Question

Newton’s law of gravity and Coulomb’s law are both inversesquare laws. Consequently, there should be a “Gauss’s law for gravity.” a. The electric field was defined as E u = F u on q /q, and we used this to find the electric field of a point charge. Using analogous reasoning, what is the gravitational field g u of a point mass?

Write your answer using the unit vector nr, but be careful with signs; the gravitational force between two “like masses” is attractive, not repulsive. b. What is Gauss’s law for gravity, the gravitational equivalent of Equation 24.18? Use ΦG for the gravitational flux, g u for the gravitational field, and Min for the enclosed mass. c. A spherical planet is discovered with mass M, radius R, and a mass density that varies with radius as r = r011 - r/2R2, where r0 is the density at the center. Determine r0 in terms of M and R. Hint: Divide the planet into infinitesimal shells of thickness dr, then sum (i.e., integrate) their masses. d. Find an expression for the gravitational field strength inside the planet at distance r 6 R.

Short answer

Hence, the gravitational field is

$\overrightarrow{g}=\frac{-GM\overrightarrow{r}}{r^3}$

The gravitational Gauss' law is given as

$\Phi G=\oint \overrightarrow{g}·\overrightarrow{da}=-4\pi GMin$

The constant of the charge density is given by

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

The gravitational field inside the sphere is found to be

$g=\frac{-6GM}{5\pi R^3}{r}^1\left(\frac{1}{3}-\frac{r}{8R}\right)$

Step-by-step solution

part (a) step 1: given information

The gravitational field as an analog to the electric field is given as

The force due to gravity is given by,

$\overrightarrow{F}=\frac{-GMm\overrightarrow{r}}{r^3}$

F is the gravitational force

r is the distance

M is the mass of the large body

m is the mass of the test body

G is the gravitational constant

If the electric field is given by $\vec{E}=\frac{\vec{F}}{q}$ where $E$ is the Electric Field, $F$ is the Electric Field and $q$ is the total charge, the by the same logic, gravitational field can be calculated by,

$\overrightarrow{g}=\frac{\overrightarrow{F}}{m}$

from the above equation, which is given by:

$\overrightarrow{g}=\frac{-GM\overrightarrow{r}}{r^3}$

part (b) step 1: given information

The gravitational Gauss' law is given as

$\Phi G=\oint \overrightarrow{g}·\overrightarrow{da}=-4\pi GMin$

$\Phi$G is the gravitational Flux

g is the gravitational Field

d a is the area

Min is the total mass

G is the gravitational constant

Explanation of Solution

The gravitational analog of Gauss' Law can be derived using the following Gauss law for electric fields.

$\Phi =\oint \overrightarrow{E}·\overrightarrow{da}=\frac{Q}{\epsilon 0}$

Where

$\Phi$is the Electric Flux

E is the Electric Field

d a is the area

Q is the total charge

$\epsilon$0 is the permittivity

For a spherical mass d a evaluates to 4$\pi r^2$. So on the left hand side we can divide by $4\pi to-GM$to recover $\overrightarrow{g}=\frac{-GM\overrightarrow{r}}{r^3}.$The gravitational Gauss' law is given as

$\Phi G=\oint \overrightarrow{g}·\overrightarrow{da}=-4\pi GMin$

part (c) step 1: given information

The constant is given by

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

$\rho$is the mass density

M is the total mass

R is the radius of sphere

Explanation of Solution

The mass inside the sphere is given by the expression:

$dm=\rho dv$

$dm$is the differential mass

$\rho$is the mass density

d v is the differential volume

Since the mass density $\rho$for a radius of sphere R is given by

$\rho =\rho 0\left(1-\frac{r}{2R}\right)$

The total charge in the sphere can be evaluated as

$dm=\rho 0\left(1-\frac{r}{2R}\right)\left(4\pi r^2\right)dr$

Integrating we get:

$M=\int dm=\int \rho 0\left(1-\frac{r}{2R}\right)\left(4\pi r^2\right)dr$

Where

M is the total mass

r is the radius

Evaluating the integral between 0 and R we get:

$M=4\pi R^3\rho 0\left(\frac{5}{24}\right)$

From the above equation we can confirm that

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