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Gauss's law is fundamental law of electrostatics. But Can we apply Gauss's law for Gravitational field also?

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    $\begingroup$ Gauß' law is simply that $\int\boldsymbol{\nabla}\cdot\boldsymbol{A}\,\mathrm{d}V = \oint\boldsymbol{A}\cdot\mathrm{d}\boldsymbol{S}$ for any vector field $\boldsymbol{A}$. $\endgroup$ – Walter Jun 10 '15 at 14:18
  • $\begingroup$ Doesn't the link you gave provide your answer? $\endgroup$ – pwf Jun 10 '15 at 18:33
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Indeed, we have that:

$$\nabla \cdot \vec{g}=-4\pi G\rho$$

Where $\rho$ is the mass density. Integrating both sides, we have:

$$\iiint_{\Sigma}(\nabla \cdot \vec{g})\:\mathrm{d}V=-4\pi G\iiint_{\Sigma}\rho(\vec{r})\:\mathrm{d}^{3}\vec{r}$$

But by Gauss' Law (the divergence theorem), we have:

$$\oint_{\partial \Sigma}\vec{g}\cdot\mathrm{d}\vec{A}=-4\pi G\iiint_{\Sigma}\rho(\vec{r})\:\mathrm{d}^{3}r$$

Which is the same as the Gauss' Law for electromagnetism, so we can apply it in the same way.


I thought it might be useful to provide a (admittedly contrived) example of where this is useful.

For instance, if we imagine a narrow tunnel drilled through the earth, and dropping something through it, we note that we can find the gravity at a radius $r$ from the center by applying Gauss' Law:

$$4\pi r^{2} \vec{g} = -4\pi G \rho_{\oplus} \cdot \frac{4 \pi r^{3}}{3}\hat{r}$$

Where $\rho_{\oplus} = \frac{M_{\oplus}}{V_{\oplus}} = \frac{3M_{\oplus}}{4\pi R_{\oplus}^{3}}$ is the density of the earth. Thus:

$$\vec{g} = -\frac{GM_{\oplus}r}{R_{\oplus}^{3}}\hat{r}$$

We note that the force on an object of mass $m$ is given by:

$$\vec{F} = m\vec{g} \implies \frac{\mathrm{d}^{2}\vec{r}}{\mathrm{d}t^{2}}=\vec{g}$$

Therefore, we have:

$$r(t)=R_{\oplus}\cos\left(\sqrt{\frac{GM_{\oplus}}{R_{\oplus}^{3}}}t\right)$$

I.e. the object will exhibit Simple Harmonic Motion about the center of the earth.

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According to my knowledge, Gauss Law can be applied to any function where the quantity(the force) is inversely proportional to distance squared.Now, is the force required to be conservative?

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    $\begingroup$ It is not a particularly good way to answer a question with a question. $\endgroup$ – Gonenc Jun 10 '15 at 16:16

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