Gravitational Flux and Poisson's equation of Gravitational Field

Gravitational Flux:

The total gravitational field flowing out of a given surface area is called gravitational flux. If E be the gravitational field intensity at a point on a surface of the surface area A then the gravitational flux is given by

\(\Phi = \vec{E}.\vec{A}\)

Gauss Law of Gravitational Field Intensity:

It states that the total gravitational flux over a closed surface enclosing certain mass M is equal to \(4\pi G\) times the mass enclosed.

\(\Phi = -4\pi GM\)

Consider a surface enclosing a point mass M. Let us take any point P on the surface at a distance r from the point mass. Thus the gravitational field intensity at a point P due to mass M is

\(E=-\frac{GM}{r^{2}}\)

Let us take an elementary surface area dA on the surface S such that the point P lies inside it. Let \(\hat{n}\) be the unitnormal to the surface dA such that E makes an angle \(\theta \) with \(\hat{n}\).

\(\therefore E_n=\vec{E}.\hat{n}\)=Ecos\(\theta\)

The gravitational flux through the surface area dA is \(d\Phi \)=\(E_ndA\)\begin{align*}=Ecos\theta dA\end{align*}

Then the total flux through the whole surface is given by

\begin{align*}\Phi =\oint d\Phi \end{align*}\begin{align*}=\oint Ecos\theta dA \end{align*}\begin{align*}=\oint -\frac{GM}{r^{2}}cos\theta dA \end{align*}\begin{align*}=-GM\oint frac{cos\theta dA}{r^{2}} \end{align*}\begin{align*}=-GM\oint d\Omega \end{align*}

Where \( d\Omega =\frac{cos\theta dA}{r^{2}}\) the solid angle subtended by the surface dA at O where \(\Omega = 4\pi \) is the solid angle subtended at O by the surface S.

\( \Phi =-4\pi GM\) Proved.\begin{align*} \end{align*}

Poisson's equation of Gravitational Field:

We have the expression of gravitational flux as given by Gauss law is

\(\Phi =-4\pi G\).Mass enclosed \begin{align*}=-4\pi G.\rho \int_v \rho .dV\rightarrow eqn I \end{align*}Again, by the definition of gravitational flux\begin{align*}\Phi =\vec{E}.\vec{A} \end{align*}\begin{align*}=\oint_s\vec{E}.\hat{n}dA \end{align*}\begin{align*}=\int_v(div\vec{E})dV\rightarrow eqnII\end{align*}Equating Eqn(I) and eqn (II) we get\begin{align*}-4\pi G\rho \int_vdV=\int_v(\bigtriangledown.\vec{E})dV \end{align*}\begin{align*}\bigtriangledown.\vec{E}=-4\pi G\rho i.e.\space Poisson's\space equation.\end{align*}

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.