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Work done on a satellite ( Physics Problem )

  1. Jul 18, 2011 #1
    1. The problem statement, all variables and given/known data
    What is the work done to place a satellite of mass 600kg which orbits the Earth at a radius of 8.00 x 10^6 m.

    3. The attempt at a solution
    1st attempt :
    W= F x d
    = GmM/(r^2) x r
    = GmM/r where r is the radius of the satellite
    The answer I got is 3 x 10^10 J which is incorrect.

    2nd attempt :
    W= F x ∆d
    = GmM/(r^2) x ( r - r.E ) where r.E is the radius of the Earth

    The answer I got is 6.2 x 10^9 J which is also incorrect.

    3rd attempt :
    W= ∆E total
    = -GmM/2r - ( -GmM/2r.E )
    = -GmM/2 x ( 1/r.E - 1/r )

    The answer I got is 3.8 x 10^9 J which is also incorrect.

    The answer for the question is 2.3 x 10^10 J.
    From my three attempt, I think I haven't understand the concept of work and energy of a satellite. Can someone help me to identify my mistakes and rectify it?
    I appreciate your kind assistance^^

    Thank you for spending your time to read this post and help me to solve the problem ^^

    Thank you once again ;):smile:
     
  2. jcsd
  3. Jul 18, 2011 #2

    Delphi51

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    Homework Helper

    The first two calcs are incorrect because they assume the force of gravity is constant as the satellite is lifted through the change in radius, and they fail to take into account the kinetic energy needed to increase the speed.

    The third one uses an energy formula that is the sum of kinetic and potential energy for circular orbits. It is useful to see its derivation here:
    http://www.pha.jhu.edu/~broholm/l24/node1.html [Broken]
    This method also has a failure. It is for a change in orbit, but the satellite is not in orbit when sitting at the surface of the earth. In particular, it is going far too slow to be in orbit at radius r.E, so the kinetic energy part of the formula is way off.

    I suggest you use E = ½ m⋅v² - GMm/r.
    Be aware - the text book answer may be based on an incorrect assumption that the kinetic energy is zero at the surface of the Earth. That may not make a lot of difference, but keep it in mind.
     
    Last edited by a moderator: May 5, 2017
  4. Jul 18, 2011 #3
    Try using conservation of energy:
    Ui + Ki + W = Uf + Kf

    Solve for Work, and find out the amount of kinetic energy the object has while it is in orbit. You can use the fact that while a satellite is in orbit it's position is constantly turning with the Earth, so you can use Newton's law to find its tangential speed as it is in constant speed with the Earth. With that speed you can find the kinetic energy of the object.
     
  5. Jul 18, 2011 #4
    Delphi51 :
    Thanks for your concern Delphi51 :)
    From your suggestion,can I substitute 1/2mv^2 with GmM/2r ?
    and by substituting the equation, I got an answer of 1.5 x 10^10 J ,still faraway from the given answer 2.3 x10^10 J. I really wonder why the value is 2.3 ? How to get to the answer.

    And one more thing to ask is that why I cant assume that the kinetic energy of the satellite on Earth's surface to be Zero since it is not moving? Is it because that the Earth's itself also move?

    Thank you once again ;)
     
  6. Jul 18, 2011 #5
    thepatient :

    Thanks for your concern thepatient ;)

    Erm,, regarding your comment I would like to ask, do you mean that I will have to assume that the satellite is geostationary?Because the speed or the angular velocity of the satellite isn't stated in the question. And,, if I would like to use your equation,, how does the working look like? ;)

    Thank you thepatient for helping me :) I appreciate your kind assistance ^^
     
  7. Jul 18, 2011 #6
    Fnet = ma

    The net force on the object when it is in orbit is equal to it's mass times its acceleration. We want the satellite to have a uniform circular motion though, so it's net force will be just the inward gravitational force, and a = v^2/R. Use that speed to find kinetic energy of the object.

    The equation that I posted is the same conservation of energy equation, but with a few terms rearranged. It states that the energy of the system before is equal to the energy in the system after the work is done. Ui stands for initial gravitational potential energy, Ki for initial kinetic energy, W for work, Uf and Kf for final potential and kinetic energies respectively.
     
  8. Jul 18, 2011 #7
    You can find velocity and angular vel. by using radius of orbit, mass of earth.

    for this, you must use the eqn, Fcentripital = FGravitational
     
  9. Jul 18, 2011 #8

    Delphi51

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    You could use your -GmM/2r formula for the energy in the orbit to save finding the velocity there.
    Use ½ m⋅v² - GMm/r for the energy at the surface of the Earth. You may be able to assume v = 0, though in fact the Earth is rotating and has a velocity at the surface. You can find it by using the circumference of the earth divided by the time for one complete turn.
     
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