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Computing The Gravitational Force At A Point

  1. Jun 26, 2013 #1
    1. The problem statement, all variables and given/known data
    (a) Compute the vector gravitational field at a point P on the perpendicular bisector of the line joining two objects of equal mass separated by a distance 2a as shown in Figure P13.26. (b) Explain physically why the field should approach zero as r ---> 0. (c) Prove mathematically that the answer to part (a) behaves in this way. (d) Explain physically why the magnitude of the field should approach 2*(GM/r^2) as r --> infinity. (e) Prove mathematically that the answer to part (a) behaves correctly in this limit.


    2. Relevant equations



    3. The attempt at a solution

    [itex]\vec{F}_0 = -2 \frac{Gmm_p}{\sqrt{r^2 + a^2}} \hat{i}[/itex] is the equation I generated.

    From the diagram, it is evident that, as r approaches zero, the point of interest becomes located in between the two fixed masses; furthermore, one can see that the two masses with pull equally in the j-hat direction, but anti-parallel, thereby causing all gravitational forces produced by the two masses to cancel. However, why is it that, when taking the limit as r --> 0, the mathematics does not show this?
     

    Attached Files:

  2. jcsd
  3. Jun 26, 2013 #2

    pasmith

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    Your expression for the force is not dimensionally consistent, and so must be wrong. (You need GMm divided by a length squared; you have GMm divided by a length.)

    It's usual to use M for the mass of fixed bodies and m for the mass of a test particle.

    For each body, you need to determine the component in the [itex]\hat i[/itex] direction of a force of magnitude GMm/(r^2 + a^2) directed along the line from (r,0) to the centre of the relevant body, and then add the two.
     
  4. Jun 26, 2013 #3
    Okay, I see that there shouldn't be a square root symbol in the denominator; however, I don't see what else is wrong with it. Both bodies pull equally in the [itex]-\hat{i}[/itex] direction.
     
  5. Jun 26, 2013 #4

    gneill

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    Staff: Mentor

    Both bodies pull directly towards themselves. So, two force vectors, neither of which is directed along the [itex]-\hat{i}[/itex] direction. What's the vector resultant?
     
  6. Jun 26, 2013 #5
    All right, let's see if I have properly determined the forces this time.

    For one mass: [itex]\vec{F}_M = - \cos \theta \frac{GMm}{a^2 + r^2} \hat{i}[/itex], where [itex]\theta = arrcos(\frac{r}{\sqrt{r^2 + a^2}})[/itex]

    Is this correct?
     
    Last edited: Jun 26, 2013
  7. Jun 26, 2013 #6

    haruspex

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    Yes.
     
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