The gravitational potential, U, can be calculated at any point, [tex]\[(adsbygoogle = window.adsbygoogle || []).push({});

{\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/physics/GravitationalPotential.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]

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# Homework Help: Computing gravitational potential for a point inside the distribution

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