[SOLVED] Gravitation?

**fardeen_gen** · May 4th 2009, 10:10 PM

Inside a fixed sphere of radius $R$ and uniform density $\rho$ , there is spherical cavity of radius $\frac{R}{2}$ such that surface of the cavity passes through the centre of the sphere. A particle of mass $m$ 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!

**Media_Man** · May 19th 2009, 09:00 AM

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 $\frac{r}{2}$ 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 $\delta$ .

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

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

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

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

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?

**the_doc** · May 19th 2009, 04:18 PM

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:

Spoiler:

**the_doc** · May 19th 2009, 04:26 PM

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!

**the_doc** · May 20th 2009, 03:24 AM

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:

Spoiler: