imported>Patrick: Created page with ''''(a)''' The initial energy is $E = \frac{1}{2} U = -\frac{1}{2} {G m_E m_s \over r} = -\frac{1}{2} {(6.67 \times 10^{-11})(6 \times 10^{24})(50) \over 2 \times 6.4 \time…'$

2011-09-20T19:44:37Z

Created page with ''''(a)''' The initial energy is <math>E = \frac{1}{2} U = -\frac{1}{2} {G m_E m_s \over r} = -\frac{1}{2} {(6.67 \times 10^{-11})(6 \times 10^{24})(50) \over 2 \times 6.4 \time…'

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'''(a)''' The initial energy is

<math>E = \frac{1}{2} U = -\frac{1}{2} {G m_E m_s \over r} = -\frac{1}{2} {(6.67 \times 10^{-11})(6 \times 10^{24})(50) \over 2 \times 6.4 \times 10^6} = -7.8 \times 10^8 J</math>
 
'''(b)''' First we have to determine the energy of the satellite in the new orbit. It is 20% less than its initial energy. Here we have to think carefully because the energy is negative. What does it mean to loose 20%? Imagine that you owe $100 and you lose some additional money. Your debt will increase! If you lose 20% of your money you may say that your debt increases by 20% to $120 or by a factor of 1.2. Similarly, the energy in the new orbit will be <math>E = 1.2 \times (-7.8 \times 10^8) = -9.4 \times 10^8 J</math>. We will use the equation for the energy and isolate <math>r</math> from this equation:

<math>E = -\frac{1}{2} {G m_E m_s \over r}</math>

<math>r = -\frac{1}{2} {G m_E m_s \over E}</math>

<math>r = -\frac{1}{2} {(6.67 \times 10^{-11})(6 \times 10^{24})(50) \over -9.4 \times 10^8} = 1.1 \times 10^7 m</math>

Note that the satellite descended from the initial distance of <math>r = 2 R_E = 1.28 \times 10^7 m</math>. 

'''(c)''' The speed of the satellite depends on the distance. From the Newton's law

<math>{G m_E \over r} = v^2</math>
 
we expect that as r decreases the speed should increase. We can use this relationship to find the speed in the new orbit <math>v_{new}</math> or we can use <math>K = -E</math>. Remember that for a circular orbit

<math>K = \frac{1}{2} {G m_E m_s \over r} = \left\vert E \right\vert</math>
 
The second relationship is simpler to use: <math>K = 9.4 \times 10^8 J</math>, which gives a speed of <math>\frac{1}{2} m {v_{new}}^2 = 9.4 \times 10^8 = \frac{1}{2} (50) {v_{new}}^2</math>. This equation yields <math>v_{new} = 6.13 km/s</math>.

The velocity in the initial orbit <math>v_{old}</math> can be determined similarly using <math>\frac{1}{2} m {v_{old}}^2 = 7.8 \times 10^8</math>. This equation yields <math>v_{old} = 5.56 km/s</math>. The percentage change in speed is approximately 10%. '''The speed increased by 10%.'''

Satellites and Binding Energy EX 4 - Revision history

imported>Patrick: Created page with ''''(a)''' The initial energy is E = \frac{1}{2} U = -\frac{1}{2} {G m_E m_s \over r} = -\frac{1}{2} {(6.67 \times 10^{-11})(6 \times 10^{24})(50) \over 2 \times 6.4 \time…'

imported>Patrick: Created page with ''''(a)''' The initial energy is $E = \frac{1}{2} U = -\frac{1}{2} {G m_E m_s \over r} = -\frac{1}{2} {(6.67 \times 10^{-11})(6 \times 10^{24})(50) \over 2 \times 6.4 \time…'$