91Ó°ÊÓ

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

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

where

g is the gravitational force

r is the distance

M is the mass of the large body

G is the gravitational constant

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

The force due to gravity is given by,

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

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 $\stackrel{→}{E}=\frac{\stackrel{→}{F}}{q}$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, $\stackrel{→}{g}=\frac{\stackrel{→}{F}}{m}$from the above equation, which is given by:

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

The gravitational Gauss' law is given as

$\mathrm{Φ}G=âˆ\circledR \stackrel{→}{g}·\stackrel{→}{da}=-4\mathrm{Ï€}GM\text{in}$

Where

$\mathrm{Φ}G$is the gravitational Flux

g is the gravitational Field

d a is the area

Min is the total mass

G is the gravitational constant

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

$\mathrm{Φ}=âˆ\circledR \stackrel{→}{E}·\stackrel{→}{da}=\frac{Q}{\mathrm{Î\mu }0}$

Where

$\mathrm{Φ}$ is the Electric Flux

E is the Electric Field

d a is the area

Q is the total charge

$\mathrm{Î\mu }$0 is the permittivity

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

$\mathrm{Φ}G=âˆ\circledR \stackrel{→}{g}·\stackrel{→}{da}=-4\mathrm{Ï€}\text{GMin}$

The constant is given by

$\mathrm{ÒÏ}$is the mass density

M is the total mass

R is the radius of sphere

The mass inside the sphere is given by the expression:

$dm=\mathrm{ÒÏ}dv$

d m 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

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

The total charge in the sphere can be evaluated as

$dm=\mathrm{ÒÏ}0\left(1-\frac{r}{2R}\right)\left(4\mathrm{Ï€}{r}^2\right)dr$

Integrating we get:

$=∫dm=∫\mathrm{ÒÏ}0\left(1-\frac{r}{2R}\right)\left(4\mathrm{Ï€}{r}^2\right)dr$

Where

M is the total mass

r is the radius

Evaluating the integral between 0 and R we get:

$M=4\mathrm{Ï€}{R}^3\mathrm{ÒÏ}0\left(\frac{5}{24}\right)$

From the above equation we can confirm that

$\mathrm{ÒÏ}0=\frac{6M}{5\mathrm{Ï€}{R}^3}$