The flux of overrightarrow-g- through this surface is given by -displaystyle oint - overrightarrow-g- cdot doverrightarrow-s- - 4pi Gmdisplaystyle oint - overrightarrow-g- cdot doverrightarrow-s- - -4pi Gmdisplaystyle oint - overrightarrow-g- cdot doverrightarrow-s- - Gmdisplaystyle oint - overrightarrow-g- cdot doverrightarrow-s- - - Gm

Question

Toppr Admin · Accepted Answer

From the inverse square dependence of forces, we can have an analogy of electric and gravitational forces.
The charge is analogous to mass.

Gravitation states $F \propto - m_{1} m_{2}$ and in Electrostatics, $F \propto q_{1} q_{2}$

The proportionality constant is $G$ in gravitation and $\frac{1}{4 π ϵ_{0}}$ in electrostatics.

Thus, $G \sim \frac{1}{4 π ϵ_{0}}$ and $m \sim q$

Gauss' law for electrostatics is $\oint \to E . \to d s = \frac{q_{encl}}{ϵ_{0}} = 4 π \frac{q_{encl}}{4 π ϵ_{0}}$

The analogous quantity for electric field is $\to g$

Thus, the Gauss' law for gravitation is $\oint \to g . \to d s = - 4 π G m_{encl}$

But, the mass enclosed in the surface is $m$
Thus, $\oint \to g . \to d S = - 4 π G m$

The flux of →g through this surface is given by :∮→g⋅d→s=4πGm∮→g⋅d→s=−4πGm∮→g⋅d→s=Gm∮→g⋅d→s=−Gm

The flux of $\to g$ through this surface is given by :
$\oint \to g \cdot d \to s = 4 π G m$
$\oint \to g \cdot d \to s = - 4 π G m$
$\oint \to g \cdot d \to s = G m$
$\oint \to g \cdot d \to s = - G m$