Gravitational self energy:

Let us consider a uniform sphere of radius \( R \), mass \( M \) and the density of the material of the sphere is \( \rho \). Let us consider the uniform sphere is formed by bringing particles one after another from infinity to the positions they occupy in the sphere. The work done in doing this process by the mutual attractive force by the particles is called gravitational self-energy of the sphere.

Since the gravitational force is attractive in nature, in order to break the sphere into parts some amount of external work is done against this gravitational attractive force. So the energy is negative.

Now let us consider at any instant of time the sphere is formed with the radius \( x \). The mass of this sphere is \( \frac{4}{3}\pi{x^3}\rho \). The final radius of the sphere is \( R \). The gravitational potential on the surface of the sphere of radius \( x \) is given by,

\( V=-G\frac{\frac{4}{3}\pi{x^3}\rho}{x} =-G\frac{4}{3}\pi{x^2}\rho \), where \( G \) is the gravitational constant.

Gravitational Potential Due To A Solid Homogeneous Sphere At A Point (i) Outside, (ii) On The Surface, And (iii) Inside A Point Of The Sphere. Read In Detail.

Now the work done in bringing another particle of mass \( dm \) from infinity on the surface of the sphere of radius \( x \) is given by,

\( dw=V\,dm=- G\frac{4}{3}\pi{x^2}\rho \cdot 4\pi{x^2}\,dx\rho \\ -\frac{16}{3}G{\pi}^2{\rho}^2{x^4}\,dx\)

Now the work done in to form the final sphere of radius \( R \) is given by,

\( W=\displaystyle\int_0^R{dw}\\= -\frac{16}{3}G{\pi}^2{\rho}^2 \displaystyle\int_0^R{x^4}\,dx\\= -\frac{16}{3}G{\pi}^2{\rho}^2\frac{R^5}{5}\\=-\frac{3}{5}G{\left(\frac{4}{3}\pi\rho{R^3}\right)}^2\cdot\frac{1}{R}\\=-\frac{3}{5}\frac{GM^2}{R} \)

Where \( M=\frac{4}{3}\pi{R^3}\rho \) =mass of the sphere of radius \( R \).