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Gravitational potential due to spherical shell

Q: Two bodies of masses \${m}\$ and \$4\ m\$ are placed at a distance \${r}.\$ The gravitational potential at a point on the line joining them where the gravitational field is zero is :

Given:Two masses \$m\$ and \$4m\$ kept at a distance \$r\$Let gravitetional potential is zero at a point \$p\$\$\dfrac{Gm}{x^2}-\dfrac{G(4m)}{(r-x)^2}=0\$\$\dfrac1x = \dfrac 2{(r-x)}\$\$r - x = 2x\$\$3x = r\$\$x = r/3\$Potential at point \$p\$\$V=-\dfrac{4Gmm}{(2r/3)}-\dfrac{Gmm}{(r/3)}\$\$\displaystyle V=-Gm(\dfrac{6}{r}+ \frac{3}{r})= -9\dfrac{Gm}{r}\$

Q: A particle of mass \$M\$ is situated at the centre of spherical shell of same mass and radius \$a\$. The magnitude of the gravitational potential at a point situated at \${a}/{2}\$ distance from the centre, will be

\$\textbf{Step 1 - Drawing free body diagram }\$\$\textbf{Step 2 - Net gravitational potential at point P is given by}\$\$V_{p} = \text{(Potential due to sphere)} + \text{(Potential due to particle)}\$As we know that Gravitational potential due to point mass\$= \dfrac {GM}{r}\$Gravitational potential inside shell \$=\dfrac{GM}{a}\$\$\because V_{p} = \dfrac {GM}{a} + \dfrac {GM}{\dfrac {a}{2}}\$\$\Rightarrow V_{p} = \dfrac {3GM}{a}\$Hence option (B) is correct.

Q: At what height from the surface of earth the gravitation potential and the value of \$g\$ are \$-5.4 \times {10}^{7}\ J {kg}^{-2}\$ and \$6.0 \ {ms}^{-2}\$ respectively? Take the radius of earth as \$6400 \ km\$.

\$V=-\dfrac{GM}{R+h}=-5.4\times 10^7\$ ....... (1)\$g=\dfrac{GM}{(R+h)^2}=6\$ ...... (2)Dividing the two equations, we get:\$\Rightarrow \dfrac{5.4\times 10^7}{(R+h)}=6\$\$\Rightarrow R+h=9000\ km\$\$\Rightarrow h=2600\ km\$

Q: Energy required to move a body of mass m from an orbit of radius 2R to 3R is:

\$\text{Energy required} =\text{(Final potential-Initial potential)m}\$\$\text{Energy required}=-\dfrac{GMm}{3R}-\dfrac{-GMm}{2R}=\dfrac{GMm}{6R}\$

Q: A particle is projected vertically upward from the surface of earth (radius\${R_e}\$) with a kinetic energy equal to half of the minimum value needed for it to escape.The height to which it rises above the surface of earth is

The kinetic energy needed to escape from the surface is given as, \$E = \dfrac{1}{2}m{v_e}^2\$ \$ = \dfrac{1}{2}m{\left( {\sqrt {\frac{{2Gm}}{{{R_e}}}} } \right)^2}\$ \$ = \dfrac{{GMm}}{{{R_e}}}\$ Since, the potential energy at the maximum height will be equal to the sum of half kinetic energy and the potential energy at the surface of the earth. It can be written as, \$ - \dfrac{{GMm}}{{{R_e}}} + \dfrac{1}{2}\dfrac{{GMm}}{{{R_e}}} = - \dfrac{{GMm}}{{{R_e} + h}}\$ \$\dfrac{1}{{2{R_e}}} = \dfrac{1}{{{R_e} + h}}\$ \$h = {R_e}\$ The maximum height is \${R_e}\$.

Q: A body of mass 'm' is taken from the earth's surface to the height equal to twice the radius (R) of the earth. The change in potential energy of body will be -

Mass = m, height = 2RAt height the acceleration due to gravity decreases , hence the value of acceleration due to gravity becomes g' = \$\dfrac{g}{1 + h/R}\$ = \$\dfrac{gR}{h + R}\$The value of potential energy \$= mgh =\dfrac{mghR}{h + R}\$Now \$h = 2R \$ (given)Henc change in potential energy = \$\dfrac{2}{3}\$mgR

Q: Define gravitational potential energy. Derive expression for gravitational potential in the gravitational field of earth at distance \$r\$ from the centre of the earth.

The work obtained in bringing a body from infinity to a point in gravitational field is called gravitational potential energy.Force of attraction between the earth and the object when an object is at the distance \$a\$ from the center of the earth.\$F = \dfrac{{GmM}}{{{x^2}}}\$Consider the \$dW\$ is the small amount of work done in bringing the body without acceleration through the very small distance \$dx\$\$dW=Fdx\$And total work done to bring the body from infinity to point \$p\$ which is at distance \$r\$ from the center of the earth.\$w = \int_\infty ^r {\dfrac{{GmM}}{{{x^2}}}dx = - \dfrac{{GMm}}{r}} \$Hence, the gravitational potential energy \${ = - \dfrac{{GMm}}{r}}\$

Q: A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would earth \$(mass=5.98 \times 10^{24} kg)\$ have to be compressed to be a black hole?

For earth to be black hole the escape velocity should be at least equal to speed of light.\$\therefore\$ escape velocity \$=\$ speed of light\$\sqrt { \dfrac { 2GM }{ R } } =C\$\$R=\dfrac { 2GM }{ { C }^{ 2 } } \$\$R=\dfrac { 2\times 6.67\times { 10 }^{ -11 }\times 5.98\times { 10 }^{ 24 } }{ 9\times { 10 }^{ 16 } } =8.86\times { 10 }^{ -3 }m={ 10 }^{ -2 }m\$

Q: Infinite number of bodies, each of mass \$2\ kg\$, are situated on \$x-axis\$ at distance \$1\ m,\ 2\ m,\ 4\ m,\ 8\ m,\ .......\$ respectively, from the origin. The resulting gravitational potential due to this system at the origin will be

Gravitational potential at distance \$x\$, \$V=-\dfrac{Gm}{x}\$\$\Rightarrow\$ Gravitational Potential at origin, \$V=-\dfrac{2G}{1} -\dfrac{2G}{2}-\dfrac{2G}{4} -\dfrac{2G}{8} - - - - to\$ \$ \infty\$ \$\Rightarrow V= -2G \left \{ \dfrac{1}{1} +\dfrac{1}{2} +\dfrac{1}{4} + \dfrac{1}{8} + + + + to \infty \right \}\$\$\Rightarrow \left \{ \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + + + + to \infty \right \}\$ is GP series Where \$a=1\$ and \$r=\dfrac{1}{2}\$\$\Rightarrow \$ Sum of GP series, \$S= \dfrac{a}{1-r}\$\$\Rightarrow V= -2G\times \dfrac{1}{1-\frac{1}{2}}=-4G\$

Q: The change in potential energy when body of mass m is raised to height \$nR\$ from earth's surface is (\$R\$=radius of the earth)

Gravitational potential energy at any point at a distance r from the centre of the earth is \$U=-\frac{GM_{E}m}{r}\$where \$M_{E}\$ and m be masses of earth and body respectively. At the surface of the earth, \$r=R_{E}\$\$\therefore U_{1}=-\dfrac{GM_{E}m}{R_{E}}\$At a height h from the surface, \$r=R_{E}+h=R_{E}+nR_{E}=\left(1+n\right)R_{E}\$\$\therefore {U}_{2}=-\dfrac{GM_{E}m}{\left(n+1\right)R_{E}}\$change in potential energy is \$\Delta U=U_{2}-U{1}\$\$=-\dfrac{GM_{E}m}{\left(n+1\right)R_{E}}-\left(-\dfrac{GM_{E}m}{R_{E}}\right)\$\$=\dfrac{GM_{E}m}{R_{E}}\left(1-\dfrac{1}{\left(n+1\right)}\right)=\dfrac{GM_{E}mn}{\left(n+1\right)R_{E}}\$\$=mgR_{E}\dfrac{n}{\left(n+1\right)} \left(\because g=\dfrac{GM_{E}}{R_{E}^{2}}\right)\$

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