Question

A spherical planet is discovered with mass M, radius R, and a mass density that varies with radius as rho=rho0(1−r/2R), where rho0 is the density at the center. Determine rho0 in terms of M and R. Hint: Divide the planet into infinitesimal shells of thickness dr, then sum (i.e., integrate) their masses.

Answer 1

Answer:

$\rho_{0}=\frac{6 M}{5 \pi R^{3}}$

Explanation:

Given:

Mass of the spherical planet = M

Radius of the spherical planet = R

$\rho = \rho_0(1-\frac{r}{2R})$

To Find :

$\rho$ in terms of M and R = ?

Solution:

Surface area integral = $\oint \vec{g} \cdot \overrightarrow{d A}=-4 \pi G M$

$g \cdot 4 \pi r^{2}=-4 \pi G \cdot M_{e n c l o s e d}$

$g \cdot r^{2}=G \cdot M_{e n c l o s e d}$

$M_{\text {enclosed}}=\oint_{0}^{r} \rho \cdot 4 \pi r^{2} \cdot d r$

$M_{e n c l o s e d}=\oint_{0}^{r} \rho_{o\left(r^{2}-\frac{r^{3}}{2 R}\right)} \cdot 4 \pi r^{2} . d r$

$M_{e n c l o s e d}=4 \pi \rho_{0} \oint_{0}^{r}\left(r^{2}-\frac{r^{3}}{2 R}\right) d r$

Substituting r = R

$M_{\text {enclosed}}=4 \pi \rho_{0}\left(\frac{R^{3}}{3}-\frac{R^{4}}{8 R}\right)$

$M_{\text {enclosed}}=4 \pi \rho_{0} R^{3}\left(\frac{1}{3}-\frac{1}{8}\right)$

$M_{\text {enclosed}}=4 \pi \rho_{0} R^{3}\left(\frac{8-3}{24}\right)$

$M_{\text {enclosed}}=4 \pi \rho_{0} R^{3}\left(\frac{5}{24}\right)$

$\rho_{0}=\frac{24}{5} \frac{M}{4 \pi R^{3}}$

$\rho_{0}=\frac{6 M}{5 \pi R^{3}}$