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Potential Energy Between Two Point Masses and System of Point Masses

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)\$

Physics

definition

Gravitational Potential Energy between two point masses

Definition: Gravitational potential energy is energy an object possesses because of its position in a gravitational field. The most common use of gravitational potential energy is for an object near the surface of the Earth where the gravitational acceleration can be assumed to be constant at about

9.8 m / s^{2}

. Since the zero of gravitational potential energy can be chosen at any point (like the choice of the zero of a coordinate system), the potential energy at a height h above that point is equal to the work which would be required to lift the object to that height with no net change in kinetic energy.

Two masses

m_{1}

and

m_{2}

separated at a distance

r

has potential energy:

U = \frac{G m _{1} m _{2}}{r}

example

Gravitational Energy of a System

Example: A planet of mass

m_{1}

revloves round the sun of mass

m_{2}

. The distance between the sun and the planet is

r

. Considering the motion of the sun find the total energy of the system assuming the orbits to be circular.

Solution:
Velocity of planet is

v = \frac{G m _{2}}{r}

And the total energy is

P . E + K . E

\Rightarrow \frac{- G m _{1} m _{2}}{R} + \frac{1}{2} m v^{2} = \frac{- G m _{1} m _{2}}{2 R}

REVISE WITH CONCEPTS

Gravitational Potential

ExampleDefinitionsFormulaes

Gravitational Potential Energy

ExampleDefinitionsFormulaes

Gravitational Potential Energy- Medium

ExampleDefinitionsFormulaes

Changing Configuration of a System

ExampleDefinitionsFormulaes

Gravitational Potential Energy - L2

ExampleDefinitionsFormulaes

LEARN WITH VIDEOS

Gravitational Potential Energy for System of Particles

8 mins

Important Questions

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 :

Medium

definition

Gravitational Potential Energy between two point masses

example

Gravitational Energy of a System

REVISE WITH CONCEPTS

Gravitational Potential

Gravitational Potential Energy

Gravitational Potential Energy- Medium

Changing Configuration of a System

Gravitational Potential Energy - L2

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 :

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

At what height from the surface of earth the gravitation potential and the value of g are −5.4×107 Jkg−2 and 6.0 ms−2 respectively? Take the radius of earth as 6400 km.

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

A particle is projected vertically upward from the surface of earth (radiusRe​) 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

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 -

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

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=5.98×1024kg) have to be compressed to be a black hole?

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

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)

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 :

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

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

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

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

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 $(m a s s = 5.98 \times 1 0^{24} k g)$ have to be compressed to be a black hole?

Infinite number of bodies, each of mass $2 k g$ , are situated on $x - a x i s$ 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

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