Question

A person with a mass m is standing up on a conical mountain. The height of the mountain is h and its radius is R. The density of the mountain is p. Calculate the gravity between the man and mountain.

Answer 1

the symmetry

Answer 2

eg if you turn it into a volume of revolution, with each disc adding an extra element to the gravitational attraction

hint

Answer 3

@Irishboy123
Hmm okay so I should find the force from one "disc" in a distance z away from m and then intergrade from 0 to h to find the total force right?
But how would the force for one disc be?

Answer 4

you can't solve this without knowing the angle at which the base meets the surface area around it. Or at least the angle on the top of the cone.

The problems come when you try to do the integral up the cone by the infinitesimal mass elements. The approach would go something like this:
1) define the infinitesimal force element over the infinitesimal mass element of the body in interest. For your case:
$$d\vec F = \Gamma \frac{mdM}{r_{12}^2}\hat r_{12}\\
\hat{r}_{12} \ \stackrel{\mathrm{def}}{=}\ \frac{\vec{r}_2 - \vec{r}_1}{\vert\vec{r}_2 - \vec{r}_1\vert}$$
That is to say in vector form, much like EDM equations the force acts on the radius vector between the two bodies.
2) use symmetry to kill the vector calculus. In this case all the "left-right-forward-backwards" elements would cancel out. Only the "top-down" forces would remain, that is the down elements to be precise. Once we recognize this we can kick the vector calc and jsut write:
$$dF = \Gamma \frac{mdM}{d^2}$$
3) Write out the mass element. We won't be integrating by disks to be frank. We'll be integrating by the infinitesimal volume elements of the conus. Imagine it as a disk, but we're integrating by small theta, dh - height change and dr - distance from the center vertical height of the cone. So in essence it's like integrating by disks, except the problem with that image is that disks radii change as you're climbing up the cone so you have to parameterize them too. Also now the distance is marked with d and not r not to confuse us about that point and is meant to mean the distance to the infinitesimal volume elements.
$$dM = \rho dV = \rho rdhd\theta dr\\
dF = \Gamma \frac{mdM}{d^2}$$
4) Parametrization of all the variables over the integration variables. We have several stuff to do here:
4.1) integration limits: [0, 2pi] for the theta, [0, R] for the radial part, except the maximal value R that r can be depends on the height and the angle on the top of the cone so that integral goes [0, h+tan(alpha)]. As far as the height goes we integrate up to height of the cone H.
4.2) but this is where our guy comes in because we need to parametrize the distance d. We're going to assume he has insignificant height and mass distribution so that he's at the height H. Draw the triangle for yourself, from the volume element up to the top of the cone and you'll see that the sides of that triangle are r (the current integration point, not R) and h (the current h, not 0 or H). So the distance here is just
$$d = \sqrt{h^2 + r^2}$$
Which means you have to be careful of the order of integration -> first do the disk up to current R (h tan alpha) and then do them up to H. In this way we make sure we did all disks with their proper R's.

And this is where we stop, because you've not made it clear we know the alpha at the top.

Answer 5

soz @beggi9 for the bum steer. i think @ljetibo got us back on track. but i simply do not get how we turn the scalar force into any kind of useable vector concoction. so this is the kinda classical physics problem that annoys the shiτ out of me

it's all down to the lack of symmetry of the cone.

i've attached below some old bollocks that i came up with; but Cartesian and cones just don't work


this link shoulda been interesting:
https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity

but, i can't get the vectors around a cone to work.... in any vector system....

Answer 6

Sorry, busy couple of days....

I don't know why you're unhappy because that's correct solution. The fact that you can't/don't know how to solve the integral doesn't mean the setup is wrong. Take a look at how I proposed the solution and you'll see that it's the same. Also take a look at the sketch (untitled-1/attachement 1) to see what's what, we use different markings and I use a lot less meticulous mathematical expressions. It's basically a substitution x0-x = h and r1-r2 = d.
$$dF = \frac{mdM}{d^2}\hat r$$
By using the R and H we have to write everything over r and h. The R' is the "maximal radius at some h" - it's only R at the bottom and less as you climb.
$$\tan\phi = \frac{R}{H} = \frac{R'}{H}\\
R' = h\tan\phi\\
d^2 = h^2 + r^2\\
\cos\alpha = \frac{h}{d} = \frac{h}{\sqrt{h^2+r^2}}\\
dV = r\,\,\,dr\,d\theta\,dh\\
dM = \rho dV = \rho r\,\,\,dr\,d\theta\,dh\\$$
Ok, so notice that I'm admiting I was wrong when I said we don't know the top angle, we do know it, we can calc it from tangens oof maximal radius at the bottom R and height of the cone H. My bad. Ok so let's start some calc.
$$\int dF = \iiint_V \frac{mdM}{d^2}\hat r\\
\int dF = \iiint_V \frac{mdM}{d^2}\cos\alpha\\
\int dF = \iiint_V \frac{m\rho r\,\,\,drd\theta dh}{\sqrt{h^2 + r^2}}\frac{h}{\sqrt{h^2+r^2}}\\
\int dF = m\rho\iiint_V \frac{hr}{(h^2 + r^2)^\frac{3}{2}}drd\theta dh$$
This is the same integral you have. Except you used a bit more precise mathematical expressions than I have, I kind of half-a$ed it on that front. Also in my version you have to integrate by r and only then by h because the borders:
$$\int dF = m\rho \int_0^H \int_0^{h\tan\phi}\int_0^{2\pi}\frac{hr}{(h^2 + r^2)^\frac{3}{2}}drd\theta dh$$
Well it's not as simple integral to solve. In uni they tend to teach you couple of methods, but in working environment you usually don't have the time to spend going all around and solving each one from the start. Therefore you just look up the shape of the integral in integral tables. I use Bronshtein and Semendyayev mathematical handbook a lot and would definitely recommend it if you plan on running into these kidns of integrals.
We notice that integral for r is an integral of irrational function and it looks similar to this form:
$$\int \frac{x}{\sqrt{(x^2+a^2)^3}}$$
You can find these general solutions on wiki too: https://en.wikipedia.org/wiki/List_of_integrals_of_irrational_functions Anyhow the solution is:
$$\int \frac{x}{\sqrt{X^3}} dx = -\frac{1}{\sqrt{X}}$$
where X = x^2+a^2. We do that and integral by theta right away.
$$\int dF = 2\pi m\rho\int_0^H h \left[ -\frac{1}{\sqrt{h^2+r^2}}\right]_0^{h\tan\phi}dh\\
= 2\pi m\rho\int_0^H h\left[ -\frac{1}{\sqrt{h^2+h^2\tan^2\phi}} + \frac{1}{\sqrt{h^2}} \right]dh\\
= 2\pi m\rho\int_0^H \left[ -\frac{1}{\sqrt{1+ \tan^2\phi}} + 1 \right] dh\\
= 2\pi m\rho\left[ -\frac{H}{\sqrt{1+ \tan^2\phi}} + H \right]$$
Ok, so time to analyze what we have. On first glance this works well with our formula. Tangents squared is an always positive function, periodical with pi. It will be 0 for 0, Pi, 2Pi...it will go to infinity every Pi/2, 3Pi/2...
If the top half angle is 0, 180, 360 etc... that doesn't make sense because the cone would turn in on itself and form a line. In those cases we neatly get 0 for our force:
$$F = 2\pi m\rho\left[ -\frac{H}{\sqrt{1+ 0}} + H \right] =2\pi m\rho\left[ -H +H \right] =0 $$
What does a cone look like if it's half-angle is 90 degrees? It's an infinite flat plane because the top angle is in total 180 degrees, notice that then:
$$ F = 2\pi m\rho H $$
Which is oddly reminiscent of the gravity force of an infinite flat plane task where you show no matter the distance you always feel the same force on you. The only difference though is that that solution has: 2pi m rho as a solution without the H.
We make a density plot (heatmap plot) of this as function of phi and H. Phi is the angle at the top of the cone (on y axis), and H is height of the cone (x axis). Phi goes from 0 to Pi, which covers an entire period and H goes over 9000. What you see is that if the mountain is very very tall, but the angle is very very small - the gravitational force becomes negligible. That's because the distance to the base of the cone, where the majority of the mass is, becomes very large and this is a force that fall off as a square of the distance. Opposite to that it seems that even if the mountain is relatively "small" as long as the angle is in between pi/4 and 3pi/4 the force you feel is pretty big. Remember the mass of the mountain is really big - "Most of the major rock-forming minerals in the Earth's crust, like quartz, feldspar and calcite, have very similar densities (around 2.5-2.7)." that's grams per cubic centimeter! of course this is evened out a bit by the gravitational constant which is very small but still.

Ok, so I'd say it looks good enough to be considered correct, except that I'm missing the gravitational constant everywhere, but just add that in.

Answer 7

Yes Sir :-|

I was thinking cylindrical would be nice, but the person is stuck in a certain point in space.

She's at \(\vec r = <x_o,y_o,z_o > = <r_o \ cos \theta_o , r_o \sin \theta_o, z_o > = ....\)

There is absolutely no symmetry here. Park her at the Origin and you have a way off-centre cone.

I would love to see a fully-pursued solution.