Gravitational Self Energy and Gravitational Self Energy of a Sphere
Gravitational self energy is the work done in bringing infinitesimal particles from their infinite separation to the present configuration. \(\rightarrow U=-\frac{1}{2}G{\sum_{i=1}^{n}\sum_{j=1}^{n}}_{i\neq j}\frac{m_im_i}{r_{ij}} \) \(\rightarrow {align*}\vec{F}=\hat{r}f(r) \) \(rightarrow U_s=-\frac{3}{5}\frac{M^{2}}{R}G \)[where M=mass of solid sphere]
Summary
Gravitational self energy is the work done in bringing infinitesimal particles from their infinite separation to the present configuration. \(\rightarrow U=-\frac{1}{2}G{\sum_{i=1}^{n}\sum_{j=1}^{n}}_{i\neq j}\frac{m_im_i}{r_{ij}} \) \(\rightarrow {align*}\vec{F}=\hat{r}f(r) \) \(rightarrow U_s=-\frac{3}{5}\frac{M^{2}}{R}G \)[where M=mass of solid sphere]
Things to Remember
Gravitational Self-energy is the work done to form a system of N number of particles.
\begin{align*}\rightarrow U_s=-\frac{3}{5}\frac{M^{2}}{R}G \end{align*} (-ve sign indicates that this much energy is actually evolveds or released during the process of assembling the solid sphere.)
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Gravitational Self Energy and Gravitational Self Energy of a Sphere
Gravitational Self Energy:

It is the work done in bringing infinitesimal particles from their infinite separation to the present configuration. It is the work done to form a system of N number of particles is given by
\begin{align*}U=-\frac{1}{2}G{\sum_{i=1}^{n}\sum_{j=1}^{n}}_{i\neq j}\frac{m_im_i}{r_{ij}} \end{align*}where \(m_i\)=mass of \(i^{th}\) particle
\(m_j\)=mass of \(j^{th}\) particle
\(r_{ij}\)= separation between \(i^{th}\) and \(j^{th}\) particle
Consider a system of three particles having masses \(m_1\), \(m_2\) and \(m_3\) with respective separation \(r_{12}\),\(r_{13}\) and \(r_{23}\) as shown in figure. Let \(m_1\) as the reference particle. The work done in bringing the mass m2 in the field of mass m1 is\begin{align*}U_{12}=-\frac{Gm_1m_2}{r_{12}} \end{align*}
Again work done if the third particle \(m_3\) is brought in the field of\(m_1\) and \(m_2\), the respective work done are \begin{align*} \end{align*}\begin{align*}U_{13}=-\frac{Gm_1m_3}{r_{13}}\space and U_{23}=-\frac{Gm_2m_3}{r_{23}}\end{align*}The total work done in forming the system is given by\begin{align*}U=U_1+U_2+U_3 \end{align*}\begin{align*}=-\frac{Gm_1m_2}{r_{12}}-\frac{Gm_1m_3}{r_{13}}-\frac{Gm_2m_3}{r_{23}} \end{align*}\begin{align*}=-G[{\frac{m_1m_2}{r_{12}}+\frac{m_1m_3}{r_{13}}+\frac{m_2m_3}{r_{23}} }] \end{align*}In summation notation,\begin{align*}U=-G{\sum_{i=1}^{3}\sum_{j=1}^{3}}_{i\neq j}\frac{m_im_j}{r_{ij}} \end{align*}Here, each term is repeated twice, so for correct expression\begin{align*}U=-\frac{1}{2}G{\sum_{i=1}^{3}\sum_{j=1}^{3}}_{i\neq j}\frac{m_im_j}{r_{ij}} \end{align*}Thus for the system of n numner of particles, the gravitational self energy is given by\begin{align*}U=-\frac{1}{2}G{\sum_{i=1}^{n}\sum_{j=1}^{n}}_{i\neq j}\frac{m_im_j}{r_{ij}} \end{align*}
Gravitational Self Energy of a Sphere:

The gavitational self energy \(U_s\) of a uniform solid sphere is equal to the amount of work done in assembling together its infinitesimal particles initially lying infinite distance apart.
Let us consider the sphere to be formed by continuous deposition of mass particles in the form of successive spherical shells around an inner spherical core of radius r until it becomes a full-fledged solid sphere of radius R as shown in figure.
If \(\rho \) be the density of the material of the sphere and hence of the spherical core, we have
mass of the inner core=volume.density=\(\frac{4}{3}\pi r^{3}\rho \)
And, if the thickness of the spherical shell deposited on it be dr, we have
Mass of the shell=its surface area \(\times \) its thickness \(\times \) density \begin{align*}4\pi r^{2}dr\rho \end{align*}
\(\therefore \) Change in P.E. or Self energy of the core due to additional mass deposited on it, say
\(dU_s\)=potential of the core(i.e. its P.E. per unit mass)\(\times \)additional mass\begin{align*}=-\frac{mass\space of\space the\space core}{radius\space of\space the\space core}G.4\pi r^{2}dr\rho \end{align*}\begin{align*}=-\frac{(\frac{4}{3}\pi r^{3}\rho )(4\pi r^{2}dr\rho)G}{r} \end{align*}\begin{align*}=-\frac{16}{3}\pi ^{2}\rho ^{2}Gr^{4}dr \end{align*}This is the increase in the self energy of the core on its radius increases from r to (r+dr).
The integral of this expression for \(dU_s\), between limit r=0 and r=R, then give the self energy of the whole sphere on being built from the very start i.e.,\begin{align*}U_s=-\int_{0}^{R}\frac{16}{3}\pi ^{2}\rho ^{2}Gr^{4}dr \end{align*}\begin{align*}=-\frac{16}{3}\pi ^{2}\rho ^{2}G[\frac{r^{5}}{5}]_{0}^{R} \end{align*}\begin{align*}=-\frac{16\pi ^{2}\rho ^{2}R^{5}G}{15} \end{align*}\begin{align*}=-\frac{3}{5}\frac{(\frac{4}{3}\pi r^{3}\rho )^2}{R}G \end{align*}\begin{align*}=-\frac{3}{5}\frac{M^{2}}{r}G[\because \frac{4}{3}\pi r^{3}\rho =M,mass\space of\space solid\space sphere] \end{align*}
\(\therefore \)gravitational self energy of the solid sphere,\begin{align*}U_s=-\frac{3}{5}\frac{M^{2}}{R}G \end{align*}-ve sign indicates that this much energy is actually evolveds or released during the process of assembling the solid sphere.
References
Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.
Feynman, Richard P. The Feynman Lectures on Physics Volume 1. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.
Mathur, D S. Mechanics. New Delhi: S. Chand & Company Pvt. Ltd., 2015.
Young, Hugh D, Roger A Freedman and A Lewis Ford. University Physics. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.
Lesson
Gravitational Potential And Field
Subject
Physics
Grade
Bachelor of Science
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