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

`\rho_(0)=(6 M)/(5 \pi R^(3))`

Step-by-step explanation:

Given:

Mass of the spherical planet = M

Radius of the spherical planet = R

`\rho = \rho_0(1-(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)-(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)-(r^(3))/(2 R)\right) d r`

Substituting r = R

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

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

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

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

`\rho_(0)=(24)/(5) (M)/(4 \pi R^(3))`

`\rho_(0)=(6 M)/(5 \pi R^(3))`