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

1. Michele_Laino

Let'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: $\large \phi \left( r \right) = \left\{ {\begin{array}{*{20}{c}} \begin{gathered} \frac{{KM}}{{2a}}\left( {{{\left( {\frac{r}{a}} \right)}^2} - 3} \right),\quad r \leqslant a \hfill \\ \hfill \\ \end{gathered} \\ { - \frac{{KM}}{r},\quad r > a} \end{array}} \right.$

2. Michele_Laino

@ganeshie8 , @Astrophysics , @IrishBoy123 , @Empty , @rvc , @Robert136 , @iambatman , @Nishant_Garg

3. anonymous

Looks Interesting! Maybe I'll get a chance to study this later in my course?:p

4. anonymous

I love it! Analysis like this motivates me to investigate further!!!

5. rvc

You r awesome as usual

6. Michele_Laino

thanks! :D @rvc

7. Michele_Laino

yes! I think so! @Nishant_Garg

8. Michele_Laino

thanks! :) @Robert136

9. Astrophysics

Thanks for sharing @Michele_Laino :)

10. Michele_Laino

thanks!! :) @Astrophysics

11. IrishBoy123

bloomin' marvellous :-))

12. Michele_Laino

thanks! :) @IrishBoy123

13. IrishBoy123
14. Jhannybean

This helped me revise what I learned in my Electromagnetism course last semester!! Thank you @Michele_Laino :D

15. Michele_Laino

Thanks!! :) @Jhannybean