A spherical shell of inner diameter R and outer diameter 3R has a uniform density . What is the magnitude of the gravitational acceleration a distance 2R from the center of the spherical shell?
A spherical shell of inner diameter R and outer diameter 3R has a uniform density . What is the magnitude of the gravitational acceleration a distance 2R from the center of the spherical shell?
a) 15πGR/7
b) 7πGR/3
c) 32πGR/5
d) 25πGR/12
e) 7πGR/6

Question

A spherical shell of inner diameter R and outer diameter 3R has a uniform density . What is the magnitude of the gravitational acceleration a distance 2R from the center of the spherical shell?
A spherical shell of inner diameter R and outer diameter 3R has a uniform density . What is the magnitude of the gravitational acceleration a distance 2R from the center of the spherical shell?
a)  15πGR/7
b) 7πGR/3
c) 32πGR/5
d) 25πGR/12
e) 7πGR/6

irishboy123 · Answer

tried using the symmetry and a Gaussian surface ?! https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity

irishboy123 · Answer

@osprey 

this one.

osprey · Answer

@IrishBoy123 Hi Shamrock ... I don't understand "this one" ? This one what ? Saw the picture on the other post. Impressive, but couldn't understand whether was happy, sad, or just "cleverly twee". It beats my version of a "smiley face", which is the only "happy" shape I've got on pptx.

irishboy123 · Answer

NB:  is there a typo in the question.  Will assume "diameter" was meant to say "radius".

|dw:1476009795243:dw|

Gauss Law for Gravity (see the Wiki link, i prefer to use less snazzy notation )

$\Large  \iint_S   \vec {g} \bullet \hat n ~ ~ da =-4 \pi G  ~ \Sigma m_{enc} \qquad \triangle $

The Gaussian surface is at r = 2R.

Due to symmetry, gravity acts radially inwards, and the normal to the surface points radially outwards).  So integrand of the LHS of $\triangle$ is:

$\large \vec g \bullet \hat n = |\vec g| cos 180 = -g$ and so the LHS of $\triangle$ becomes: 

$\large -g \int \int_S   ~ da = -g ~ 4 \pi (2R)^2=- g . 16 \pi R^2$

On the RHS of $\triangle$, the mass enclosed by the surface $\Sigma m_{enc} $ is 

$\large \Sigma m_{enc}  = \frac{4}{3} \pi \left( (2R)^3 - R^3 \right ) \rho$ for uniform  density $\rho$

$\large = \frac{28}{3} \pi R^3\rho$

$\triangle$ becomes

$\large  - g . 16 \pi R^2   =  - 4 \pi G \frac{28}{3} \pi R^3\rho$

$\large g  =    \frac{7}{3}  \pi G   R \color{red}{\rho}$

so it's "nearly" b) 


@osprey 
any whoppers in there?!

PS that was a bemused emoticon.

irishboy123 · Answer

I think this works too

from the differential form of Gauss' Law for gravitation

$\large \nabla \bullet \vec g = - 4 \pi G \rho$

and spherical div from Wiki is 

$\large \nabla \bullet \vec A =  {1 \over r^2}{\partial \left( r^2 A_r \right) \over \partial r}
+ {1 \over r\sin\theta}{\partial \over \partial \theta} \left(  A_\theta\sin\theta \right)
+ {1 \over r\sin\theta}{\partial A_\varphi \over \partial \varphi}$

Which shortens due to  fact its a radial field to

$\large   \nabla \bullet \vec g = {1 \over r^2}{\partial \left( r^2 g_r \right) \over \partial r}$


So it's really single variable calculus... 

$\large  {1 \over r^2}{\partial \left( r^2 g_r \right) \over \partial r} = - 4 \pi G \rho$


$\large  {\partial \left( r^2 g_r \right) \over \partial r} = - 4 \pi G \rho r^2$

Integrating


$\large  r^2 g_r   = - \frac{4}{3} \pi G \rho r^3 + C$

$\large  g_r   = - \frac{4}{3} \pi G \rho r + C \frac{1}{r^2}$

Apply the IV

$\large  g_r (R) = 0    = - \frac{4}{3} \pi G \rho R + C \frac{1}{R^2}$

$\large \implies C =  \frac{4}{3} \pi G \rho R^3$

$\large  g_r(r)   =  \frac{4}{3} \pi G \rho \left (  \frac{R^3}{r^2} - r  \right ) $

$\large  g_r(2R)   =  \frac{4}{3} \pi G \rho \left (  \frac{R^3}{(2R)^2} - 2R  \right ) $

$\large =  \frac{4}{3} \pi G \rho R \left (  \frac{1}{4} - 2  \right ) $

$\large = - \frac{7}{3} \pi G \color{red}{\rho} R $

nearly b) again...

osprey · Answer

@IrishBoy123 Wow. I'll trust to your judgement on this one (cop out clause). But I do have a couple of points for you. Firstly, thanks for "niggling me" by mentioning, fairly often, the Gaussian surface idea. Herr G was a pretty smart bloke, and his maths surface has just possibly cleared a mental block, or several thousand, that I've got over gravitation, and various expleteving fluxes. 
Also, I see that you and a few others have what looks like a "smart score" of 99 ... but no one has a score higher than this. My understanding of how this site works is about as valuable as £1, re previous exchange. So, what IS this "score", and how do you get to it it ?
Happy "de-gaussing". (Ugh !)

irishboy123 · Answer

@osprey

i've tried finding a sensible explanation for the SS and so on - can't really see anything but i think it's pretty much all about medals.  i have this:

|dw:1476201440571:dw|

still not sure what it really means, but there is one OS'er with a SS= 100,  it's supposedly summat to do with centiles, so they say.

you may also get credit for testimonials, BTW i'll write you one if you wanna provide the text

thing about physics is there are far fewer questions and far fewer people to dish out medals than maths.  and the mods don't really come over here and they can multiple-medal in a thread.  well, the generous ones do but some are as tight as they are uptight..... you know, duck's arse, gnat's chuff......here and elsewhere.

so whilst aiming for a high SS or medals seems a reasonable thing, you may need to go over to the maths forum and help on school-level stuff, which can be gratifying IF it works. 

then, da-da, if you get 3000 medals you can get a title Professor of Whatever.  but i think only in an area where you help out 

and that means you have spent a lot of time & effort helping people,....,and not that you're ready to kick Hawking or Terry Tao out onto the street

[mah, didn't say that......you do get to choose a Profship @ MIT or Caltech or Zurich, or a decent school such as Imperial or Cambridge if you have any discernment , ahem]

osprey · Answer

Thanks for the treatise. :)
I sort of figured that I was ** against the wind when the "increments" seemed to grind to a halt in the 50s. I wonder what mr and mrs gates are trying to achieve with the idea - kudos, as they put it. Question, though, really, for me is what am I trying to achieve. Answer is helping someone who's in a spot of science bovver, and to try to stop people getting totally bamboozled by things. A a bit of a lot of humour can help of course. 
I've never actually seen a red herring, but this seems to be one. MIT et al await the survivors of the site maybe.
Thanks very much for offering a testimonial. I've thought about this and reckon that it's probably a good idea.  Something along the lines of my interests - physics, laplace transforms and control theory, filter theory, and ballistics. It's a pity that "real" ballistics isn't done much in the already rare physics bit. By real, I mean coriolis, drag, yaw etc.