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   "source": [
    "作业3\n",
    "计算球状星团（$10^5 M_{\\odot}$），椭圆星系（$10^{11} M_{\\odot}$）,星系团（$10^{14} M_{\\odot}$)中的速度弥散度，穿越时标和弛豫时标"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### SSY 星系团的粒子数是星系数目"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "解：\n",
    "速度弥散：\n",
    "\n",
    "   $\\frac{GM}{R} \\sim \\sigma^2$ \n",
    "\n",
    "   $G=4.301 \\times 10^{-9} km^2 Mpc M_{\\odot} s^{-2}$\n",
    "\n",
    "   球状星团：$M=10^5M_{\\odot}$, $R=10pc$\n",
    "   \n",
    "   $\\sigma^2 = \\frac{4.301 \\times 10^{-9} \\times 10^5 km^2 Mpc s^{-2}}{10 pc} = 43.01 km^2 s^{-2}$\n",
    "   \n",
    "   $\\sigma = \\sqrt{43.01 km^2 s^{-2}} = 6.6 km s^{-1}$ \n",
    "\n",
    "   椭圆星系：$M=10^{11}M_{\\odot}$, $R=10kpc$\n",
    "\n",
    "   $\\sigma = 2 \\times 10^2 km s^{-1}$\n",
    "\n",
    "   星系团：$M=10^{14}M_{\\odot}$, $R=1Mpc$\n",
    "\n",
    "   $\\sigma = 0.6 \\times 10^3 km s^{-1}$\n",
    "\n",
    "穿越时标：\n",
    "\n",
    "   球状星团：$t_{cross} = \\frac{R}{\\sigma} = \\frac{10 \\times 3.0857 \\times 10^{13} km}{6.6 km s^{-1}} = 4.7 \\times 10^{13}s = 0.147 Myr$\n",
    "\n",
    "   椭圆星系：$t_{cross} = \\frac{R}{\\sigma} = \\frac{10^4 \\times 3.0857 \\times 10^{13} km}{200 km s^{-1}} = 1.5 \\times 10^{15}s = 47 Myr$\n",
    "\n",
    "   星系团：$t_{cross} = \\frac{R}{\\sigma} = \\frac{10^6 \\times 3.0857 \\times 10^{13} km}{1000 km s^{-1}} = 3.1 \\times 10^{16}s = 1 Gyr$\n",
    "\n",
    "弛豫时标：\n",
    "\n",
    "   $\\frac{t_{relax}}{t_{cross}} \\sim \\frac{N}{6ln(N/2)}$,\n",
    "   $t_{relax} = \\frac{N}{6ln(N/2)} \\times {t_{cross}}$\n",
    " \n",
    "   球状星团：$N \\sim 10^{6} stars$, $t_{relax} = 0.147 Myr \\times \\frac{10^{6}}{6ln(0.5 \\times 10^{6})} = 1867 Myr = 1.9 Gyr$\n",
    "\n",
    "   椭圆星系：$N \\sim 10^{11} stars$, $t_{relax} = 47 Myr \\times \\frac{10^{11}}{6ln(0.5 \\times 10^{11})} = 10^7 Gyr$\n",
    "\n",
    "   星系团：$N \\sim 10^{14} stars$, $t_{relax} = 1 Gyr \\times \\frac{10^{14}}{6ln(0.5 \\times 10^{14})} = 10^9 Gyr$\n",
    "\n",
    "\n",
    "\n",
    "    \n",
    "   "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "作业4\n",
    "估算银河系的角动量自旋参数$\\lambda$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "解：$\\lambda = J \\cdot |E|^{-\\frac{1}{2}} \\cdot G^{-1} \\cdot M^{-\\frac{5}{2}}$\n",
    "\n",
    "   $|E|^{\\frac{1}{2}} = \\frac{M^{\\frac{1}{2}} \\cdot V_c}{2^{\\frac{1}{2}}}$\n",
    "\n",
    "   $\\lambda = J \\cdot G^{-1} \\cdot M^{-\\frac{5}{2}} \\cdot \\frac{M^{\\frac{1}{2}} \\cdot V_c}{2^{\\frac{1}{2}}}$\n",
    "\n",
    "   $J = \\frac{J_d}{j_d}$\n",
    "\n",
    "   $m_d = \\frac{M_d}{M}$\n",
    "\n",
    "   $m_d \\sim j_d$\n",
    "\n",
    "   $V_c = 200 km/s$, $R_d = 3 kpc$\n",
    "\n",
    "   假设银河系的质量$M = 10^{12}M_{\\odot}$, $m_d = 0.2$\n",
    "\n",
    "   所以$j_d = 0.2$, $M_d = m_d \\cdot M = 0.2 \\times 10^{12}M_{\\odot}$\n",
    "\n",
    "   $J_d = M_d \\times V_c \\times R_d = 0.2 \\times 10^{12}M_{\\odot} \\times 200km/s \\times 3kpc = 3.72 \\times 10^{29}km^2s^{-1}M_{\\odot}$\n",
    "\n",
    "   $J = \\frac{J_d}{j_d} = \\frac{3.72 \\times 10^{29}km^2s^{-1}M_{\\odot}}{0.2} = 1.86 \\times 10^{30}km^2s^{-1}M_{\\odot}$\n",
    "\n",
    "   $\\lambda = 1.86 \\times 10^{30}km^2s^{-1}M_{\\odot} \\times 2.3 \\times 10^8 km^{-2} Mpc^{-1} M_{\\odot}s^{2} \\times 10^{-24} M_{\\odot}^{-2} \\times 200 km s^{-1} \\times 2^{{-\\frac{1}{2}}} = 2 \\times 10^{-5}$\n",
    "\n",
    "\n"
   ]
  }
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