Computing gravitational potential for a point inside the distributionby Mike2 Tags: computing, distribution, gravitational, inside, point, potential 

#1
Nov205, 09:02 AM

P: 1,308

The gravitational potential, U, can be calculated at any point, [tex]\[
{\rm{\vec r}} \][/tex], for a mass density distribution, [tex]\[ {\rm{\rho (r)}} \][/tex], using the formula: [tex]\[ {\rm{U =  }}\int_{{\rm{all space}}} {G\frac{{{\rm{\rho }}({\rm{\vec r')}}}}{{\left {{\rm{ \vec r  \vec r' }}} \right}}} \,\,\,d^3 {\rm{r'}} \][/tex]. See: http://scienceworld.wolfram.com/phys...Potential.html My question is how is this calculated for points inside the distribution. For points outside the distribution, [tex]\[ {\rm{\rho (r)}} \][/tex] is zero, and there is no problem. But inside the distribution where [tex]\[ {\rm{\rho (r)}} \][/tex] is not zero, there will be points where [tex]\[ {{\rm{\vec r  \vec r' }}} \] [/tex] does become zero and send the integrand to infinity. Wouldn't this cause a problem? And how would you work around it? Thanks. Could it be as simple as adding a small constant in the denominator that goes to zero after the integration, such as: [tex]\[ {\rm{U(\vec r) = }}\mathop {\lim }\limits_{\varepsilon \to 0} {\rm{(  }}\int_{{\rm{all space}}} {G\frac{{{\rm{\rho }}({\rm{\vec r')}}}}{{\left {{\rm{ \vec r  \vec r' }}} \right + \varepsilon }}} \,\,\,d^3 {\rm{r')}} \][/tex] 



#2
Nov205, 02:19 PM

Mentor
P: 28,783

Er... no. One uses techniques such as residues to do such integration that contains poles.
Zz. 



#3
Nov205, 04:30 PM

P: 1,308

Are you referring to a Stoke's Theorm or a Divergence Theorm to turn integration throughout a volume which has an infinity into a surface integral that surrounds the infinity and so does not integrate through the infinity? 



#4
Nov205, 04:54 PM

Mentor
P: 28,783

Computing gravitational potential for a point inside the distributionZz. 



#5
Nov505, 08:24 PM

P: 1,308




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