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Energy involving satellite of the Earth

  1. May 8, 2015 #1
    1. The problem statement, all variables and given/known data
    A satellite of the Earth has a mass of 1550 kg. It orbits Earth with a mean radius of orbit of 7.00 x 10^6 m.

    a) What is the gravitational potential energy of the satellite with respect to Earth?

    b) What is the kinetic energy and the velocity of the satellite in Earth’s orbit?

    c) What is the binding energy of the satellite to Earth?

    d) What is the total mechanical energy of the satellite in its orbit?

    2. Relevant equations
    PEg = -Gm1m2/R
    Vescape = root(2)(G)(m1)/Rm
    E binding = 1/2<PEg> in orbit


    3. The attempt at a solution

    a)PEg = -(6.67 X 10^-11)(5.98 X 10^24)(1550)/(1.338 X 10^7)
    = -4.62 X 10^10J

    b) Vescape = root(2)(6.67 x 10^-11)(5.98 x 10^24)/(6.38 x 10^6)
    = 1.12 x 10^4 m/s
    KE = (1/2)(m)(V)^2
    KE = (1/2)(1550)(1.12 x 10^4)^2
    KE = 9.72 x 10^10J
    c) E binding = 1/2<PEg> in orbit
    (0.5)(4.62 X 10^10J) = 2.31 x 10^10J

    d) I'm not certain which equation I should use for this one
     
  2. jcsd
  3. May 8, 2015 #2

    ehild

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    The total mechanical energy consist of the kinetic energy + the potential energy.
    You calculated the kinetic energy form the escape velocity. That is wrong. The satellite is on orbit, so its speed is less.
     
  4. May 8, 2015 #3
    So for question d) I will just use equation Me = Ke + Pe
    and for b) Ke = 1/2<PEg> which is 2.31 x 10^10J or I might be completely off
     
  5. May 8, 2015 #4

    ehild

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    KE is not half of the PE. They do not have the same sign.
    I do not understand what radius did you use when calculating the PE.
     
  6. May 8, 2015 #5
    The radius I used to calculated was (6.38 x 10^6m) The earth's radius plus 7.00 x 10^6 m from the question. (7.00 x 10^6m)+(6.38 x 10^6) = 1.338 x 10^7 m
    Is this what you are referring to? Sorry for making mistake, I actually trying my best
     
  7. May 8, 2015 #6

    haruspex

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    The 7 106 is not the height above the Earth's surface, it's the radius of orbit, so is measured from the Earth's centre.
     
  8. May 8, 2015 #7
    Oh so to figure out how far it's distance from earth I had to subtract (6.38 x 10^6m) from 7.00 x 10^6
     
  9. May 8, 2015 #8

    ehild

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    The distance from the Earth surface is irrelevant. The satellite orbits Earth with a mean radius of orbit of 7.00 x 10^6 m. The force between the Earth and satellite and the potential energy depend on the distance from the Earth centre.
     
  10. May 13, 2015 #9
    So I fully realized my error, I should known it was the radius of orbit. I did re-calculation for all questions.
    a)PEg = -(6.67x10^-11)(5.98x10^24)(1550)/(7.00 x 10^6)
    = -8.83 x10^10J
    b) I didn't had to use Vescape formula since it was in orbit
    V= root (G)(Me)/R
    v= root (6.67x10^-11)(5.98 x10^24)/(7.00 x 10^6)
    V = 7.55 x 10^3 m/s
    Ke = (0.5)(1550)(7.55 x 10^3)^2
    Ke = 4.42 x 10^10J

    If this isn't right, I'm dumb :(
     
  11. May 13, 2015 #10

    haruspex

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    Read the question carefully:
     
  12. May 13, 2015 #11
    My small mistake, Thanks haruspex
     
  13. May 13, 2015 #12
    Can you give me a little hint :( My answers keep getting off
     
  14. May 13, 2015 #13

    SammyS

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    Did you recalculate the gravitational potential energy?

    That velocity you have for part (b) looks more like orbital velocity than escape velocity.
     
  15. May 13, 2015 #14
    Since it is in orbit, yes I calculated part (b) as orbital velocity.
     
  16. May 13, 2015 #15

    SammyS

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    Oh ! yes, of course, that's right.

    Have you corrected you answer to part (a) so the answer is relative to Earth?

    What was the satellite's gravitational potential energy when it sat on Earth's surface?
     
  17. May 13, 2015 #16
    Gravitational potential energy?
    PEg = mgh
    PEg = (1550kg)(7.00 x 10^6m)(g)
     
  18. May 13, 2015 #17

    SammyS

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    That's good for objects very near earth's surface, and is relative to being exactly at earth's surface, where h = 0.

    At earth's surface, how far is an object from the center of the earth?
     
  19. May 13, 2015 #18
    Yeah because gravity would not affect much when it's it orbit. Due to distance
    At Earth's surface, how far is an object from the centre of the earth?
    (7.00 x 10^6m) - (Earth's radius)
     
  20. May 13, 2015 #19

    SammyS

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    Let's review part (a) of your question.

    You have found the gravitational potential energy of the satellite in its orbital position, but that is not with respect to earth. (What is that with respect to?)

    What was the satellite's gravitational potential energy when it sat on Earth's surface? That's not in orbit.
     
  21. May 13, 2015 #20
    Would you mind if you can briefly explain how "relative" thing works? I need to put those into my head first :smile:
     
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