2018年IAO理论低年组第5题-奥尔特云

英文题目

Oort cloud. It is currently considered that tha source of long-periodic and non-periodic comets is th outer Oort cloud, the inner and ouyer radii of which are estimated as 0.2 and 0.8 light years, respectively. Comet bodies in this cloud move erratically and sometimes collide with each other. As a result, every century, people of the Earth observe from 10 to 20 bright comets coming into the internal regions of the Solar System. The average sizn of the nuclei of such comets is about 2-3kilometers. Roughly estimate (in order of magnitude) the total number of such comet bodies in the outer Oort cloud, the average distance between them, and their total maww. Compare the results with distance and masses of bidies in our Solar System.

中文题目

奥尔特云 目前认为长周期和非周期彗星的来源是太阳系外部的奥尔特云，其内球和外球半径分别估计为0.2和0.8光年。 云中的彗星体不规则地移动，有时会相互碰撞。 因此，每个世纪，地球人都会观察到10到20颗明亮的彗星进入太阳系的内部区域。 这种彗星的彗核平均尺寸约为2-3千米。 粗略地估计（在数量级上）外部奥尔特云中这种彗星体的总数，它们之间的平均距离，以及它们的总量。 将结果与我们的太阳系中的天体（比如行星）的距离和质量大小进行比较。

英文解答

Oort cloud. The fact that every century we observe from 10 to 20 bright comets means that approximately once every 6 years 8 months (T ≈ 2.1·10⁸ s) two comet bodies collide in the outer Oort cloud, resulting in one of them rushes into the inner regions of the solar system. Collisions occur because all these bodies move chaotically with typical velocities less than the second cosmic velocity (otherwise they would leave this area). To estimate the average their velocities, we take the first cosmic velocity for the average distance (R+r)/2 = 0.5 ly:

$$u = (GM/[(R+r)/2])^{1/2}\ ≈\ 170\ m/s$$,

where M is the mass of the Sun, R and the outer and inner radii of the cloud. (One can also use the formula u ～ R^-1/2, 0.5 ly is about 30,000 times larger than the distance from the Earth to the Sun, therefore, the speed of the Earth in its orbit shoule be divided by the square root of 30,000). So we obtain u ≈ 170 m/s

Let us find how the frequency of collision is related to the number of comet bodies in the Oort cloud. If there are N comet bodies in total, then their concentration in space is

$$n\ =\ N\ /\ [4/3\ {\pi}\ (R^{3}-r^{3})]\ ≈\ N\ /\ 4/3\ {\pi}R^{3}$$ (bodies per m³).

(The face r³ « R³ was taken into account.) We will use the collision model in which each comet body moves relative to the others with the same average velocity u. During the time T, the comet body will cover the distance u·T and in its movemert will captuer the volume

$$V\ =\ u·T·S\ =\ u·T·{\pi}d^{2}/4$$,

where d is the average diameter of the comet body, we sill cosider it equal to 2.5 km. If another comet body appears in this volume, then a collision occurs, and the probility of such an event is equal to

$${\alpha}\ =\ n·V_{1} = n·u·T·{\pi}d^{2}/4$$,

Since any of the N comet bodies can collide, the total number of collisions will be equal to:

$$K\ =\ {\alpha}·N\ =\ n·u·T·{\pi}d^{3}/4·n·4/3·{\pi}R^{3}\ =\ u·T·{\pi}^{2}n^{2}d^{2}R^{3}$$,

$$K\ =\ {\alpha}·N\ =\ N\ /\ (4/3\ {\pi}R^{3})·u·T·{\pi}·d^{2}/4·N\ =\ 3·u·T·N^{2}d^{2}/16R^{3}$$,

Each collision leads to the fact that with some probabilty β our body will be sent to the inner regions of the Solar System. This probability can be estimated from different models, let us estimate it as an approximate ratio of solid angles:

$${\beta}\ =\ {\pi}r^{2}/4{\pi}R^{2}\ ≈\ 0.015$$,

Thus, in orger that one bodu goes to the inner regions of the solar system, it is necessary to fulfill hte condition

$$K·{\beta}\ =\ 1$$,

$$3·u·T·N^{2}d^{2}/16R^{3}·{\beta}\ =\ 1$$,

$$N\ =\ 4R^{3/2}/d(3{\beta}uT)^{1/2}$$.

Calculations give: $$N\ =\ 2.6·10^{16}$$,

with an accuracy of one significant digit: $$N\ ≈\ 3·10^{16}$$.

If the concentration of the comet bodies is n (bodies per m³), that each body has a volume of space 1/n, and, respectively, the average distance will (1/n)^1/3.

$$n\ ≈\ N\ /\ 4/3\ {\pi}R^{3}\ ≈\ 3^{1/2}/{\pi}d({\beta}uTR^{3})^{1/2}$$,

$$L\ ≈\ 1/n^{1/3}\ ≈\ R^{1/2}({\pi}d)^{1/3}({\beta}uT/3)^{1/6}$$,

$$L\ ≈\ 4·10^{10}\ m\ ≈\ 40\ mln.km.$$

This is roughly the minimum distance between the Eater and Venus.

To estimate the total mass of the comet bodies, we assume that their average density is equal to the density of ice (ρ ≈ 900 km/m³), and average volume is v = 1/6 πd³. Thus, the average mass of one comet body is:

$$m\ =\ {\rho}·v\ =\ 1/6 {\pi}d^{3}\ {\rho}\ ≈\ 7·10^{12}kg $$,

and the sum of the mass of all the bodies:

$$M\ =\ N·m\ ≈\ 2·10^{29}\ kg$$.

This is one hundred masses of Jupiter or one-theth of the mass of the Sun.