{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "#  Chapter 8 : Gravitation"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "## Example 8.1 , page : 185"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " The planet will take a longer time to traverse BAC than CPB\n"
     ]
    }
   ],
   "source": [
    "# Importing module\n",
    "\n",
    "import math\n",
    "\n",
    "# Variable declaration\n",
    "\n",
    "mp=1                     # For convenience,mass is assumed to be unity \n",
    "rp=1                     # For convenience,sun-planet distance at perihelton is assumed to be unity \n",
    "vp=1                     # For convenience,speed of the planet at perihelton is assumed to be unity \n",
    "ra=1                     # For convenience,sun-planet distance at aphelton is assumed to be unity \n",
    "va=1                     # For convenience,speed of the planet at aphelton is assumed to be unity \n",
    "Lp=mp*rp*vp              # Angular momentum at perihelton\n",
    "La=mp*ra*va              # Angular momentum at ahelton\n",
    "\n",
    "# Result\n",
    "\n",
    "# From angular momentum conservation, mp*rp*vp = mp*ra*va or vp/va = rp/ra\n",
    "# From Kepler’s second law, equal areas are swept in equal times\n",
    "print(\" The planet will take a longer time to traverse BAC than CPB\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 8.2 , page : 187 "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(a) The force acting = [0.0, 2.5849394142282115e-26, 0.0] ≈ 0\n",
      "(b) The force acting = 2 Gm²\n"
     ]
    }
   ],
   "source": [
    "# Importing module\n",
    "\n",
    "import math\n",
    "\n",
    "# Variable declaration\n",
    "\n",
    "G=6.67*pow(10,-11)       # Gravitational constant\n",
    "m=1                      # For convenience,mass is assumed to be unity \n",
    "x=30                     # The angle between GC and the positive x-axis is 30° and so is the angle between GB and the negative x-axis\n",
    "y=math.radians(x)        # The angle in radians\n",
    "a=math.cos(y)\n",
    "b=math.sin(y)\n",
    "v1=(0,1,0)\n",
    "v2=(-a,-b,0)\n",
    "v3=(a,-b,0)\n",
    "c=(2*G*pow(m,2))/1       # 2Gm²/1\n",
    "\n",
    "# Calculation\n",
    "\n",
    "#(a)\n",
    "F1=[y * c for y in v1]   # F(GA)\n",
    "F2=[y * c for y in v2]   # F(GB)\n",
    "F3=[y * c for y in v3]   # F(GC)\n",
    "# From the principle of superposition and the law of vector addition, the resultant gravitational force FR on (2m) is given by\n",
    "Fa=[sum(x) for x in zip(F1,F2,F3)]\n",
    "\n",
    "#(b)\n",
    "# By symmetry the x-component of the force cancels out and the y-component survives\n",
    "Fb=4-2                   # 4Gm² j - 2Gm² j\n",
    "\n",
    "# Result\n",
    "\n",
    "print(\"(a) The force acting =\",Fa,\"≈ 0\")\n",
    "print(\"(b) The force acting =\",Fb,\"Gm²\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 8.3 , page : 192 "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Potential energy of a system of four particles = -5.414213562373095 Gm²/l\n",
      "The gravitational potential at the centre of the square = -5.65685424949238 Gm²/l\n"
     ]
    }
   ],
   "source": [
    "# Importing module\n",
    "\n",
    "import math\n",
    "\n",
    "# Variable declaration\n",
    "\n",
    "G=6.67*pow(10,-11)       # Gravitational constant\n",
    "m=1                      # For convenience,mass is assumed to be unity \n",
    "l=1                      # For convenience,side of the square is assumed to be unity \n",
    "c=(G*pow(m,2))/l\n",
    "n=4                      # Number of particles\n",
    "\n",
    "# Calculation\n",
    "\n",
    "d=math.sqrt(2)\n",
    "# If the side of a square is l then the diagonal distance is  √2l\n",
    "# We have four mass pairs at distance l and two diagonal pairs at distance √2l \n",
    "# Since the Potential Energy of a system of four particles is -4Gm²/l) - 2Gm²/dl\n",
    "w=(-n-(2/d))  \n",
    "# If the side of a square is l then the diagonal distance from the centre to corner is \n",
    "# Since the Gravitational Potential at the centre of the square\n",
    "u=-n*(2/d)\n",
    "\n",
    "# Result\n",
    "\n",
    "print (\"Potential energy of a system of four particles =\",w,\"Gm²/l\")\n",
    "print(\"The gravitational potential at the centre of the square =\",u,\"Gm²/l\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 8.4 , page : 193 "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Minimum speed of the projectile to reach the surface of the second sphere = ( 0.6 GM/R ) ^(1/2)\n"
     ]
    }
   ],
   "source": [
    "# Importing module\n",
    "\n",
    "import math\n",
    "\n",
    "# Variable declaration\n",
    "\n",
    "R=1                      # For convenience,radii of both the spheres is assumed to be unity \n",
    "M=1                      # For convenience,mass is assumed to be unity \n",
    "m1=M                     # Mass of the first sphere\n",
    "m2=6*M                   # Mass of the second sphere\n",
    "m=1                      # Since the mass of the projectile is unknown,take it as unity\n",
    "d=6*R                    # Distance between the centres of both the spheres\n",
    "r=1                      # The distance from the centre of first sphere to the neutral point N\n",
    "\n",
    "G=6.67*pow(10,-11)       # Gravitational constant\n",
    "\n",
    "# Calculation\n",
    "\n",
    "# Since N is the neutral point; GMm/r² = 4GMm/(6R-r)²  and we get\n",
    "r=2*R\n",
    "# The mechanical energy at the surface of M is; Et = m(v^2)/2 - GMm/R - 4GMm/5R\n",
    "# The mechanical energy at N is; En = -GMm/2R - 4GMm/4R\n",
    "# From the principle of conservation of mechanical energy; Et = En and we get\n",
    "v_sqr=2*((4/5)-(1/2))\n",
    "\n",
    "# Result\n",
    "\n",
    "print(\"Minimum speed of the projectile to reach the surface of the second sphere =\",\"(\",round(v_sqr,5),\"GM/R\",\")\",\"^(1/2)\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 8.5 , page : 195 "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "(i)  Mass of Mars = 6.475139697520706e+23 kg\n",
      "(ii) Period of revolution of Mars = 684.0033777694376 days\n"
     ]
    }
   ],
   "source": [
    "# Importing module\n",
    "\n",
    "import math\n",
    "\n",
    "# Variable declaration\n",
    "\n",
    "π=3.14                   # Constant pi\n",
    "G=6.67*pow(10,-11)       # Gravitational constant\n",
    "R=9.4*pow(10,3)          # Orbital radius of Mars in km\n",
    "T=459*60\n",
    "Te=365                   # Period of revolution of Earth\n",
    "r=1.52                   # Ratio of  Rms/Res, where Rms is the mars-sun distance and Res is the earth-sun distance. \n",
    "\n",
    "# Calculation\n",
    "\n",
    "# (i) \n",
    "R=R*pow(10,3)\n",
    "# Using Kepler's 3rd law:T²=4π²(R^3)/GMm\n",
    "Mm=(4*pow(π,2)*pow(R,3))/(G*pow(T,2))\n",
    "\n",
    "# (ii)\n",
    "# Using Kepler's 3rd law: Tm²/Te² = (Rms^3/Res^3)\n",
    "Tm=pow(r,(3/2))*365\n",
    "\n",
    "\n",
    "# Result\n",
    "\n",
    "print(\"(i)  Mass of Mars =\",Mm,\"kg\")\n",
    "print(\"(ii) Period of revolution of Mars =\",Tm,\"days\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 8.6 , page : 195 "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Mass of the Earth = 5.967906881559221e+24 kg\n",
      "Mass of the Earth = 6.017752855396305e+24 kg\n"
     ]
    }
   ],
   "source": [
    "# Importing module\n",
    "\n",
    "import math\n",
    "\n",
    "# Variable declaration\n",
    "\n",
    "g=9.81                   # Acceleration due to gravity\n",
    "G=6.67*pow(10,-11)       # Gravitational constant\n",
    "Re=6.37*pow(10,6)        # Radius of Earth in m\n",
    "R=3.84*pow(10,8)         # Distance of Moon from Earth in m\n",
    "T=27.3                   # Period of revolution of Moon in days\n",
    "π=3.14                   # Constant pi\n",
    "\n",
    "# Calculation\n",
    "\n",
    "# I Method\n",
    "# Using Newton's 2nd law of motion:g = F/m = GMe/Re²\n",
    "Me1=(g*pow(Re,2))/G\n",
    "\n",
    "# II Method\n",
    "# Using Kepler's 3rd law: T²= 4π²(R^3)/GMe\n",
    "T1=T*24*60*60\n",
    "Me2=(4*pow(π,2)*pow(R,3))/(G*pow(T1,2))\n",
    "\n",
    "#Result\n",
    "\n",
    "print(\"Mass of the Earth =\",Me1,\"kg\")\n",
    "print(\"Mass of the Earth =\",Me2,\"kg\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 8.7 , page : 195 "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Period of revolution of Moon = 27.5 days\n"
     ]
    }
   ],
   "source": [
    "# Importing module\n",
    "\n",
    "import math\n",
    "\n",
    "# Variable declaration\n",
    "\n",
    "k=pow(10,-13)           # A constant = 4π² / GME\n",
    "Re=3.84*pow(10,5)       # Distance of the Moon from the Earth in m\n",
    "\n",
    "# Calculation\n",
    "\n",
    "k=pow(10,-13)*(pow(1/(24*60*60),2))*(1/pow((1/1000),3))\n",
    "T2=k*pow(Re,3)\n",
    "T=math.sqrt(T2)         # Period of revolution of Moon in days\n",
    "\n",
    "# Result\n",
    "\n",
    "print(\"Period of revolution of Moon =\",round(T,1),\"days\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 8.8 , page : 196  "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Change in Kinetic Energy = 3124485000.0 J\n",
      "Change in Potential Energy = 6248970000.0 J\n"
     ]
    }
   ],
   "source": [
    "# Importing module\n",
    "\n",
    "import math\n",
    "\n",
    "# Variable declaration\n",
    "\n",
    "m=400                   # Mass of satellite in kg\n",
    "Re=6.37*pow(10,6)       # Radius of Earth in m\n",
    "g=9.81                  # Acceleration due to gravity\n",
    "\n",
    "# Calculation\n",
    "\n",
    "# Change in energy is E=Ef-Ei\n",
    "ΔE=(g*m*Re)/8           # Change in Total energy\n",
    "# Since Potential Energy is twice as the change in Total Energy (V = Vf - Vi)\n",
    "ΔV=2*ΔE                 # Change in Potential Energy in J\n",
    "\n",
    "# Result\n",
    "\n",
    "print(\"Change in Kinetic Energy =\",round(ΔE,4),\"J\")\n",
    "print(\"Change in Potential Energy =\",round(ΔV,4),\"J\")"
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.4.3"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 0
}