91影视

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

The force due to gravity is given by,

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

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,

$\stackrel{鈫�}{g}=\frac{\stackrel{鈫�}{F}}{m}$

from the above equation, which is given by:

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

The gravitational Gauss' law is given as

$桅G=鈭�\stackrel{鈫�}{g}路\stackrel{鈫�}{da}=-4蟺GMin$

$桅$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.

$桅=鈭�\stackrel{鈫�}{E}路\stackrel{鈫�}{da}=\frac{Q}{蔚0}$

Where

$桅$is the Electric Flux

E is the Electric Field

d a is the area

Q is the total charge

$蔚$0 is the permittivity

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

$桅G=鈭�\stackrel{鈫�}{g}路\stackrel{鈫�}{da}=-4蟺GMin$

The constant is given by

$蚁0=\frac{6M}{5蟺R^3}$

$蚁$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:

$dm=蚁dv$

$dm$is the differential mass

$蚁$is the mass density

d v is the differential volume

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

$蚁=蚁0\left(1-\frac{r}{2R}\right)$

The total charge in the sphere can be evaluated as

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

Integrating we get:

$M=鈭�dm=鈭�蚁0\left(1-\frac{r}{2R}\right)\left(4蟺r^2\right)dr$

Where

M is the total mass

r is the radius

Evaluating the integral between 0 and R we get:

$M=4蟺R^3蚁0\left(\frac{5}{24}\right)$

From the above equation we can confirm that

$蚁0=\frac{6M}{5蟺R^3}$