Question

QuestionDetails:

Answer 1

Answer:

The sketch for the Gravitational force F and the potential energy U are attached to this answer.

Explanation:

To obtain the gravitational force, we can consider the gravitational field GF(r) as:

$\vec{F_{g}}=m_{par} \cdot \vec{G_F(r)}$

To calculate the gravitational field we can use the Gauss theorem. By considering a homogeneous mass of the sphere (constant density) and the spherical symmetry, we can determinate than the gravitational field direction is $-\hat{r}$.

Considering a constant density:

$Vol=\pi\frac{4}{3}(a^3-b^3)$

$\rho=\frac{m_{sph}}{Vol} =\frac{3m_{sph}}{4\pi(a^3-b^3)}$

Applying a spherical gaussian surface for different radius r:

for R<b:

$\displaystyle\oint_{S} \vec{G_{f}}\, \vec{ds}=-4m_{int}G=0N/kg \rightarrow G_{f}=0N/kg \rightarrow F_{g}=0N$

for b<R<a:

$\displaystyle\oint_{S} \vec{G_{f}}\, \vec{ds}=-4\pi m_{int}G\\\displaystyle\oint_{S} -G_{f}\, ds=-4\pi G\displaystyle\int_{} \int_{} \int_{V} \rho\, dV\,\,\\\\-4\pi r^2 G_{f}(r)=-4\pi Gm_{sph}\frac{(r^3-b^3)}{(a^3-b^3)} \\G_{f}(r)=Gm_{sph}\frac{(r^3-b^3)}{r^2(a^3-b^3)} \\F_g =Gm_{par}m_{sph}\frac{(r^3-b^3)}{r^2(a^3-b^3)}(-\hat{r})$

for a<R:

$\displaystyle\oint_{S} \vec{G_{f}}\, \vec{ds}=-4\pi m_{int}G\\\\-4\pi r^2 G_{f}(r)=-4\pi Gm_{sph} \\ F_g =Gm_{par}m_{sph}\frac{1}{r^2}(-\hat{r})$

For the potential energy you can integrate the field to obtain the gravitational potential and the multiplying for the particle mass:

for a<R:

$U(r)=G\frac{m_{sph}m_{par}}{r}$

for b≤R≤a:

$U(r)=G\frac{m_{sph}m_{par}}{a} +G\frac{m_{sph}m_{par}}{(a^3-b^3)} (\frac{a^{2}}{2}+\frac{b^{2}}{a}-\frac{r^{2}}{2}-\frac{b^{2}}{r})$

for R≤b:

$U(r)=G\frac{m_{sph}m_{par}}{a} +G\frac{m_{sph}m_{par}}{(a^3-b^3)} (\frac{a^{2}}{2}+\frac{b^{2}}{a}-\frac{b^{2}}{2}-b)$