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Derive an expression for energy of satellite.

Answer» <html><body><p></p>Solution :The total energy of the satellite is the sum of its kinetic energy and the gravitational potential energy. The potential energy of the satellite is, <br/> ` U = - (GM_s M_E)/((R_E + h))`...(1) <br/>Here `M_s`-mass of the satellite, `M_E` -mass of the <a href="https://interviewquestions.tuteehub.com/tag/earth-13129" style="font-weight:bold;" target="_blank" title="Click to know more about EARTH">EARTH</a>, `R_E`-radius of the Earth. The Kinetic energy of the satellite is <br/>`K.E = 1/2 M_(s) v^2`.....(2) <br/> Here v is the orbital speed of the satellite and is equal to <br/>` v = sqrt( (GM_E)/((R_E + h)) ` <br/>Substituting the value of v in (2) the kinetic energy of the satellite <a href="https://interviewquestions.tuteehub.com/tag/becomes-1994370" style="font-weight:bold;" target="_blank" title="Click to know more about BECOMES">BECOMES</a>, <br/>`K.E = 1/2 ( GM_E M_s)/((R_E + h))` <br/> Therefore the total energy of the satellite is <br/>`E = 1/2 (GM_E M_s)/((R_E + h)) - (GM_s M_E)/((R_E + h))` <br/> ` E = (GM_s M_E)/((R_E + h)) ` <br/>The total energy <a href="https://interviewquestions.tuteehub.com/tag/implies-1037962" style="font-weight:bold;" target="_blank" title="Click to know more about IMPLIES">IMPLIES</a> that the satellite is bound to the Earth by means of the attractive gravitational force. Note: As h approaches `<a href="https://interviewquestions.tuteehub.com/tag/oo-583684" style="font-weight:bold;" target="_blank" title="Click to know more about OO">OO</a>` , the total energy <a href="https://interviewquestions.tuteehub.com/tag/tends-7717049" style="font-weight:bold;" target="_blank" title="Click to know more about TENDS">TENDS</a> to zero. Its physical meaning is that the satellite is completely free from the influence of Earth.s gravity and is not bound to Earth at large distance.</body></html>


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