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+{
+"cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 14: Gravitation"
+ ]
+ },
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 14.1: Sample_Problem_1.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"exec ('Gravitation.sci',-1)\n",
+"\n",
+"//Given that\n",
+"m1 = 6 //kg\n",
+"m2 = 4 //kg\n",
+"m3 = 4 //kg\n",
+"a = 2 * (10^-2)\n",
+"\n",
+"//Sample Problem 14-1\n",
+"printf('**Sample Problem 14-1**\n')\n",
+"//F1 = F12 + F13\n",
+"F12 = [0,-GForce(m1,m2,a)]\n",
+"F13 = [GForce(m1,m3,2*a),0]\n",
+"F1 = F12 + F13\n",
+"printf('The magnitude of net force is approximately equal to %e N', norm(F1))"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 14.2: Sample_Problem_2.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"exec('Gravitation.sci',-1)\n",
+"exec('degree_rad.sci', -1)\n",
+"\n",
+"//Given that\n",
+"//masses in kg\n",
+"m1 = 8\n",
+"m2 = 2\n",
+"m3 = 2\n",
+"m4 = 2\n",
+"m5 = 2\n",
+"a = 2*(10^-2); //in meter\n",
+"Theta = dtor(30) //in radians\n",
+"\n",
+"//Sample Problem 14-2\n",
+"printf('**Sample Problem 14-2**\n')\n",
+"//The net force will be equal to the vector eum of all the forces acting on the particle due to the rest of the particles i.e F1 = F12 + F13 + F14 + F15\n",
+"F12 = [GForce(m1,m2,(2*a))*sin(Theta), GForce(m1,m2,(2*a))*cos(Theta)]\n",
+"F13 = [GForce(m1,m3,a)*sin(Theta), -GForce(m1,m3,a)*cos(Theta)]\n",
+"F14 = [-GForce(m1,m4,(2*a))*sin(Theta), -GForce(m1,m4,(2*a))*cos(Theta)]\n",
+"F15 = [-GForce(m1,m5,a)*sin(Theta),-GForce(m1,m5,a)*cos(Theta)]\n",
+"F1 = F12 + F13 + F14 + F15\n",
+"printf('The net force on particle 1 is approimately equal to %e N', norm(F1))"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 14.3_a: Sample_Problem_3a.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"exec ('Gravitation.sci',-1)\n",
+"\n",
+"//Given that\n",
+"r = 6.77 * 10^6 //in meter\n",
+"dr = 1.7 //in meter\n",
+"\n",
+"//Sample Problem 3a\n",
+"printf('**Sample Problem 3a**\n')\n",
+"dg = -2 * G * Me * dr /(r^3)\n",
+"printf('The difference in acceleration is approximately equal to %e m/sec*sec', dg)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 14.3_b: Sample_Problem_3b.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"exec ('Gravitation.sci',-1)\n",
+"\n",
+"//variavles with their values\n",
+"Mh = 1.99 * 10^31 //in kg\n",
+"R = 6.77 * 10^6 //in meter\n",
+"DR = 1.7 //in meter\n",
+"\n",
+"//Sample Problem 3b\n",
+"printf('**Sample Problem 3b**\n')\n",
+"//the difference in gravitational acceleration is given by\n",
+"DG = -2 * G * Mh * DR /(R^3)\n",
+"printf('The difference in acceleration is approximately equal to %em/s^2', DG)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 14.5: Sample_Problem_5.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"exec ('Gravitation.sci',-1)\n",
+"//Given that\n",
+"Vi = 1.2 * 10^4 //in m/sec\n",
+"d = 10*Re;\n",
+"m = 10 //let say it will mass cancel out later\n",
+"//Sample Problem 5\n",
+"printf('**Sample Problem 5**\n')\n",
+"//we know that E(initial) = E(final)\n",
+"//=> Ki + Ui = Kf + Uf\n",
+"//K = .5*m*Vi*Vi (Kinetic Energy)\n",
+"//U = gravitational potential (Potential Energy)\n",
+"Ki = .5*m*Vi*Vi; \n",
+"Ui = GPotential(m,Me,d);\n",
+"Uf = GPotential(m,Me,Re);\n",
+"Kf = Ki + Ui -Uf;\n",
+"Vf = sqrt(2*Kf/m);\n",
+"printf('The final velocity of the asteroid is equal to %e m/sec', Vf)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 14.6: Sample_Problem_6.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"exec ('Gravitation.sci',-1)\n",
+"\n",
+"//Given that\n",
+"T = 76 * 365 * 24 * 60 * 60 //time period in seconds (converting from years)\n",
+"\n",
+"//Sample Problem 6a\n",
+"printf('**Sample Problem 6a**\n')\n",
+"//We know that Ra + Re = 2*a\n",
+"Rp = 8.9 * 10^10 //in meter\n",
+"a = KeplerRadius(Ms,T)\n",
+"//therefore\n",
+"Ra = 2*a -Rp //in meter\n",
+"printf('The Aphelion distance is equal to %em\n', Ra)\n",
+"\n",
+"//Sample Problem 6b\n",
+"printf('\n**Sample Problem 6b**\n')\n",
+"//we know that e*a = a - Rp\n",
+"e = 1 - Rp/a\n",
+"printf('The eccentricity of the path is %e ', e)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 14.7: Sample_Problem_7.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"exec ('Gravitation.sci',-1)\n",
+"\n",
+"//Given that\n",
+"//Both the stars are moving around the centre of mass of the two particale system\n",
+"//m1 = mass of visible star\n",
+"//m2 = mass of invisible star\n",
+"//r1 = distance of m1 from center of mass\n",
+"//r2 = distance of m2 from center of mass\n",
+"//r = r1+r2 distance between both the stars\n",
+"//we have G*m1*m2/(r*r) = m1*v1*v1/r1 = m2*v2*v2/r2 ....1\n",
+"v1 = 270*10^3 //in meter/sec\n",
+"T = 1.7 * 24 * 60 * 60 //in s\n",
+"m1 = 6* Ms \n",
+"\n",
+"//Sample Problem 7\n",
+"printf('**Sample Problem 7**\n')\n",
+"//m2 = ?\n",
+"//using definition of center of mass\n",
+"// we have r = r1 * (m1 + m2)/m2 ....2\n",
+"//& 2*pi*r1/v1 = T ....3\n",
+"//therefore\n",
+"r1 = v1*T/(2*%pi); //from equation 3\n",
+"//from equation 1 & 2\n",
+"//G*(m2^3)/((r1*(m1+m2))^2) = v1*v1/r1\n",
+"//we have a polynomial equation in order 3 \n",
+"//(m2^3)/(m1+m2)^2 = v1*v1*r1/G\n",
+"temp = v1*v1*r1/G; //say\n",
+"//=> -m2^3 + temp*m2^2 + 2*m1*temp*m2+ m1*m1*temp\n",
+"solpoly = (poly([-m1*m1,-2*m1,-1,1/temp],'x','c'));\n",
+"sol = roots(solpoly,'e');\n",
+"printf('The mass of the invisible star is equal to %e kg\n', sol(1))\n",
+"printf('The mass of the invisible star is equal to %f times the mass of Sun', sol(1)/Ms)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 14.8: Sample_Problem_8.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"exec ('Gravitation.sci',-1)\n",
+"\n",
+"//Given that\n",
+"m = 7.20 //in kg\n",
+"h = 350 * 10^3 //altitude in meter\n",
+"\n",
+"//Sample Problem 8a\n",
+"printf('**Sample Problem 8a**\n')\n",
+"//mechanical energy E = K + U\n",
+"//E = - G * M * m /(2* r)\n",
+"E = .5*GPotential(m,Me,(h+Re))\n",
+"printf('The total energy at the given altitude is %e joule\n',E)\n",
+"\n",
+"\n",
+"//Sample Problem 8b\n",
+"printf('\n**Sample Problem 8b**\n')\n",
+"//here the k = 0\n",
+"E0 = GPotential(m,Me,Re)\n",
+"printf('The total energy on the launchpad is %e joule\n',E0)\n",
+"deltaE = E - E0;\n",
+"printf('The differencein both the energy %e joule',deltaE)"
+ ]
+ }
+],
+"metadata": {
+ "kernelspec": {
+ "display_name": "Scilab",
+ "language": "scilab",
+ "name": "scilab"
+ },
+ "language_info": {
+ "file_extension": ".sce",
+ "help_links": [
+ {
+ "text": "MetaKernel Magics",
+ "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+ }
+ ],
+ "mimetype": "text/x-octave",
+ "name": "scilab",
+ "version": "0.7.1"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
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