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Newton's law of gravitation

  1. Sep 3, 2009 #1
    Hi there,

    I don't understand why the equal of Newton's law of gravitation F = Gm1m2/d(squared) is different for no spherical objects.

    In the image of my attachment there are one spherical object and a no spherical object. So why if there are not very distant the equal used is F = Gm1m2/d(d+L).

    How It's used (d).(d+L) on the contrary (d).(d), which is the original equal?


    Thanks
    Tony

    PS: Sorry about my english.
     

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  2. jcsd
  3. Sep 3, 2009 #2

    Doc Al

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    Your question isn't quite clear. Your diagram shows (I believe) a spherical body (centered at location x = 0, say) and an extended body that goes from x = D to x = D + L. Given that, I do not understand your first equation for the gravitational force between the two objects.(1) The second equation makes sense, since it says that when the objects are far enough apart they can be treated as point masses and you can ignore the small size of the object.

    Strictly speaking, Newton's law of gravity only applies for point masses (and certain special geometries, such as uniform spheres and spherical shells). To get the net force between the two objects you'd have to integrate the force on each mass element of the extended body.

    (1)Edit: My bad, as I was too lazy to do the integral. :uhh: As D H points out, your first equation is perfectly fine. Note that this is allowed because your first object is spherical and can thus be treated as a point mass at x = 0.
     
    Last edited: Sep 3, 2009
  4. Sep 3, 2009 #3

    D H

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    Strictly speaking, Newton's law of gravitation only applies to point masses. It can be applied to non-point masses by integrating over the volume. In the case of this (presumably constant density) bar,

    [tex]a = \int_d^{d+L} \frac{G\rho}{x^2}dx[/tex]

    where [itex]\rho[/itex] is the density of the bar: [itex]\rho=M_{\text{bar}}/L[/itex]

    This yields

    [tex]a = \frac{GM_\text{bar}}{d\cdot(d+L)}[/tex]

    Note well: This result is valid for all distances d. Another way to write the above is

    [tex]a = \frac{GM_\text{bar}}{d^2}\frac 1 {1+L/d}[/tex]

    In the case that [itex]d\gg L[/itex], the final factor is very close to one. In other words,

    [tex]a \approx \frac{GM_\text{bar}}{d^2}\,,\,d\gg L[/tex]
     
  5. Sep 3, 2009 #4
    For d >> L we can validly write the equation as:

    [tex]F =~ \frac{G m_1 m_2}{d^2}[/tex] because think about it:

    If d = 10000000000000000m and L = 1m than d+L = 10000000000000001, so when we square it the numbers are pretty much the same. Using d by itself in this case turns out to only give an error of ~2*10^-8% from the real value.
     
  6. Sep 3, 2009 #5

    D H

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    You can't (or shouldn't) do that.

    Edit
    That was a bit too terse. The reason you should not do that is because it is wrong. The gravitational attraction between two non-point masses is the sum of the gravitational attractions amongst all the bits of matter that comprise the two objects (or you can integrate if you ignore that matter is not discrete). The only way this would summed / integrated form would yield the same result as the center of mass to center of mass calculation is if gravitation was a linear function of distance. It is not of course; gravity is an inverse square law.
     
    Last edited: Sep 3, 2009
  7. Sep 3, 2009 #6
    I'm surprise with the number of answers. You really help me!!

    Thanks Doc Al, D H and Feldoh, your answers are brilliant.

    And thanks to adivice the matters of integrate.


    Thanks
    Tony
     
  8. Sep 3, 2009 #7
    Good explanation and reminder......That explains why the first equation "looks strange" at first.....I had initially thopught "Should the second term be (d+L)/2??".....before I read further...
     
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