Gravitation Concept Page - 5

Diagram

Gravitational Field due to Spherical Shell

Gravitational Field due to Spherical Shell is:
E=GMr2$E={\displaystyle \frac{GM}{r^2}}$ (For outside the shell)
E=0$E=0$ (For inside the shell)

Definition

Graph for gravitational field due to thin Spherical Shell

Graph for gravitational field:

Definition

Gravitational Potential due to a point mass

Gravitational Potential due to a point mass is given by:

Example

Gravitational Potential due to a uniform ring

Example: Two rings having masses M$M$ and 2M$2M$, respectively, having the same radius are placed coaxially as shown in the figure. If the mass distribution on both the rings is non-uniform, then what is the gravitational potential at point P$P$ ?

Solution:Gravitational potential due to ring (M,R$M,R$) at an axial point which is x unit away from the center,    V=−GMx2+R2−−−−−−−√$V=-{\displaystyle \frac{GM}{\sqrt{x^2+R^2}}}$Thus gravitational potential at P,    VP=(−GMR2+R2−−−−−−−√)+(−G(2M)(2R)2+R2−−−−−−−−−√)$V_P=(-{\displaystyle \frac{GM}{\sqrt{R^2+R^2}}})+(-{\displaystyle \frac{G(2M)}{\sqrt{(2R{)}^2+R^2}}})$ 
VP=−GM[(12R2−−−√)+(25R2−−−√)]$V_P=-GM[({\displaystyle \frac{1}{\sqrt{2R^2}}})+({\displaystyle \frac{2}{\sqrt{5R^2}}})]$VP=−GMR[12√+25√]$V_P=-{\displaystyle \frac{GM}{R}}[{\displaystyle \frac{1}{\sqrt{2}}}+{\displaystyle \frac{2}{\sqrt{5}}}]$

Example

Gravitational Potential inside and outside of a thin spherical shell

Gravitational field inside the shell:
Eg=−GM(3R2−r2)2R3$E_g={\displaystyle \frac{-GM(3R^2-r^2)}{2R^3}}$

Gravitational field outside the shell:
Eg=−GMr$E_g={\displaystyle \frac{-GM}{r}}$

Example

Gravitational Potential due to a solid sphere

Example The earth does not have a uniform density; it is most dense at its centre and least dense at its surface. An approximation of its density is  ρ(r)=(A−Br)$\rho (r)=(A-Br)$, where A=12700kg/m3,B=1.50×10−3kg/m3$A=12700kg/m^3,B=1.50\times {10}^{-3}kg/m^3$ and  r$r$  is the distance from the centre of earth. Use R=6.4×106$R=6.4\times {10}^6$ 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=ρ(r)dV$dm=\rho (r)dV$. 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$r$. If B =$=$ 0, then find gravitational potential at the centre?

Solution:
We have,
ρ=A−Br$\rho =A-Br$
dm=ρdv$dm=\rho dv$
      =(A−Br)dv$=(A-Br)dv$
Now,
v=43πr3$v={\displaystyle \frac{4}{3}\pi r^3}$
∴dv=4πr2dr$\therefore dv=4\pi r^{2}dr$
Thus,
dm=(A−Br)⋅4πr2dr$dm=(A-Br)\cdot 4\pi r^{2}dr$
We know,
dU=−Gdmr${\displaystyle dU=-\frac{Gdm}{r}}$
=−G(A−Br)r⋅4πr2dr$=-{\displaystyle \frac{G(A-Br)}{r}\cdot 4\pi r^{2}dr}$
=−4π⋅G(Ar−Br2)dr$=-4\pi \cdot G(Ar-B{r}^2)dr$
Integrating from 0 to R,
∴U=−4π⋅∫R0(Ar−Br2)dr$\therefore U=-4\pi \cdot {\displaystyle {\int }_0^R(Ar-B{r}^2)dr}$
=−4πG[Ar22−Br33]R0$={\displaystyle -4\pi G{[\frac{A{r}^2}{2}-\frac{B{r}^3}{3}]}_0^R}$
=−4πGAR22$=-4\pi {\displaystyle \frac{GA{R}^2}{2}}$                        (∵B=0)$(\because B=0)$
∴=−2πGAR2$\therefore =-2\pi GA{R}^2$

Example

Change in gravitational potential energy as work done by gravitational force

Example: If R$R$ is radius of the earth and W$W$ is work done in lifting a body from the ground to an altitude R$R$, the calculate the amount of work which should be done in lifting it further to twice that altitude?

Solution:work done is equal to change in PE
W=−ΔU=U1−U2=−GMmR−(−GMm2R)=−GMm2R$W=-\mathrm{\Delta }U=U_1-U_2=-{\displaystyle \frac{GMm}{R}}-(-{\displaystyle \frac{GMm}{2R}})=-{\displaystyle \frac{GMm}{2R}}$
work done in lifting body from 2R$2R$ to 4R$4R$
W′=−ΔU=U1−U2=−GMm2R−(−GMm4R)=−GMm4R=W2$W^{\prime }=-\mathrm{\Delta }U=U_1-U_2=-{\displaystyle \frac{GMm}{2R}}-(-{\displaystyle \frac{GMm}{4R}})=-{\displaystyle \frac{GMm}{4R}}={\displaystyle \frac{W}{2}}$

Example

Gravitational field as a gradient of gravitational potential

Example: Gravitational field is uniform and the gravitational potential difference between surface of a planet and a point 100m$100m$ above is 50J/Kg$50J/Kg$. What is the work done in moving a man 5kg$5kg$ from surface to a point 10m$10m$ above?

Solution:
By definition:
V=∫Edr$V=\int Edr$
V=EΔr$V=E\mathrm{\Delta }r$
Hence,
E=ΔV/Δr$E=\mathrm{\Delta }V/\mathrm{\Delta }r$
Thus,
E=50/100=0.5$E=50/100=0.5$Work done in moving a 5$5$ kg from the surface to a point 10$10$ m above isW=mEd$W=mEd$     W=0.5×5×10=25J$W=0.5\times 5\times 10=25J$

Example

Potential as path integral of gravitational field

Example: The gravitational field in x direction due to some mass distribution is E=kx3$E={\displaystyle \frac{k}{x^3}}$, where k$k$ is a constant. Assuming the gravitational potential to be zero at infinity, then what will be its value at a distance x$x$ ?

Solution:Gravitational potential energy is given by the relation,
V=−∫E.dx$V=-\int E.\mathrm{d}x$
V=−∫(kx3).dx$V=-\int (\frac{k}{x^3}).\mathrm{d}x$
V=k2x2$V={\displaystyle \frac{k}{2x^2}}$

Example

Gravitational energy density per unit volume

Example: Planet Y$Y$ has density D$D$, and surface gravitational acceleration, g$g$. The radius of planet Y$Y$ is suddenly doubled while its density remains the same. Compared to the original g$g$, what would be the new value of the surface gravitational acceleration?

Solution: Accelaration due to gravity g=GMr2$g=\frac{GM}{r^2}$
here M=43ρπr3,ρ=D.$M=\frac{4}{3}\rho \pi r^3,\rho =D.$
So, g=G43Dπr3r2=43GDπr......................(1)$g=\frac{G{\displaystyle \frac{4}{3}}D\pi r^3}{r^2}={\displaystyle \frac{4}{3}}GD\pi r......................(1)$. 
Now, the radius is doubled with all the quantities kept constant.
New g,gn=43GDπ(2r)=2g($g,g_n={\displaystyle \frac{4}{3}}GD\pi (2r)=2g($from (1))$(1))$

BookMarks

Page 1  Page 2  Page 3  Page 4  Page 5  Page 6  Page 7  Page 8  Page 9  Page 10