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Does Newton's law of gravitation explain this?

  1. May 3, 2010 #1
    Being possibly the newest member here, I would like to put a question that has troubled my mind for quite a while..

    If I have a uniform ring (having a certain mass) and a point mass in some idealised gravity free space & I orient them such that the point mass lies exactly at center of the ring(mutual gravitational interactions possible). (my confusion begins here..)

    What would be the force between these 2 objects in this configuration?

    If I try to find it by directly applying the formula for field at the center of a ring, the answer would definitely be 0..
    Taking Symmetry considerations, we can answer 0 once again..
    But looking at the expression in the universal law of gravitation, we find that the answer comes out to be infinite as centers of mass of the 2 bodies are coinciding..
    How can this be explained? Further the field expressions can be derived from this law. Isn't this paradoxical? Is a concept of infinite force between two rigid bodies of any physical significance?

    Hoping for a quick reply...
     
  2. jcsd
  3. May 3, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi Starwanderer1! Welcome to PF! :wink:
    No, that formula only applies to spherical masses (and to points outside the sphere). :smile:

    (For any other shape, we have to treat the body as made up of lots of very small spherical or point masses, and add their individual forces, as vectors)
     
  4. May 3, 2010 #3

    D H

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    Welcome to Physics Forums, Starwanderer1.

    You are misapplying Newton's law of gravitation. Strictly speaking, Newton's law of gravitation applies to point masses only. One can employ this law to compute the gravitational force between non-point masses by integrating the law over the differential masses that comprise the masses. Conceptually, the gravitational force on object 1 by object 2 is

    [tex]F_{12} = \int_{M_1} \int_{M_2} G \frac {\mathbf r_2 - \mathbf r_1}{||\mathbf r_2 - \mathbf r_1||^3} dm_2 \,dm_1[/tex]

    where [itex]\mathbf r_i[/itex] is the distance from the origin of some inertial frame to the differential mass [itex]dm_i[/itex].
     
  5. May 3, 2010 #4
    yes you are right but what about the distance of centre of masses? means here the distance between the centre of masses is 0 because they are coincideed?
     
  6. May 3, 2010 #5
    Ok, I get that. But the concept of center of mass can be applied to the ring treating the whole ring to be a point mass centered at its center of mass and then applying the law ,we are doing the same thing as you are telling. Isn't it true that thinking very basically and applying the law we can come across this paradoxical situation? Can Newton's law be valid even when the center to center radius vector has zero magnitude? You cant deny that the c.o.m. of the ring lies in the free space treating the whole of it as arigid body (before treating it as a collection of differential elements when the individual contributions get cancelled vectorially)..
     
  7. May 3, 2010 #6

    Doc Al

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    No, you can't apply the concept of center of mass in that manner here. That would be useful in describing the force on the ring if it were in a uniform gravitational field, but that's not what's going on here.

    As D H stated, Newton's law of gravity applies between point masses. You cannot find the gravitational force between two extended objects by plugging in the distance between their centers of mass, except in certain special cases of high symmetry.
     
  8. May 3, 2010 #7

    D H

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    The concept of center of mass cannot be used in that way.

    Before I go further, is this homework? We have rules against telling people the answers to homework problems.
     
  9. May 3, 2010 #8
    No, this isn't homework (it wouldn't be very fair to expect a member to violate rules from day 1).I have read all your rules and I would perfectly abide by those. Actually I have asked this to many people and received replies which dont really satisfy me. Thank You all for your efforts but I think that I need to really think about this with my own mind (but ofcourse with references to my experinces).
     
  10. May 3, 2010 #9
    Actually this may come out as a derivative of the original question.I want to ask the mentor on the lines of the comment made: What are these "certain special cases of high symmetry" where I can use the c.o.m. concept as I had used in my query??

    Hoping for a quick reply..
     
  11. May 3, 2010 #10

    Dale

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    In general the center of mass will only equal the center of gravity in the case of a uniform gravitational field. Also, the center of gravity in a nonuniform field can in general depend on the object's position and orientation within the field.
     
  12. May 3, 2010 #11

    D H

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    It might not be fair, but unfortunately that is what we have to do. A significant portion of the people who violate our rules are students who violate the rules from day one. Most new members apparently do not read the rules. They just scroll down until they can see the accept button and click on it.

    Excellent. That puts you well ahead of most of our users.


    Since this is not homework, I can show you how to solve this problem.

    Imagine breaking the ring into smaller and smaller pieces. Adding the gravitational force from each of this components will yield the total force between the ring and the point mass. The sum becomes an integral in the limit of infinitesimally small ring pieces. One way to split a ring into tiny chunks is via angular displacement: [itex]dM = M/(2\pi)\,d\theta[/itex]. Newton's law does apply to these infinitesimally small pieces of the ring.

    The distance between any point on the ring and the center of the ring is constant; it is just the radius of the ring. Thus the magnitude of the gravitational force between the point mass and one of these infinitesimally small pieces of the ring will be

    [tex]dF = \frac G{r^2} m dM = m \frac G{r^2} \frac M {2\pi} d\theta[/tex]

    (Here I am using little m to denote the mass of the point mass at the center of the ring and big M to denote the mass of the ring.)

    While the magnitude of the force is constant, the direction is not. In terms of angle theta, the unit vector from the center toward one of these infinitesimal pieces of ring mass is

    [tex]\hat F = \cos \theta \hat x + \sin\theta \hat y[/tex]

    Integrating around the ring,

    [tex]\mathbf F =\frac{GmM}{2\pi r^2} \int_0^{2\pi} (\cos \theta \hat x + \sin\theta \hat y) d\theta[/tex]

    and this integral is of course zero.


    Things get a bit more interesting when the point mass is located somewhere along the ring's axis rather than at the center of the ring. If we denote z as the position of the point mass with respect to the center of the ring, the force is

    [tex]\mathbf F =-\,\frac{GmMz}{(r^2 + z^2)^{3/2}}\hat z [/tex]

    I can go through the derivation if you like. Note that the force goes to zero (not infinity) as z goes to zero.
     
  13. May 4, 2010 #12
    Sorry for the long silence.I couldn't get near the comp yesterday due to certain reasons.but I am interested to learn about the derivation..
    (This may not be relevant but I am finding all the sections given in the LaTex patterns very blurred and difficult to read. Is it just my machine or do they appear as such naturally?)
     
  14. May 5, 2010 #13

    D H

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    If you have an old (very old) version of Internet Exploder, it's your browser that is making the generated LaTeX figures hard to read.

    I'll show two derivations. One uses on the concept of gravitational potential energy while the other develops the force directly. The goal is to develop the gravitational force between a ring of mass with mass M and radius r and a point mass with mass m. The point mass is located somewhere along the axis of the ring. First, define some coordinates. The center of the ring is an obvious choice for the origin. The x and y axes are any two orthogonal axes in the plane of the ring. The z axis is the axis of the ring, with the direction defined by the cross product of the x and y axes. The point mass is on the ring axis, so its coordinates are [itex]z\hat z[/itex].

    The gravitational potential energy due to a pair of point masses m1 and m2 separated by a distance d is ϕ=-Gm1m2/d -- plus some arbitrary constant. We don't have point masses here; we have a point mass and a ring. We'll have to integrate over infinitesimal ring elements. The distance d between the point mass and any point on the ring is

    [tex]d = \sqrt{r^2 + z^2}[/tex]

    This is a constant, so the integration is trivial.

    [tex]\phi = -\frac {GmM}{\sqrt{r^2 + z^2}[/tex]

    The force is simply the additive inverse of the gradient of this potential:

    [tex]\vec F = - \nabla \phi = -\frac{GmMz}{(r^2 + z^2)^{3/2}}\hat z[/tex]


    Now for finding the force directly. The magnitude of the force between an infinitesimal ring element and the point mass is constant since the distance between the point mass and any point on the ring is constant.

    [tex]||d\vec F|| = \frac {GmM}{2\pi (r^2 + z^2)}\,d\theta[/tex]

    The direction of the force is not constant. Let α be the angle between the ring axis and a line connecting the point mass and some point on the ring. The unit vector from the point mass toward some infinitesimal ring element will be

    [tex]\hat F = \sin \alpha \hat r \pm \cos \alpha \hat z[/tex]

    In terms of z and r, this unit vector is

    [tex]\hat F = \frac{r}{\sqrt{^2 + z^2}} \hat r - \frac{z}{\sqrt{^2 + z^2}} \hat z[/tex]

    The infinitesimal force is thus

    [tex]d\vec F = \frac {GmM}{2\pi (r^2 + z^2)^{3/2}}(r\hat r - z\hat z)\,d\theta[/tex]

    Integrating around the ring yields the total force. While the radial component obviously vanishes with this integration, the axial component does not. The axial component is independent of azimuthal angle, so the integration is once again trivial:

    [tex]\vec F = -\frac {GmMz}{(r^2 + z^2)^{3/2}}\hat z[/tex]

    This is the same result as obtained using potential.


    One final note. When the point mass is very far removed from the ring, the radical is approximately equal to |z|3. Thus for |z| >> r,

    [tex]\vec F \approx -\frac {GmM}{z^2}\frac z{|z|}\hat z[/tex]

    In other words, the ring looks more and more like a point mass as the distance between the point mass and the ring increases.
     
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