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Homework Help: Heat energy

  1. Jun 7, 2009 #1
    how do you do heat energy questions? is it the energy lost?

    An unpowered 1600 kg object has an upward velocity of 7.0 x 10^3 m s at an
    altitude of 100 km above the earth. The object reaches a maximum altitude of 400 km.
    100 km
    400 km
    v = 7.0 × 10^3 m s

    What is the heat energy generated during the object’s increase in altitude from 100 km
    to 400 km?
    A. 3.3 x 10^10 J
    B. 3.4 x 10^10 J
    C. 3.5 x 10^10 J
    D. 5.5 x 10^10 J

    this is from a sample exam, my teacher didnt teach us any of it.

    I used EP= Gm1m2/(r^2) for each position, and ek = .5 mv^2 and subtracted the two and got a totally wrong answer.... Please help me.
  2. jcsd
  3. Jun 15, 2009 #2

    heres a pic....

    also would solving this be similar to how you would find the heat lost from a resistor in a circuit because I had a question like that and I couldnt solve it.
  4. Jun 16, 2009 #3
    I read that the change in potential is = work/energy and since velocity is constant EK is constant.
    but when I use the formula for ep I dont get the right answer.
  5. Jun 16, 2009 #4


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    Well for one,

    [tex]E_P= \frac{GM}{r}[/tex]

    at the max height, kinetic energy=0.
  6. Jun 16, 2009 #5
    so at the start is velocity the 7E3 or is that the average velocity

    Ive tried finding Etotal at the start and at the end and subtract one from the other but I always get the wrong answer.
  7. Jun 17, 2009 #6

    Andrew Mason

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    Write out the equation for the total energy of the object as a function of R if there was no loss of energy. Determine what its KE would be at 400 km altitude. Compare that to the actual energy at 400 km (KE = 0). Where did that energy go?

  8. Jun 17, 2009 #7

    Etotal=1/2mv^2 + (Gm1m2)/r


    EK=4.82E12 ?

    whats wrong?
  9. Jun 17, 2009 #8


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    You need to use the radius from the center of Earth
  10. Jun 17, 2009 #9
    yeah I forgot about the earth radius

    I keep getting 4.36E10 though.
  11. Jun 17, 2009 #10


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    I approach it as what should the velocity be.

    1/2*Vi2 - μ/ri = 1/2*Vf2 - μ/rf

    Using the Standard parameter for earth of 400,000 (which is in km3)

    1/2*(49) - 400,000/6370 = 1/2*Vf2 - 400,000/ 6770

    That yields 1/2*Vf2 = 24.5 - 62.79 + 59.084 = 20.79

    Since that's already 1/2*Vf2 then multiply by the mass giving 33,270. But those units are in km2/sec2 so you need to properly convert to get J I'd think.
  12. Jun 17, 2009 #11

    Andrew Mason

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    You are using R to be 100 and 400 km. This is not correct. What is R? And where is it measured from?

  13. Jun 17, 2009 #12
    Ya from the center of the earth, somebody else mentioned it but I still get the wrong answer, its getting closer though....but still wrong.
  14. Jun 18, 2009 #13

    Andrew Mason

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    Show us your calculations and we will tell you where you are going wrong. We can't figure it out any other way.

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