2017年IOAA理论第11题-本星系群的质量
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英文原题
(T11) Mass of the Local Group [50 marks]
The dynamics of M31 (Andromeda) and the Milky Way (MW) can be used to estimate the total mass of the Local Group (LG). The basic idea is that galaxies currently in a binary system were at approximately the same point in space shortly after the Big Bang. To a reasonable approximation, the mass of the local group is dominated by the masses of the MW and M31. Via Doppler shifts of the spectral lines, it was found that M31 is moving towards the MW with a speed of 118 km·s−1. This may be surprising, given that most galaxies are moving away from each other with the general Hubble flow. The fact that M31 is moving towards the MW is presumably because their mutual gravitational attraction has eventually reversed their initial velocities. In principle, if the pair of galaxies is well-represented by isolated point masses, their total mass may be determined by measuring their separation, relative velocity and the time since the universe began. Kahn and Woltjer (1959) used this argument to estimate the mass in the LG.
In this problem we will follow this argument through our calculation as follows.
a) Consider an isolated system with negligible angular momentum of two gravitating point masses $$m_1$$ and $$m_2$$ (as observed by an inertial observer at the centre of mass).
Write down the expression of the total mechanical energy ($$E$$) of this system in mathematical form connecting $$m_1$$ , $$m_2$$ , $$r_1$$ , $$r_2$$ , $$v_1$$, $$v_2$$ , and the universal gravitational constant $$G$$, where $$v_1$$ and $$v_2$$ are the radial velocities of $$m_1$$ and $$m_2$$ , respectively.[5]
b) Re-write the equation in a) in terms of $$r$$, $$v$$, $$\mu$$, $$M$$, and $$G$$, where $$r\equiv r_1 +r_2$$ is the separation distance between $$m_1$$ and $$m_2$$ , $$v$$ is the changing rate of the separation distance,$$\mu \equiv \frac{m_1 m_2}{m_1 + m_2}$$ is the reduced mass of the system, and $$M \equiv m_1+m_2$$ is the total mass of the system. [10]
c) Show that the equation in b) yields \[ v^2 = (2GM)(\frac{1}{r}-\frac{1}{r_0}) \text{, where }r_0 \text{ is a new constant.}\]
Find $$r_0$$ in terms of $$\mu$$ , $$M$$ , $$G$$ and $$E$$. [5]
The solution of the equation in b) is given below in parametric form, under the initial condition $$r=0 \text{ at } t=0$$ : \[r(\theta)=\frac{r_0}{2}(1-\cos\theta)\] \[t(\theta)=(\frac{r_0^3}{8GM})^{\frac{1}{2}}(\theta-\sin\theta)\] where $$\theta$$ is in radians.
d) From the above parametric equations, show that an expression for $$\frac{vt}{r}$$ is
$$\frac{vt}{r}=\frac{(\sin\theta)(\theta-\sin\theta)}{(1-\cos\theta)^2}$$ [1]
e) Now we consider $$m_1$$ and $$m_2$$ as the MW and M31, respectively, such that the current values of $$v$$ and $$r$$ are $$v=–118\text{km s}^{-1}$$ and $$r=710 \text{kpc}$$,and $$t$$ may be taken to be the age of the Universe (13700 million years ). Find $$\theta$$ using numerical iteration. [10]
f) Use the value of $$\theta$$ from e) to calculate the maximum distance between M31 and the MW,$$r$$ , and hence also obtain the value of $$M$$ in solar masses. [10]
中文题目
本星系群的质量 [50 分]
M31（仙女座星系）以及 银河系 之间的动力学过程可以用来估算整个本星系群的质量。可以做个这样的基本假设：现存星系对里的两个星系在宇宙大爆炸后没多久，大致都在相同的位置上。为了合理估算整个本星系群的质量，这里还得假设本星系群总质量绝大部分都是由 M31 和 银河系 贡献出来的。 通过谱线测量出的多普勒红移，可看出 M31 现在正以 118 km·s−1 朝 银河系 运动。也许你会比较惊讶，毕竟通常而言星系会因为哈勃流导致相互远离。但实际上 M31 正朝向 银河系 运动也不是不可能，因为（星系群里）引力的相互作用的效果（本动速度）要高于哈勃流（退行速度）带来的效果。 把星系对中的俩星系视为俩质点，它们总质量就能通过两者间距离、相对运动速度以及宇宙年龄来得出。 Kahn & Woltjer (1959) 就是这样计算本星系群质量的。
请顺着该思路计算本星系群的质量。
