imported>Patrick at 17:16, 15 August 2011

2011-08-15T17:16:27Z

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Since the radius of the orbit is <math>r = 1.5 R_E</math>, we have to use the general formula for the change in gravitational potential energy of the system satellite-Earth. The initial distance between the satellite and the Earth is equal to the radius of the Earth, or <math>r_i = R_E = 6.37 \times 10^6 m</math>. When the satellite orbits at the altitude <math>7 \times 10^5 m</math>, the final distance between the Earth and the satellite becomes <math>r_f = (7 \times 10^5) + (6.37 \times 10^6) = 7.07 \times 10^6 m</math>. Now we can compute the change in gravitational potential energy:

<math>\Delta U = {-G m_1 m_2 \over r_f} - \left ({-G m_1 m_2 \over r_i}\right ) = G m_1 m_2 \left (\frac{1}{r_i} - \frac{1}{r_f}\right )</math>
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Since <math>m_1 = M_E</math> and <math>m_2 = m_S</math>, we substitute for <math>r_i</math> and <math>r_f</math> and write

<math>\Delta U = G M_E m_S \left (\frac{1}{6.37 \times 10^6} - \frac{1}{7.07 \times 10^6}\right )</math>

<math>\Delta U = (6.67 \times 10^{-11})(6 \times 10^{24})(1000) \left (\frac{1}{6.37 \times 10^6} - \frac{1}{7.07 \times 10^6}\right )</math>

<math>\Delta U = 6.23 \times 10^9 J</math>

Potential Energy EX 9 - Revision history

imported>Patrick at 17:16, 15 August 2011