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Change in mechanical energy (gravitation)

  1. Aug 24, 2015 #1
    I got a little bit confused while studying gravitational potential energy, particularly the expression for long distances: (-GMm/r). An exercise asks me for the velocity, period and radial acceleration for a 1000kg satellite that I wish to put into orbit. That's okay. But after that, I'm questioned for the amount of work needed to put this satellite into orbit. The book says that the necessary work is given by the difference between the total mechanical energy when the satellite is in orbit and the total mechanical energy when the satellite is grounded. The book uses the equation: (E=-GMm/2r) for mechanical energy - where M is the mass of the Earth and r is the distance between the satellite and the center of the Earth.
    That confuses me. I know that, by the work-energy theorem that work equals the change in kinetic energy, and I know that the change in gravitational potential energy equals minus Work (delta U = -W). But calculating work by change in mechanical energy is new for me. Where does it come from?
     
  2. jcsd
  3. Aug 24, 2015 #2
    Try using the work energy theorem here.
     
  4. Aug 24, 2015 #3
    The total mechanical energy in the initial position (grounded):
    E1 = K1 + U1 ; K is kinetic energy and U is gravitational potential energy. Since it is grounded, K1 = 0
    While it is in orbit, the total mechanical energy is:
    E2 = K2 + U2
    The difference in energy has to come from somewhere. That is external work. It could be because of the exhaust gases, or Superman throws the satellite into orbit, or whatever. So
    E2 - E1 = W (1)
    The same thing can be done starting from the work - kinetic energy theorem:
    ΔK = K2 - K1 = Work done by gravity + Work by other forces (gases, Superman, whatever) (2)
    Work done by gravity = - ΔU = - (U2 - U1) (3)
    substitute (3) into (2) and rearrange the terms to get (1)
     
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