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Newton's argument, gravitational force inside a sphere

  1. Mar 28, 2015 #1
    This is an excerpt from "Introduction to Mechanics" by Kleppner and Kolenkow:

    "The reason why gravitational force vanishes inside a spherical shell can be seen by a simple argument due to Newton. Consider the two small mass elements marked out by a conical surface with its apex at ##m##.

    The amount of mass in each element is proportional to its surface area. The area increases as (distance)2. However, the strength of the force varies as 1/(distance)2, where the distance is measured from the apex to the shell. Thus the forces of the two mass elements are equal and opposite, and cancel. We can pair up all the elements of the shell this way and so the total force on ##m## is zero."

    Are the cones similar because of the spherical symmetry of the shell?
  2. jcsd
  3. Mar 29, 2015 #2
    Yes, in a word, it is due to symmetry.
    In a few more words:
    Consider a 1 Kg mass at point P some distance inside the spherical shell, not at the center.

    Area A1 = θr1^2, for thickness t Volume = θr1^2t, Mass = θr1^2tp
    The gravitational Force exerted on the 1 kg mass by this section of shell is
    F = G(θr1^2tp) / r1^2 = Gθtp
    Using the same method the mass at Area A2 = θr2^2tp
    F = G(θr2^2tp) / r2^2 = Gθtp
    The gravitational forces are exactly equal and cancel out and this holds true for all of the outer shell material taking all of the forces in pairs.
    But be careful to note this only applies to a spherical shell. Inside a solid sphere there is also a solid inner sphere that will exert a gravitational force on the object that varies with the distance from the center. But if that point is exactly at the center so that there is no inner sphere then the total gravitational force is zero.
  4. Mar 29, 2015 #3
    So the area of the spherical segment is ##r^2 \theta##?
    Also, what is the gravitational force experienced at ##r = R## (for a thin shell of radius ##R##)?
  5. Mar 29, 2015 #4
    Yes but keep in mind that θ in this case is the solid angle, measured in steradians. Since the surface area of a sphere = 4πR^2 you can deduce there are exactly 4π steradians of solid angle in a sphere.
    For all points inside the shell the gravitational force is zero, including the point at the center where r = R.
  6. Mar 30, 2015 #5


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    At the shell, the force is not well defined, but this does not matter much for any type of application. In reality, you cannot have this situation, your assumption that the shell is thin means quite some idealisation, as does the assumption of a point like test particle.

    The center is r=0, r=R is the shell itself.
  7. Mar 30, 2015 #6
    I was referring to the drawing I posted earlier, in which case the distances r1 and r2 are not measured from the center of the sphere but from the object inside the sphere to the outer shell. In that case, when r = R the object is at the center of the sphere. When stating any measurement, point of reference should be clearly stated also.
  8. Mar 30, 2015 #7


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    Yes, but I do not believe this is what OP was referring to, there is nothing special about that point (other than being the centre of the mass distribution and having complete rotational symmetry around it, meaning the force there is always zero).
  9. Mar 30, 2015 #8


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    The quickest modern argument is that Newtonian gravitation is described by Poisson's equation
    $$\Delta g=4 \pi \gamma \rho.$$
    For a spherically symmetric mass distribution around the origin of the coordinate system, with ##\rho(\vec{x})=0## for ##r=|\vec{x}|<R## the unique solution in this region is $$g(\vec{x})=\text{const}$$.

    Outside of the mass distribution, i.e., if ##\rho(\vec{x})=0## also for ##r \geq R'##, the unique solution there is
    $$g(\vec{x})=-\frac{\gamma M}{r}, \quad M=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \rho(\vec{x}).$$
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