{
 "cells": [
  {
   "cell_type": "code",
   "execution_count": 6,
   "id": "0fe8c54c-bf90-43ae-b74c-dacc1629265f",
   "metadata": {},
   "outputs": [],
   "source": [
    "from math import pi, sqrt\n",
    "import pandas as pd"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "3470c0a9-cc70-4e62-a50f-8e959ce4a7db",
   "metadata": {
    "tags": []
   },
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   "source": [
    "c = 299792458 # speed of light in metres / second"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "9795d017",
   "metadata": {},
   "outputs": [],
   "source": [
    "# Newton's gravitational constant\n",
    "G=6.674e-11"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "44e5cc18-0530-449f-8871-d3e2ccfd9284",
   "metadata": {},
   "source": [
    "Wikipedia gives the following values for the size of the observable universe.\n",
    "https://en.wikipedia.org/wiki/Observable_universe"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "2947f6f2",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "<div>\n",
       "<style scoped>\n",
       "    .dataframe tbody tr th:only-of-type {\n",
       "        vertical-align: middle;\n",
       "    }\n",
       "\n",
       "    .dataframe tbody tr th {\n",
       "        vertical-align: top;\n",
       "    }\n",
       "\n",
       "    .dataframe thead th {\n",
       "        text-align: right;\n",
       "    }\n",
       "</style>\n",
       "<table border=\"1\" class=\"dataframe\">\n",
       "  <thead>\n",
       "    <tr style=\"text-align: right;\">\n",
       "      <th></th>\n",
       "      <th>Wikipedia</th>\n",
       "      <th>Reed</th>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>Name</th>\n",
       "      <th></th>\n",
       "      <th></th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <th>Mass</th>\n",
       "      <td>1.500000e+53</td>\n",
       "      <td>1.519868e+53</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>Radius</th>\n",
       "      <td>4.400000e+26</td>\n",
       "      <td>2.257255e+26</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>Volume</th>\n",
       "      <td>3.566000e+80</td>\n",
       "      <td>4.817600e+79</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table>\n",
       "</div>"
      ],
      "text/plain": [
       "           Wikipedia          Reed\n",
       "Name                              \n",
       "Mass    1.500000e+53  1.519868e+53\n",
       "Radius  4.400000e+26  2.257255e+26\n",
       "Volume  3.566000e+80  4.817600e+79"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "data = {\n",
    "    'Name':['Mass', 'Radius', 'Volume'],\n",
    "    'Wikipedia':[1.5e53, 4.4e26, 3.566e80],\n",
    "    'Reed':[4*pi*c**5/(3*G), 8*pi*c**3/3, 4*pi*(8*pi*c**3/3)**3/3]\n",
    "}\n",
    "df = pd.DataFrame(data)\n",
    "df.set_index('Name', inplace=True)\n",
    "df"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "d414e5e0",
   "metadata": {},
   "source": [
    "The volume of any sphere is $ V = \\frac{4}{3} \\pi R^3 $ so the radius R is therefore $ R  = \\sqrt[3] {\\frac{3V}{4 \\pi}} $"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "id": "68b653e7",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "3.5681790480452396e+80"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# Let's check if Wikipedia's radius is self consistent with its volume\n",
    "4*pi*df['Wikipedia']['Radius']**3/3"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f83985f5",
   "metadata": {},
   "source": [
    "Looks like it agrees to about 3 significant places."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "440980f0",
   "metadata": {},
   "source": [
    "Reed's formula:\n",
    "$$ c = \\sqrt{ \\frac{2 ( 4 \\pi c^5/3 )}{ (8 \\pi c^3/3)}} $$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e0d2c56e",
   "metadata": {},
   "source": [
    "Reed further claims that $ GM = 4 \\pi c^5 / 3 $ and $ R = 8 \\pi c^3/3 $ in spite of the dimensions being hopeless inconsistent yet the values do seem to approximate the wikipedia values in some cases."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "41ef11a3",
   "metadata": {},
   "source": [
    "This is based on the notion of calculation escape velocity as $ E = \\frac{1}{2}mv^2 + \\int_R^{ \\infty } \\frac{GMm}{r^2} dr = \\frac{1}{2}mv^2  + GMm \\int_R^{ \\infty } \\frac{1}{r^2} dr  = \\frac{1}{2}mv^2  - \\frac{GMm}{R} = 0 $\n",
    "\n",
    "From what I understand equating the energy expression above to zero is the same as saying that this is the amount of energy left over after the projectile reaches infinity.  That is it just escapes and no more. This leads to the formula for the escape velocity:\n",
    "\n",
    "$$ v = \\sqrt{ \\frac {2GM}{R} } $$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "edfbebb7",
   "metadata": {},
   "outputs": [],
   "source": []
  }
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