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Michele_Laino
 one year ago
New Tutorial:
On the Gravitational Field of an Object with Spherical Symmetry
(Advanced Level)
Michele_Laino
 one year ago
New Tutorial: On the Gravitational Field of an Object with Spherical Symmetry (Advanced Level)

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.11Let's start from the Poisson equation for gravity: \[\large \Delta \phi = 4\pi K\rho \qquad (1)\] where \( \large \phi \) is the potential function, namely the potential energy of an object with unitary mass, \( \large K\) is the Newton's gravitational constant, and \( \large \rho \) is the density of ponderable matter. By definition of operator \( \large \Delta={\nabla ^2} \), we can rewrite equation \( \large (1)\) as below: \[\large {\text{div}}\left( {{\text{grad}}\phi } \right) = 4\pi K\rho \] Now we can define a new vector field \( \large {\mathbf{g}}\) as below: \[\large {\mathbf{g}} = {\text{grad}}\phi \] therefore, after a substitution into equation \( \large (1)\) we have: \[\large {\text{div }}{\mathbf{g}} = 4\pi K\rho \qquad (2) \] As we can see the equation $(3)$ is similar to the Maxwell equation for electric field ${\mathbf{E}}$ and electric charge density $\rho $: \[\large {\text{div }}{\mathbf{E}} = 4\pi \rho \] Next, if we apply the \(divergence \; theorem\) to equation \( \large (2)\), we get: \[\large \Phi \left( {\mathbf{g}} \right) = 4\pi KM \qquad (3)\] where \( \large M \) is the mass of our spherical object. We can think to the equation \( \large (3)\) like the \(Gauss \; Theorem\) for gravitational field \(\large {\mathbf{g}}\). Let's suppose that our spherical object has radius \( \large r=a\), and mass \( \large M\), then we can subdivide the entire space in two regions; namely the region \( \large \#1\) of the points of our sphere, and the region \( \large \#2\) of the points external to our sphere. dw:1441532029014:dw Next let's consider a sphere inside region \( \large \#1\) whose radius \( \large r_1\) : by geometrical consideration, we can say that the field \( \large {\mathbf{g}}\) is a field with a central symmetry, so, developing equation \( \large (3) \), we get: \[\large 4\pi r_1^2g\left( r \right) = 4\pi K\frac{{4\pi }}{3}\rho r_1^3\] where \[\large \rho = \frac{M}{{\frac{{4\pi }}{3}{a^3}}}\] is the density of our spherical object, and the subsequent quantity: \[\large \frac{{4\pi }}{3}\rho r_1^3\] is the mass contained inside the sphere of radius \( \large r=r_1\). After a simplification, we get: \[\large g\left( r \right) = \frac{{4\pi }}{3}K\rho r, \qquad r \leqslant a\] namely the gravitational field, at each point inside our spherical object, depends linearly on the distance from its center. Now, let's consider a concentric sphere with our object, whose radius is \( \large r=r_2\), again developing equation \( \large (3)\), we get: \[\large 4\pi r_2^2g\left( r \right) = 4\pi KM\] and finally: \[\large g\left( r \right) = \frac{{KM}}{{{r^2}}}, \qquad r>a\] namely, the gravitational field, at each point outside our spherical object, goes like \( \large 1/r^2\), being \( \large r \) the distance of such point from the center of our spherical object, as predicted by \(Newton\) law. those results can be summarized in the subsequent picture: dw:1441532564231:dw What about of the potential function \( \large \phi \left( r \right)\)? Developing the equation: \[\large {\mathbf{g}} = {\text{grad}}\phi \] using the formula of \(grad\) operator in polar coordinates, and the expression of gravitational field for points inside our spherical object, we get: \[\large \frac{{\partial {\phi _1}}}{{\partial r}} = \frac{{4\pi }}{3}K\rho r = \frac{{KM}}{{{a^3}}}r\] and, after one integration, we can write: \[\large \phi_1 \left( r \right) = \frac{{KM}}{{{a^3}}}\frac{{{r^2}}}{2} + {c_1},\quad r \leqslant a\] being \(c_1\) the corresponding constant of integration. Whereas, if we repeat the same procedure fusing the expression for the field \(\large {\mathbf{g}}\) for point outside our spherical object, we have: \[\large \frac{{\partial {\phi _2}}}{{\partial r}} = \frac{{KM}}{{{r^2}}}\] again, after a simple integration, we get this: \[\large {\phi _2}\left( r \right) =  \frac{{KM}}{r} + {c_2},\quad r > a\] where \( \large c_2\) is another constant of integration. It is reasonable to request that, in the limit of large distances: $r \to \infty $, the potential function, goes to zero, hence we can write: \[\large {c_2} = 0\] so the potential function at points outside our sphere, can be rewritten as follows: \[\large {\phi _2}\left( r \right) =  \frac{{KM}}{r}\] It is reasonable to request that the potential function has to be a continuous function, namely at points at the boundary of our spherical object (\( \large r=a\) ) the subsequent condition holds: \[\large {\phi _1}\left( a \right) = {\phi _2}\left( a \right)\] hence: \[\large \frac{{KM}}{{{a^3}}}\frac{{{a^2}}}{2} + {c_1} =  \frac{{KM}}{a}\] therefore: \[\large {c_1} =  \frac{3}{2}\frac{{KM}}{a}\] with that value for constant \( \large c_1\), the potential function \( \large {\phi _1}\) becomes: \[\large {\phi _1}\left( r \right) = \frac{{KM}}{{2a}}\left( {{{\left( {\frac{r}{a}} \right)}^2}  3} \right),\quad r \leqslant a\] Summarizing our results, we can write the potential function of our spherical object, like below: \[\large \phi \left( r \right) = \left\{ {\begin{array}{*{20}{c}} \begin{gathered} \frac{{KM}}{{2a}}\left( {{{\left( {\frac{r}{a}} \right)}^2}  3} \right),\quad r \leqslant a \hfill \\ \hfill \\ \end{gathered} \\ {  \frac{{KM}}{r},\quad r > a} \end{array}} \right.\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.11@ganeshie8 , @Astrophysics , @IrishBoy123 , @Empty , @rvc , @Robert136 , @iambatman , @Nishant_Garg

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Looks Interesting! Maybe I'll get a chance to study this later in my course?:p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I love it! Analysis like this motivates me to investigate further!!!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.11thanks! :D @rvc

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.11yes! I think so! @Nishant_Garg

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.11thanks! :) @Robert136

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for sharing @Michele_Laino :)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.11thanks!! :) @Astrophysics

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0bloomin' marvellous :))

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.11thanks! :) @IrishBoy123

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/study#/updates/55eedde4e4b0445dfd1a5322

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0This helped me revise what I learned in my Electromagnetism course last semester!! Thank you @Michele_Laino :D

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.11Thanks!! :) @Jhannybean
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