# Force from a HEMIspherical shell [closed]

I am working on this problem and just trying to figure out what my prof/ TA did in the solution sheet -- but also to make sure I understand what I am doing and some of the mathematics involved.

A spherical shell will exert no net force on a particle inside it. But what of a HEMIspherical shell? That is, (working from the question asked on here before about gravity on a hemispherical planet) how to show rigorously what kind of forces will act on a particle in that situation. If you do this problem and ad on the force for the other "half" of course, it should all come out to zero. This is also a bit different from the straight up hemisphere since we aren't assuming that its exactly 1/2.

So, given a spherical shell of radius R and surface density σ, and a particle at position a from the center where a is between 0 and R. Assume the position a is along the z (vertical) axis.

What is the force on the particle from the smaller hemisphere, described by the part that extends from the top (at z=R) down to where the particle is (at z=a) and where is the center of mass of that hemisphere?

I am using θ as the angle between the origin and the point on the shell that intersects a line drawn through a that is perpendicular to the z axis. (I wish I knew how to post a diagram here, that would make it easier I think, but I hope people get the idea).

Repeat for the larger part of the shell.

Now, I know intuitively (and from the symmetry principle) that the rings, each with a mass of dM = 2πRsinθRdθ will exert a net zero force in the x-y plane as we are assuming symmetry here. So, I only need the force in the z-direction.

Anyhow, first I want to know the total mass of the smaller part of the shell. That's going to be an integral from 0 to θ. Since the radius of the strip is R cosθ, and the force vector is along the z-axis, I should get this:

$$\int dF_z = \int_0^{\theta} \frac{Gm2\pi R \sin \theta R d \theta}{r^2}\cos\theta$$

where r is the distance from my particle at a to the piece of the shell I want. So far so good.

Now, the distance r is going to be determined by the law of cosines, if I want to get it in terms of θ. So I can say $r^2 = a^2 + R^2 - 2aR \cos \theta$.

I'm also multiplying the Force by cosθ because it is in the z direction, which in terms of R and a is $\frac{R^2+a^2-r^2}{2aR}$

If I differentiate $r^2$ -- because I am setting up an integral here -- I get $2rdr = 2aR \sin \theta d \theta$. A little algebra gets me $\sin \theta d \theta = \frac{2rdr}{2aR} = \frac{rdr}{aR}$ and I should be able to plug that back into the integral and get:

$$\int dF_z = \sigma Gm2\pi R^2 \int_0^{\theta} \frac{rdr}{aR} \frac{1}{r^2} \frac{R^2+a^2-r^2}{2aR}$$

and we have to change the limits, which here would be arccos(a/R) so I have:

$$\int dF_z = \sigma Gm2\pi R^2 \int_0^{\cos^{-1} \frac{a}{R}} \frac{rdr}{aR} \frac{1}{r^2} \frac{R^2+a^2-r^2}{2aR} = \sigma Gm\pi \int_0^{\cos^{-1} \frac{a}{R}} \frac{rdr}{a^2} \frac{1}{r^2} (R^2+a^2-r^2)$$ $$= \sigma Gm\pi \int_0^{\cos^{-1} \frac{a}{R}} \frac{R^2+a^2-r^2}{a^2} \frac{1}{r} dr = \frac{\sigma Gm\pi}{a^2} \int_0^{\cos^{-1} \frac{a}{R}} \frac{R^2+a^2-r^2}{r} dr$$ and because $\sigma = \frac{M}{4 \pi R^2}$ the integral looks like:

$$= \frac{M Gm}{4R^2a^2} \int_0^{\cos^{-1} \frac{a}{R}} \frac{R^2+a^2-r^2}{r} dr$$

So my question is, how am I doing so far? I feel like there is something I am missing -- because the answer sheet says that I should get $$\frac{M Gm}{4Ra^2}2(\sqrt{R^2-a^2}-2R$$

and I am not entirely sure if I got the final integral right. thanks in advance.

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## closed as off-topic by David Z♦Dec 9 '13 at 0:21

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – David Z
If this question can be reworded to fit the rules in the help center, please edit the question.

I don't understand why this was put on hold. Sure, it's a bit rambling and takes some time to get one's head around (that's unavoidable I think), but it certainly shows considerable work by the asker to answer their question. –  WetSavannaAnimal aka Rod Vance Dec 9 '13 at 0:31
I tried editing it a little, and I don't mean to have it be too long but I wanted to reduce the ambiguity as much as possible. I feel like there's one piece that is missing, and I am trying to see if I am having a conceptual problem or a math problem. If this doesn't fit it doesn't, and so be it. –  Jesse Dec 9 '13 at 0:43