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Total Time for Newton's Law of Gravitation

  1. Dec 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the total time, t, that an object takes to reach the surface of the Earth from a distance, D, using the Law of Gravitation: $$F_{g} = \frac{GMm}{x^2}$$

    R is radius of Earth
    D is distance from surface
    R+D is total distance from center of masses

    ****** One Dimension ******

    2. Relevant equations
    F = ma
    $$\frac{GMm}{x^2} = ma, a = \frac{GM}{x^2}$$


    3. The attempt at a solution
    $$\frac{a}{v} = \frac{dv}{dx}$$

    $$a dx = v dv \frac{-GM}{x^{2}} dx = v dv$$

    $$-GM\int_{R}^{R+D} \frac{1}{x^2} dx = \int_{0}^{v} v dv$$

    $$v = \sqrt{2GM(\frac{1}{R+D} - \frac{1}{R})}$$

    $$\frac{dx}{dt} = \sqrt{2GM(\frac{1}{R+D} - \frac{1}{R})}$$

    $$dt = \frac{1}{\sqrt{2GM(\frac{1}{R+D} - \frac{1}{R})}} dx$$

    $$\int_{0}^{t} dt = \int_{R}^{R+D} \frac{1}{\sqrt{2GM(\frac{1}{R+D} - \frac{1}{R})}} dx$$


    $$t = \frac{D}{\sqrt{2GM(\frac{1}{R+D} - \frac{1}{R})}}$$






    Is this the correct solution to the problem?
     
  2. jcsd
  3. Dec 14, 2016 #2

    mfb

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    A simple cross-check: for D>0, you take the square root of a negative number. That cannot be right.

    An unrelated issue: Your speed does not depend on x (the current position) at all. Does that sound realistic?
     
  4. Dec 14, 2016 #3
    a = GM/r^2

    If you treat Earth at the x value of 0 then r is a function of x..?

    r(x) = x?




    I do understand D>0 causing a problem, but I don't know where I went wrong.
     
  5. Dec 14, 2016 #4

    mfb

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    R is the constant radius of Earth.

    What is r now? The same as x?
    Probably a sign error somewhere. Using conservation of energy, you can directly find v(x) and avoid a few steps.
     
  6. Dec 14, 2016 #5

    At time = 0, r = R + D.

    I was going to use conservation of energy, but hesitated due to the fact that you use "g", an approximation.



    In the first integral, would you use "R" as the top bound instead of "R + D"?
     
  7. Dec 14, 2016 #6

    vela

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    Why'd you throw the minus sign in? I'm not saying it's wrong, but above you wrote ##a = \frac{GM}{x^2}##.


    The bottom limit of the integral is where you're starting from, and the top limit is where you're ending at. That's the convention you followed on the velocity integral, but you did it backwards on the LHS.

    Note that if you say your end point is at ##x=R##, the ##v## you find is ##v(x=R)##, the speed when the object is at a distance ##R##. Is that what you want to find?
     
  8. Dec 14, 2016 #7
    I added the negative just because the drawing that I drew had the object moving left.

    I'm trying to find the total time it takes the object to move from (x = R+D) to (x=R).

    I realized the mistake and the new equations are:
    $$-GM\int_{R+D}^{R} \frac{1}{x^{2}} dx = \frac{v^{2}}{2}$$

    $$v = \sqrt{2GM(\frac{1}{R} - \frac{1}{R+D})} = \frac{\mathrm{dx} }{\mathrm{dt}}$$

    $$t = \int_{R+D}^{R} \frac{1}{\sqrt{2GM(\frac{1}{R} - \frac{1}{R+D})}} dx$$

    $$t = \frac{D}{\sqrt{2GM(\frac{1}{R} - \frac{1}{R+D})}}$$
     
  9. Dec 14, 2016 #8

    vela

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    As mfb noted earlier, your expression for ##v## doesn't depend on ##x##. It's a constant, so you might notice your last calculation amounts to time = distance/speed (with a sign error).

    In this problem, you need to pay a bit more attention to the signs. Which way does the velocity point as the object nears the Earth?
     
  10. Dec 14, 2016 #9

    Velocity is a function of x. As the object starts off (with a large x value) the velocity is very small. However, as the object gets closer to Earth, the velocity is very great...

    The object is traveling to the left, toward the Earth.
     
  11. Dec 14, 2016 #10

    vela

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    Right, so your expression for ##v## should be a function of ##x##. Yours isn't.

    According to your sign convention, did you pick the right sign for ##v## when you solved ##v^2=##...?
     
  12. Dec 14, 2016 #11

    I understand now... However, I don't know how to make it a function of x.

    And is this right?
    $$\frac{-v^{2}}{2}$$
     
  13. Dec 14, 2016 #12

    vela

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    Think about what I said in post 6.

    No. If you had the equation, ##x^2 = 4##, there are two solutions, right? Same thing here. You have ##v^2=\text{something}##. Make sure you pick the right solution.
     
  14. Dec 14, 2016 #13
    I'm sorry, I still don't understand
     
  15. Dec 15, 2016 #14

    mfb

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    If you know that x is negative, and x2=4, what is x?
     
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