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IOAA 2022, Теория — Задача 8. Ring of a planet

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International Olympiad on Astronomy and Astrophysics 2022 Georgia — Theoretical Competition · 16 август 2022 г. · 20 т.

Условие

We are given a flat disk of mass M with inner radius $r$ and the outer radius $R$ .

(a) A point mass $m$ is located on the symmetry axis of the disk at a distance $x$ from the plane of the disk (We assume throughout that mass $m$ stays on the symmetry axis, and is free to move in the vertical direction.). What is the gravitational force exerted on the point mass?

A grey flat annulus labelled M seen in perspective, with inner radius r and outer radius R drawn from the centre; a point mass m on the vertical symmetry axis at height x above the centre. — Figure 2: Flat disk around a point mass

Hint 1: You may denote surface density by the symbol $\sigma$ and calculate force exerted by a small surface element $\Delta S$ subtending a solid angle $\Delta\Omega$ with $m$

Hint 2: The area cut by cone with opening angle $2\theta$ on a sphere of radius $R$ :

S = 2\pi R^2(1 - cos(\theta))

[10 т.]

A light grey sphere with a darker spherical cap labelled S on top bounded by a blue circle; dash-dotted lines from the centre to the axis and to the rim of the cap show the half-opening angle θ. — Figure 3: Area cut by a cone on a sphere

(b) what will be the frequency of the small oscillations of the system ( $x \ll r$ ). [10 т.]

Отговори

Решение

(a) Consider small surface element ( $\Delta S$ ) of the disk. The gravitational field ( $\Delta E$ ) generated by this surface element at at a point $O$ located at a distance $a$ from $\Delta S$ (see the figure below):

\Delta E = G\frac{\sigma\Delta S}{a^2}

where $\sigma$ is a surface density of the mass.

An irregular grey surface labelled σ with a small element dS; a dashed normal rises from dS, and a thin cone dΩ goes from dS to a point O above right; the angle α is marked between the normal and the direction to O.

where $\alpha$ is the angle between surface normal and the vector $\vec{r}$ and $\Delta\Omega$ is infinitesimal solid angle corresponding to the area $\Delta S$ .

For a ring, the surface density will be given by $\sigma = M/\pi(R^2 - r^2)$ .

As the horizontal component will get cancelled due to symmetry considerations, we only consider the normal component:

\Delta E_{\perp} = \Delta E cos(\alpha) = \frac{G\sigma}{a^2}\Delta S cos(\alpha) = \frac{G\sigma}{a^2}\Delta S_{\perp} = G\sigma\Delta\Omega \qquad \mathbf{5.0}

By Summation, we can obtain total normal field of the surface:

E_{\perp} = G\sigma\Omega \qquad \mathbf{1.0}

here $\Omega$ is the total solid angle subtended by the disk. In our case, when we observe disk from the point of view of the test particle:

\begin{aligned} \Omega &= 2\pi\left(1 - \cos\theta_{out}\right) - 2\pi\left(1 - \cos\theta_{in}\right) \\ &= 2\pi\left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right) - 2\pi\left(1 - \frac{x}{\sqrt{x^2 + r^2}}\right) \\ \Omega &= 2\pi x\left(\frac{1}{\sqrt{x^2 + r^2}} - \frac{1}{\sqrt{x^2 + R^2}}\right) && \mathbf{3.0} \\ \therefore F_{\perp} &= G\sigma\Omega m \\ F_{\perp} &= \frac{2GMmx}{R^2 - r^2}\left(\frac{1}{\sqrt{x^2 + r^2}} - \frac{1}{\sqrt{x^2 + R^2}}\right) && \mathbf{1.0} \end{aligned}

(b) In small displacement approximation:

F_{\perp} \approx \frac{2GMmx}{R^2 - r^2}\left(\frac{1}{r} - \frac{1}{R}\right) = \left(\frac{2GMm}{Rr(R + r)}\right)x = kx \qquad \mathbf{3.0}

when the distance between the disk and the point mass is $x$ , the distance of the center of mass of the system from the center of the disk will be:

x_1 = \frac{mx}{m + M} \qquad \mathbf{2.5}

If we choose the center of mass of the system as an origin of our coordinate system, then equation of motion of the disk will have a form:

\begin{aligned} Ma_1 &= -kx \\ Ma_1 + k\frac{m + M}{m}x_1 &= 0 && \mathbf{2.5} \end{aligned}

For this simple harmonic oscillator, the period of small oscillations will be:

T = 2\pi\sqrt{\frac{Rr(R + r)}{2G(M + m)}} \qquad \mathbf{2.0}

Оригинал в Архива: Theory Problems.pdf · официални решения: Theory Solutions.pdf