Calculating Work to Place Satellite in Synchronous Orbit

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To calculate the work needed to place a 650 kg satellite into synchronous orbit, the formula used is W=GMm(1/r of Earth - 1/r of synchronous satellite), resulting in an initial calculation of approximately 3.44 x 10^10 J. However, the correct value is 3.7 x 10^10 J, prompting a discussion about significant digits and energy types. The gravitational potential energy calculated is -4.1 x 10^10 J, while the total energy is -3.1 x 10^9 J, with kinetic energy being 3.1 x 10^9 J. The final work calculation incorporates these energy types, confirming the correct work needed is 3.75 x 10^10 J. This highlights the importance of considering both potential and kinetic energy in orbital mechanics.
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here is another problem...
a 650kg satellite is to be placed into synchronus orbit around Earth.
What amount of work must be done onthe satellite to raise it into synchronous orbit?

i did..
W=GMm(1/r of Earth - 1/r of synchronous satellite)
= 6.673x10^-11N m^2/kg^2 (5.98x10^6kg)(650kg)(1/6.38x10^6m - 1/4.22x10^7m)
W = 3.442998737X10^10 J

the right answer is 3.7x10^10 J... is ther something wrong with the significant digits?? can anyone tell me the mistake that i did?
 
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six789,

What kind of energy did you calculate for the satellite? Does it have any other kind when it's in orbit?
 
a) the gravitational potential energy is -4.1x10^10 J
b) the total enrgy is -3.1x10^9 J

d) the kinetic enrgy is 3.1x10^9 J
 
The kinetic energy in this case is not the same as the potential energy. The only time kinetic energy = Potential energy is when the change in kinetic energy is caused by a change in potential energy. In this case, the potential energy is always cosntant as in a circular orbit, the satelite is always the same distance from the earth.
 
i get the right answer from doc al... he said that...
change of energy = Work-Energ(total)

change=work
therefore.. Work = Work-E
=3.442998737X10^10 J - (-3.1x10^9 J)
=3.75x10^10J..
so the answer is right!

thanks again for the help...
 
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