We consider a spherical star cluster and assume it is isotropic, meaning the mass density \rho at a distance r from the center depends only on r, i.e., \rho = \rho(r). Thus, the total mass within a spherical region of radius r is: M(r) = \int_0^r 4\pi x^2 \rho(x) dx
Imagine a star with a relatively small mass (in fact, compared to the total mass of the cluster, the mass of any single star is very small) moving under the gravitational influence of the cluster (much like the Solar System moves around the Milky Way), assuming the star experiences no resistance from other matter (such as interstellar dust). We have previously proven that the gravitational force inside an isotropic spherical shell is zero. In this case, the motion is equivalent to the star being attracted only by the mass within the spherical region between it and the center. According to the law of universal gravitation, choosing the center of the cluster as the reference frame, we can derive: \ddot{\vec{r}} = -GM(r) \frac{\vec{r}}{r^3}
Substituting M(r) into the equation, we get: \ddot{\vec{r}} = -G \left[ \int_0^r 4\pi x^2 \rho(x) dx \right] \frac{\vec{r}}{r^3}
This is the general case. If we assume the globular cluster is uniform, meaning the density \rho is a constant, then the above equation becomes: \ddot{\vec{r}} = -\frac{4}{3} \pi G \rho \vec{r}
Remarkably, this is a force proportional to the distance! This is undoubtedly the simplest case. If we represent the vector \vec{r} as a complex number z, the equation can be written as: \ddot{z} = -\left( \frac{4}{3} \pi G \rho \right) z = -\mu z
This is a second-order linear differential equation. Its characteristic equation is \lambda^2 + \mu = 0, which yields \lambda = \pm \sqrt{\mu} i. The general solution to the equation is: z = C_1 \exp(i \sqrt{\mu} t) + C_2 \exp(-i \sqrt{\mu} t)
Alternatively, this can be rewritten in terms of trigonometric functions for the x and y coordinates, though I prefer the complex form as it makes calculations more convenient. The general solution indicates a periodic elliptical motion with an angular velocity of \omega = \sqrt{\mu}. Thus, the period of motion is: T = \frac{2 \pi}{\sqrt{\mu}} = \sqrt{\frac{3\pi}{G\rho}}
In other words, all stellar bodies within the cluster have approximately the same orbital period!
For general galaxies, we can use this formula for a basic density estimation. The orbital period of the Solar System around the Milky Way is approximately 250 million years, which is about 8 \times 10^{15} seconds. Substituting this into the formula, we calculate: \rho = 2.2 \times 10^{-21} \text{ kg/m}^3
According to data found online, the average density of interstellar matter in the Milky Way is approximately 1 hydrogen atom per cubic centimeter. The total mass of interstellar matter accounts for about 10% of the Milky Way’s total mass. The mass of a hydrogen atom is 1.674 \times 10^{-27} kg. Therefore, the average density of the Milky Way should be approximately: \rho = 1.674 \times 10^{-20} \text{ kg/m}^3
Compared to the estimated value, there is a difference of one order of magnitude. However, considering the rough nature of the calculations in this article, this is already a quite accurate result.
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