Gravitation Concept Page - 4

Example

Gravitational field due to a continuous mass system with variable mass density

Example: The earth does not have a uniform density; it is most dense at its centre and least dense at its surface. Anapproximation of its density is p(r)=(A−Br)$p(r)=(A-Br)$ , where A=12700$A=12700$kg/m3,B=1.5×10−3kg/m3${}^3,B=1.5\times {10}^{-3}kg/m^3$ and r$r$ is the  distance from the centre of earth. Use R=6.4×106m$R=6.4\times {10}^{6}m$ for the radius of earth approximated as a sphere,Imagine dividing the earth into concentric, elementary spherical shells. Each shell has radius r$r$, thickness dr$dr$,volume dV$dV$ and mass dm=p(r)dV$dm=p(r)dV$. By integrating dm$dm$ from zero to R$R$  the mass of earth can be found. Knowing the fact that a uniform spherical shell gives no contribution to acceleration due to gravity inside it,we can also find g$g$ as a function of r$r$. What is the gravitational field due this mass of the earth?

Solution:
Mass of earth:
M=∫R0(A−Br)4πr2dr=43πR3(A−34BR)$M={\int }_0^R(A-Br)4\pi r^{2}dr={\displaystyle \frac{4}{3}}\pi R^3(A-{\displaystyle \frac{3}{4}}BR)$

Gravitational field:
Eg=GMr2$E_g={\displaystyle \frac{GM}{r^2}}$

Example

Gravitational potential due to a continuous mass

Example:
A thin uniform annular disc (see figure) of mass M$M$ has outer radius 4R$4R$ and inner radius 3R$3R$. Find the work required to take a unit mass from point P$P$ on its axis to infinity.Solution:
We know that the work required to take a unit mass from P to infinity =−Vp$-V_p$, where −Vp$-V_p$ is the gravitational potential at P due to the disc. To find Vp$V_p$, we divide the disc into small elements, each of thickness dr$dr$. Consider one such element at a distance r$r$ from the center of the disc as shown.
Mass of the element dm=M(2πrdr)π(4R)2−π(3R)2$dm={\displaystyle \frac{M(2\pi rdr)}{\pi (4R{)}^2-\pi (3R{)}^2}}$
=2Mrdr7R2$={\displaystyle \frac{2Mrdr}{7R^2}}$
Thus
Vp=−∫4R3RGdmr2+16R2−−−−−−−−√$V_p=-{\displaystyle {\int }_{3R}^{4R}{\displaystyle \frac{Gdm}{\sqrt{r^2+16R^2}}}}$

=−2MG7R2∫4R3Rrdr(r2+16R2)1/2$=-{\displaystyle \frac{2MG}{7R^2}}{\displaystyle {\int }_{3R}^{4R}{\displaystyle \frac{rdr}{(r^2+16R^2{)}^{1/2}}}}$

Putting r2+16R2=x2$r^2+16R^2=x^2$,we get 2rdr=2xdx$2rdr=2xdx$ or rdr=xdx$rdr=xdx$

When r=3R,x=9R2+16R2−−−−−−−−−−√=5R$r=3R,x=\sqrt{9R^2+16R^2}=5R$

When r=4R,x=16R2+16R2−−−−−−−−−−√=42√R$r=4R,x=\sqrt{16R^2+16R^2}=4\sqrt{2}R$ 

Vp=−2MG7R2∫42√R5Rdx=−2MG7R2(42√−5)R$V_p=-{\displaystyle \frac{2MG}{7R^2}}{\displaystyle {\int }_{5R}^{4\sqrt{2}R}dx=-{\displaystyle \frac{2MG}{7R^2}}(4\sqrt{2}-5)R}$
or
−Vp=2GM7R(42√−5)$-V_p={\displaystyle \frac{2GM}{7R}}(4\sqrt{2}-5)$

Example

Force between continuous bodies

Example:
Find the gravitational force of attraction between a uniform sphere of mass M$M$ and a uniform rod of length l$l$ and mass m$m$ oriented as shown.
Solution:The force of attraction between sphere and shaded position dF=GM(mldx)x2${\displaystyle dF=GM{\displaystyle \frac{\left({\displaystyle \frac{m}{l}}dx\right)}{x^2}}}$
F=∫r+1rGMmlx2dx=GMml∫r+1r1x2dx${\displaystyle F={\int }_r^{r+1}{\displaystyle \frac{GMm}{l{x}^2}}dx={\displaystyle \frac{GMm}{l}}{\int }_r^{r+1}\frac{1}{x^2}dx}$

=GMml∫r+1rx−2dx=GMml[x−2+1−2+1]r+1r${\displaystyle ={\displaystyle \frac{GMm}{l}}{\int }_r^{r+1}{x}^{-2}dx=\frac{GMm}{l}{\left[\frac{x^{-2+1}}{-2+1}\right]}_r^{r+1}}$

=−GMml[x−1]r+1r=−GMml[1x]r+1r=GMmr(r+1)${\displaystyle =-\frac{GMm}{l}{\left[x^{-1}\right]}_r^{r+1}=-\frac{GMm}{l}{\left[\frac{1}{x}\right]}_r^{r+1}=\frac{GMm}{r(r+1)}}$

Definition

Gravitational Field due to a point mass

Gravitational field due to a point mass is given by the following relation:

Definition

Gravitational Field due to solid sphere

Gravitational Field at an external point :
E=GMr2$E={\displaystyle \frac{GM}{r^2}}$
Gravitational Field at an internal point :
E=GMR3×r$E={\displaystyle \frac{GM}{R^3}}\times r$ (Where 'R' is radius of earth.)

Diagram

Graph for gravitational field due to Solid Sphere

Graph for gravitational field due to solid sphere:

Example

Vector sum of gravitational forces on a mass due to a system of masses

Let's consider a mass m$m$ is situated at the centre of a circle. Two more equal masses 2m$2m$ are at the diametric opposite points on the circle. Then net force on mass m$m$ will be:
G.m.2mr2${\displaystyle \frac{G.m.2m}{r^2}}$ - G.m.2mr2=0${\displaystyle \frac{G.m.2m}{r^2}}=0$

Equal forces in opposite directions will cancel out each other.

Definition

Gravitational Field due to a uniform ring

Gravitational Field due to a uniform ring is given by:
E=GMr(a2+r2)3/2$E={\displaystyle \frac{GMr}{(a^2+r^2{)}^{3/2}}}$

Definition

Define and use analogous gauss's law for gravitation

Gauss's Law for Gravity:
The total electric flux out of a region was related to the charge in that region:
ϕ=∫E⋅n^dA$\varphi =\int E\cdot \hat{n}dA$ =4πkeq$=4\pi k_{e}q$
There is an exactly analogous formula for the gravitational field:
ϕg=∫g⋅n^dA=−4πGM${\varphi }_g=\int g\cdot \hat{n}dA=-4\pi GM$
If inside a spherical region there are three masses m1,m2$m_1,m_2$ and m3$m_3$ then what's the gravitational flux from the closed region in this case?
ϕg=−4πG(m1+m2+m3)${\varphi }_g=-4\pi G(m_1+m_2+m_3)$

m1,m2,m3$m_1,m_2,m_3$ are masses closed inside the region.

Definition

Gravitational Potential (V)

The gravitational potential (V) is the gravitational potential energy (U) per unit mass: where m is the mass of the object. Potential energy is equal (in magnitude, but negative) to the work done by the gravitational field moving a body to its given position in space from infinity.

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