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Gravitational Force

  1. Apr 29, 2009 #1
    Hi Guys... This topic was discussed in the class. Professor gave out the hand out. I tried to pay attention but couldn't follow it.

    Can someone explain this to me in simple English? I'm having hard time understanding this.

    I'm attaching a handout that I scanned.

    http://img115.imageshack.us/img115/4284/gravitationalforce.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 29, 2009 #2

    jtbell

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    Staff: Mentor

    Starting from the beginning, what is the first point where you become stuck, and why?
     
  4. Apr 29, 2009 #3
    I understand the Newton's law. I just don't know how to prove, that if the mass m1 and m2 are the points located at p1 and p2, the distance of those two points are equals r. I don't know how to prove this.
     
  5. Apr 29, 2009 #4

    Integral

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    Staff Emeritus
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    Gold Member

    If I understand your question, that is a defintion, you don't have to prove it.
     
  6. Apr 29, 2009 #5
    Right. I understand. This is not a theory which i have to prove (if this then that).

    But I have (Gm1m2 / r^2 )

    then i have to prove F = (Gm1m2 / p1 - p2) * (p1 - p2)

    In the hand out, it says # 3 and shows the above formula. How did they get it? (and rest of the formula) thats what i want to know....
     
    Last edited: Apr 29, 2009
  7. Apr 29, 2009 #6
    That formula you wrote out there is not nearly correct. You are missing parentheses, you are missing a third power, and you are missing the most important factor: vectors!

    Answer yourself these questions. Draw the points P1 and P2 on paper (arbitrary positions) and then try to draw it out:

    What is
    [tex]\vec{P_1} - \vec{P_2}[/tex]
    (Hint, it is still a vector!)

    Then what is
    [tex]\left| \vec{P_1} - \vec{P_2} \right|[/tex]
    (Hint, this is not a vector anymore!)

    Finally, what is
    [tex]\frac{\vec{P_1} - \vec{P_2}}{\left| \vec{P_1} - \vec{P_2} \right|^3}[/tex]

    Do you now understand the third power also?
     
  8. Apr 29, 2009 #7
    Sorry about my typing. I had no idea, how to code this formula and stuff and yes
    I do know the difference between vectors and non-vectors (absolute value).

    However, I have no idea the output of
    [tex]
    \frac{\vec{P_1} - \vec{P_2}}{\left| \vec{P_1} - \vec{P_2} \right|^3}
    [/tex]



    My question is how to get
    [tex]
    {\vec F}({\vec p}) = - \frac { c } { \left | \vec {P} \right | ^3} } { \vec P }
    [/tex]

    ps: wow it took me 20 min to type it. glad i got it though.
     
  9. Apr 30, 2009 #8
    If you define the vector r as:
    [tex]\vec{r} = \vec{P_1} - \vec{P_2}[/tex]

    With length:
    [tex]r = \left| \vec{P_1} - \vec{P_2} \right|[/tex]

    Then your vector equation for the force becomes:
    [tex]\vec{F} = \frac{G m_1 m_2 \, \vec{r}}{r^3}[/tex]

    Then finally you have to think about the direction of the force. The direction of the gravitational force is along the vector r. It has the same direction as the vector r. You can also see that in the equation.
    Because the direction of the force should have nothing to do with it's magnitude, you have to normalize the vector r:
    [tex]\vec{e_r} = \frac{\vec{r}}{|\vec{r}|} = \frac{\vec{r}}{r}[/tex]
    This is often called the unit vector in the direction of r. The length of this vector is 1, because we have divided the vector r by it's length (hence the name unit vector).

    Then finally, plugging this into the equation, we get:
    [tex]\vec{F} = \frac{G m_1 m_2}{r^2} \frac{\vec{r}}{r} = \frac{G m_1 m_2}{r^2} \vec{e_r}[/tex]
    Which is simply the magnitude of the force (the first equation on your paper), pointing in the direction of r.


    To get the very last equation, you simply take P1 to be the origin, which is point (0,0,0) in three dimensions, or (0,0) in two dimensions. In each case, the vector P1 is now the zero vector. So:
    [tex]\vec{P_1} - \vec{p} = \vec{O} - \vec{p} = -\vec{p}[/tex]
    And because you take the magnitude of the vector, the minus sign does not matter.
     
  10. Apr 30, 2009 #9
    That's not the difference between vectors and non-vectors. The non-vectors Nick89 was referring to are scalars.

    Scalar = a number
    Vector = a list of numbers, usually to represent coordinates in some space

    A vector can be used to represent many things. It could be a 3D point in that space. Or it could be a direction in that space, if you consider the direction between (0,0,0) and the given point. Or it could be a velocity, if you consider the direction and the length of the vector to be the speed.

    ie,in Euclidean space [tex] v = (1,0,0) [/tex] is a vector pointing in the direction of X axis, or it is a point located on the X-axis.

    [tex]||v||[/tex] = [tex] \sqrt{ v.x^2 + v.y^2 + v.z^2 }[/tex] = length of v = a number (scalar)
     
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