Gravitational Force

Discussion in 'Calculus' started by nycmathguy, May 22, 2022.

  1. nycmathguy

    nycmathguy

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    Calculus
    Section 2.5

    Screenshot_20220516-084301_Samsung Notes.jpg
     
    nycmathguy, May 22, 2022
    #1
  2. nycmathguy

    MathLover1

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    graph:

    upload_2022-5-21_21-9-40.jpeg


    The function is certainly continuous on r element [0, R) and r element (R,∞).
    The only question is whether it is continuous at r = R.
    Checking the limit from each side, we find that

    [​IMG]
    R^-)( piecewise | (G M r)/R^3 | r<=R (G M)/r^2 | r>=R) = (G M)/R^2" style="-x-ignore: 1" data-paste-id="1">
    [​IMG]

    R^+)( piecewise | (G M r)/R^3 | r<=R (G M)/r^2 | r>=R) = (G M)/R^2" style="-x-ignore: 1" data-paste-id="2">
    Both limits are equal, and the value of F(r) is equal to the limit because of the r ≥ R
    clause in the function definition. Therefore the function is continuous for any r > 0.

     

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    MathLover1, May 22, 2022
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  3. nycmathguy

    nycmathguy

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    Thanks. Did you try uploading several images? If so, only one image is in display.
     
    nycmathguy, May 22, 2022
    #3
  4. nycmathguy

    MathLover1

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    [​IMG]

    [​IMG]
    did you see these imgs
     
    MathLover1, May 22, 2022
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  5. nycmathguy

    nycmathguy

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    Now I can see the images. Thank you.

    Note:

    It is really humid in NYC. This morning the humidity was 80%. I don't think it has gone down. I do not have an air conditioner in my room. My fan is blowing hot air.

    I took two showers today but to no avail. So, in light of this uncomfortable weather condition, I will not do the Intermediate Value Theorem tonight. I will leave it for another day.
     
    nycmathguy, May 22, 2022
    #5
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