a rough sketch of the hodograph circle in the answer sheet. [4 т.]

Отговори

Решение

12.1

For a satellite that describes a uniform circular motion of radius r around the star and mass M, we have:

v=[R_{Hodograph}]=\sqrt{\frac{GM}{r}}

[1 point]

12.2

\vec a=-\frac{Gm}{r^2}\hat r

[1 point]

L=mr^2\omega

[1 point]

\rightarrow\vec a=-\frac{GMm\omega}{L}\hat r=\frac{\Delta\vec v}{\Delta t}

\frac{GMm}{L}\frac{\Delta\theta}{\Delta t}=\left|\frac{\Delta v}{\Delta t}\right|

\Delta v=\pm\frac{GMm}{L}\Delta\theta

k=\frac{GMm}{L}

[2 points]

12.3

L=mv_cR

[1 point]

k_c=\frac{GMm}{L}=v_c=\sqrt{\frac{GM}{R}}

[1 point]

12.4

In the following scheme, the hodographic circumference has been constructed as follows:

For $\theta=0$ , the velocity vector is drawn in arbitrary units (3, for instance) corresponding to the velocity in the periastron, $v_P$ . Its head determines the point $A$ .
For $\theta=\pi$ , the velocity vector is drawn diametrically opposite in the apastron, $v_A$ , also in arbitrary units. Its head determines point $B$ . The hodograph must pass through points $A$ and $B$ . Therefore, the radius of the hodograph must be equal to:

R_{Hodograph}=\frac{v_P+v_A}{2}

[1 point]

And the “distance” $d$ from the star to the center of the circumference will be:

d=\frac{v_P-v_A}{2}

[1 point]

Elliptical orbit and its offset circular hodograph on a grid.

[2 points]

12.5

In the following scheme, the hodographic circumference has been constructed as follows:

For $\theta=0$ , the velocity vector is drawn in arbitrary units (4, for instance) corresponding to the velocity in the periastron, that is $v_P$ . Its head determines the point $A$ . Then the velocity vector is drawn diametrically opposite in the apastron, that is, $v_A=0$ . Its head determines point B that coincides with the star. The hodograph must pass through points A and B. Therefore, the radius of the hodograph must be equal to:

R_{Hodograph}=\frac{v_P}{2}

[1 point]

And the distance $d$ from the star to the center of the hodographic circumference

d=\frac{v_P}{2}

[1 point]

Parabolic orbit and its circular hodograph passing through the star.

[2 points]

Official source discrepancy

The first acceleration formula in official solution 12.2 prints lowercase m, whereas its subsequent derivation uses uppercase M for the stellar mass. The printed formulas are retained.

Оригинал в Архива: TQ-12-Q.pdf · официални решения: TQ12.pdf · допълнителни данни: TQ-12-A.pdf · допълнителни данни: Standard-Table-of-constants.docx.pdf · допълнителни данни: TQ-G.pdf · допълнителни данни: data.pdf