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Thread: [SOLVED] Gravitation?

  1. #1
    Super Member fardeen_gen's Avatar
    Jun 2008

    [SOLVED] Gravitation?

    Inside a fixed sphere of radius $\displaystyle R$ and uniform density $\displaystyle \rho$, there is spherical cavity of radius $\displaystyle \frac{R}{2}$ such that surface of the cavity passes through the centre of the sphere. A particle of mass $\displaystyle m$ is released from rest at centre B of the cavity. Calculate velocity with which particle strikes the centre A of the sphere. Neglect earth's gravity. Initially sphere and particle are at rest.

    For diagram, refer gravitation on Flickr - Photo Sharing!
    Last edited by fardeen_gen; May 4th 2009 at 10:21 PM.
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  2. #2
    Senior Member
    Apr 2009
    Atlanta, GA

    Check me...

    Okay, I've made a few educated assumptions here, that I need someone to check, and I'll be as descriptive as possible in leu of a picture. We have a sphere of radius r centered at point A, with a spherical cavity of radius $\displaystyle \frac{r}{2}$ situated in such a way that the cavity's poles connect A's surface to A's center. Call the single intersection between spheres A and B A's North pole, N, and its opposite the South pole, S. By symmetry, A's center of mass will be somewhere along the line segment AS.

    Assumption 1: The center of mass, C, will occur at the same point as the "center of area" of the cross section of the figure. Call the distance between A and C $\displaystyle \delta$.

    If this assumption is correct, then a line through C perpendicular to AS will cut the cross-section perfectly in half. Using simple geometry, the area of the north half is $\displaystyle A_N=(\pi-\theta)r^2+\delta\alpha-\frac{\pi}{4}r^2$ , where $\displaystyle \theta$ represents the central angle opened by the cut and $\displaystyle \alpha$ represents half the length of this line. A right triangle is formed by connecting A and C to the edge of the circle, so $\displaystyle \theta=2cos^{-1}(\frac{\delta}{r})$ and $\displaystyle \alpha=\sqrt{r^2-\delta^2}$ . Similarly, the south half of the cross section, $\displaystyle A_S=\frac{\theta}{2}r^2+\delta\alpha$ . Setting $\displaystyle A_N=A_S$ and solving for $\displaystyle \delta$ gives $\displaystyle \delta=rcos(\frac{3\pi}{8})$ . The distance between our small mass m and the center of mass C is therefore $\displaystyle x_0=\frac{r}{2}+\delta=\frac{r}{2}+rcos(\frac{3\pi }{8})$ . The mass of the solid is $\displaystyle \rho V=\rho (V(A)-V(B))=\frac{7}{6}\pi\rho r^3$ . The force of gravity connecting these two centers of mass is given by Newton's $\displaystyle F=G\frac{m_1m_2}{x^2}$ .

    Assumption 2: The Work done by gravity pulling m to A will be equivalent to the total kinetic energy of the system when m hits A.

    If correct, then $\displaystyle W=\int_{x_0}^{\delta} G\frac{m_1m_2}{x^2} dx = \frac{1}{2}(m_1+m_2)v^2$ for x representing the distance from m to C at any given point in time.

    Integrating, $\displaystyle -G\frac{m_1m_2}{x}|_{x_0}^{\delta} = \frac{1}{2}(m_1+m_2)v^2$ .

    Solving, $\displaystyle v=\sqrt{\frac{7Gm\rho \pi r^4}{3m+3.5\rho \pi r^3}(\frac{1}{cos(\frac{3\pi}{8})}-\frac{1}{1/2+cos(\frac{3\pi}{8})})}$

    Further simplification would only be necessary if given actual figures to calculate a final answer. Aside from conceivable algebra errors, do you see any problem in this logic?
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  3. #3
    May 2009
    Mediaman, your calculation is incorrect.

    First I'd like to say that there is a very easy method to solve this that I posted previously but for some reason it either didn't get posted properly or it was deleted (for whatever reason) by a mod. So this time I'm going to do it both the easy way (in this post) and the harder, possibly more convincing, way in the next post.

    OK so here's the easy way:


    To begin with it's a well known result in gravitation that the sphere has geometry such that for gravitational calculations all it's mass can be considered to be concentrated at its centre for objects outside of its radius.

    Second there is another result that states that if inside a spherical shell then there is no gravitational influence i.e. net gravitational force on an obejct there is zero. This can be easily shown using geometry since the force on either side is proportional to $\displaystyle 1/r^2$ and the area subtended on either side is proportional to $\displaystyle r^2$. Hence the forces cancel and so within the cavity the net contribution is zero.

    Now consider a solid sphere of radius $\displaystyle R$ with a 'tunnel' of negligible width (compared to the size of the sphere) going into the sphere along a diameter. At any point at radius $\displaystyle r$, $\displaystyle r<R$, in this tunnel the gravitational influence can be considered to be the sum of that of a spherical shell with inner radius $\displaystyle r$ and a solid sphere of radius r. As already explained the shell then has zero contribution and the influence of the inner sphere on a mass $\displaystyle m $ is then

    $\displaystyle \color[rgb]{0,0,1} F = \frac{Gm}{r^2} \frac{4}{3} \pi r^3 \rho = \frac{4}{3}\pi \rho G m r$ . (Gravitational force inside a sphere)

    Ok, so now we know this how do we use this result for this punctured sphere?
    Well, this is the equivalent to whatever the net gravitational effect of the large sphere (radius $\displaystyle R$) minus the net gravitational effect of the small sphere (radius $\displaystyle R/2$).

    I'll call the big solid sphere A and the small one B so then the net graviational force will be:

    $\displaystyle \color[rgb]{0,0,1} F_{\mathrm{net}} = F_A - F_B$.

    We know the net force will be towards A so let us define the force in the direction of BA as the positive direction and let x be the distance of our mass m along AB from point B. Hence from the above result we have that

    $\displaystyle \color[rgb]{0,0,1} F_A = \frac{4}{3}\pi \rho G m \left(\frac{R}{2} -x \right)$


    $\displaystyle \color[rgb]{0,0,1} F_B = - \frac{4}{3}\pi \rho G m x$

    so that

    $\displaystyle \color[rgb]{1,0,0} F_{\mathrm{net}} = \frac{2}{3} \pi \rho G m
    $ .

    This means this is a constant acceleration problem and since the sphere is fixed we can simply use

    $\displaystyle \color[rgb]{0,0,1}v^2 = u^2 + 2as$

    to find that the velocity at impact at A is:

    $\displaystyle \color[rgb]{0,0,1}v^2 = 2 \cdot \left(\frac{2}{3} \pi \rho G \right) \cdot \left( \frac{R}{2} \right) $


    $\displaystyle \color[rgb]{1,0,0} v^2 = \frac{2}{3} \pi \rho G R$ .

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  4. #4
    May 2009
    It's late so will post the `hard' way of getting the same result tomorrow when I get a chance. Basically I'm going to integrate the forces in spherical polar geometry so will take some time to latex!
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  5. #5
    May 2009
    Now for the 'hard' way! By hard I just mean that it actually involves carrying out some integration but I am actually going to do this the easiest way possible (as far as I know).

    So here goes:


    To begin with you should note that the problem has rotational symmetry about the axis through AB. So the two most appropriate geometries to integrate within are then either spherical or cylindrical geometry. Spherical is the one to choose for a number of reasons (which I'll let you figure out for yourself) but the bottom line is it's the easiest to work with.

    So now we define the coordinates in our spherical geometry:

    Let $\displaystyle \phi$ be the angle of rotation about AB. The point from which we're rotating is unimportant since it will be a full rotation back to the starting point.

    Let $\displaystyle r$ be the separation between the particle of mass m and the infinitesimal mass we're integrating.

    Let $\displaystyle \theta$ be the angle that the vector $\displaystyle \mathbf{r}$ makes with the axis AB on the side where A is.

    Ideally I would have drawn a diagram but I don't have the time to do so.

    From Newton's law of gravitation the force dF exerted by an infinitesimal mass dM in the hollowed sphere on the particle at distance x from B (on the same side of B as A) is:

    $\displaystyle \mathbf{\mathrm{d}F} = \frac{Gm}{r^2} \mathrm{d}M $

    but we're only interested in the component of this force along AB, $\displaystyle F_{AB}$, so then:

    $\displaystyle \mathrm{d}F_{AB} = \frac{Gm}{r^2} \, \cos \theta \, \mathrm{d}M \, \hat{\mathbf{e}}_r$ .

    Putting this in the final integral form we have:

    $\displaystyle F_{AB} = \int_{0}^{\pi} \int_{C_1}^{C_2} \int_{0}^{2 \pi} \frac{G m \rho}{r^2} \cos \theta \, r^2 \sin \theta \, \mathrm{d} \phi \, \mathrm{d} r \, \mathrm{d} \theta $


    $\displaystyle F_{AB} = 2 \pi \rho G m \int_{0}^{\pi} \cos \theta \, \sin \theta \left[r\right]_{C_1}^{C2} \, \mathrm{d} \theta $

    where $\displaystyle C_1$ is $\displaystyle r(\theta)$ on the surface of sphere B and $\displaystyle C_2$ is $\displaystyle r(\theta)$ on the surface of sphere A.

    I'll leave it to the reader to draw diagrams from which it will be possible to use the cosine rule in each case to find a quadratic in $\displaystyle r$ which when solved yields:

    $\displaystyle C_1 = \sqrt{\left(\frac{R}{2}\right)^2 - x^2 \sin^2 \theta}-x \cos \theta $


    $\displaystyle C_2 = \frac{R-2x}{2} \cos \theta +\sqrt{1 - \left(\frac{R-2x}{2R}\right)^2 \sin^2 \theta} $.

    Plugging these limits in for $\displaystyle r$ we find that:

    $\displaystyle F_{AB} = 2 \pi \rho G m \left( \frac{R}{2} I_1 +I_2 -I_3 \right)$


    $\displaystyle I_2 = \int_{0}^{\pi} \sin \theta \cos \theta \sqrt{1 - \left(\frac{R-2x}{2R}\right)^2 \sin^2 \theta} \,\mathrm{d}\theta = 0$ ,

    $\displaystyle I_3 = \int_{0}^{\pi} \sin \theta \cos \theta \sqrt{\left(\frac{R}{2}\right)^2 - x^2 \sin^2 \theta} \,\mathrm{d}\theta = 0$ ,


    $\displaystyle I_1 = \int_{0}^{\pi} \sin \theta \cos^2 \theta \, \mathrm{d}\theta = \frac{2}{3}$ .

    The integrals $\displaystyle I_2$ and $\displaystyle I_3$ were computed using the sine and cosine double angle formulae and $\displaystyle I_1$ should be obvious!


    $\displaystyle F_AB = \frac{2}{3} \pi \rho G m$

    as shown in the previous post and so the remaining part of the solution is identical to above.

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