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Solved: Gauss’s Law for Gravitation. The gravitational
Chapter 22, Problem 59P(choose chapter or problem)
Gauss’s Law for Gravitation. The gravitational force between two point masses separated by a distance r is proportional to \(1 / r^{2}\), just like the electric force between two point charges. Because of this similarity between gravitational and electric interactions, there is also a Gauss’s law for gravitation.
(a) Let \(\overrightarrow{\boldsymbol{g}}\) be the acceleration due to gravity caused by a point mass m at the origin, so that \(\overrightarrow{\boldsymbol{g}}=-\left(G m / r^{2}\right) \hat{\boldsymbol{r}}\). Consider a spherical Gaussian surface with radius r centered on this point mass, and show that the flux of \(\overrightarrow{\boldsymbol{g}}\) through this surface is given by
\(\oint \overrightarrow{\boldsymbol{g}} \cdot d \overrightarrow{\boldsymbol{A}}=-4 \pi G m\)
(b) By following the same logical steps used in Section 22.3 to obtain Gauss’s law for the electric field, show that the flux of \(\overrightarrow{\boldsymbol{g}}\) through any closed surface is given by
\(\oint \overrightarrow{\boldsymbol{g}} \cdot d \overrightarrow{\boldsymbol{A}}=-4 \pi G M_{\mathrm{encl}}\)
where \(M_{\text {encl }}\) is the total mass enclosed within the closed surface.
Questions & Answers
QUESTION:
Gauss’s Law for Gravitation. The gravitational force between two point masses separated by a distance r is proportional to \(1 / r^{2}\), just like the electric force between two point charges. Because of this similarity between gravitational and electric interactions, there is also a Gauss’s law for gravitation.
(a) Let \(\overrightarrow{\boldsymbol{g}}\) be the acceleration due to gravity caused by a point mass m at the origin, so that \(\overrightarrow{\boldsymbol{g}}=-\left(G m / r^{2}\right) \hat{\boldsymbol{r}}\). Consider a spherical Gaussian surface with radius r centered on this point mass, and show that the flux of \(\overrightarrow{\boldsymbol{g}}\) through this surface is given by
\(\oint \overrightarrow{\boldsymbol{g}} \cdot d \overrightarrow{\boldsymbol{A}}=-4 \pi G m\)
(b) By following the same logical steps used in Section 22.3 to obtain Gauss’s law for the electric field, show that the flux of \(\overrightarrow{\boldsymbol{g}}\) through any closed surface is given by
\(\oint \overrightarrow{\boldsymbol{g}} \cdot d \overrightarrow{\boldsymbol{A}}=-4 \pi G M_{\mathrm{encl}}\)
where \(M_{\text {encl }}\) is the total mass enclosed within the closed surface.
ANSWER:
Solution 59P
Step 1
