GRAVITATIONAL POTENTIAL ENERGY

The amount of work done in bringing a body from infinity to a point within the gravitational field is called the gravitational potential energy of the body at that point.

$U=-W_{\infty \to r}^{}=-{\int }_{\infty }^r\overrightarrow{F}\cdot \overrightarrow{dr}$

$orU={\int }_{\infty }^r-\frac{GMm}{r^2}\hat{r}\cdot \overrightarrow{dr}=-{\int }_{\infty }^r-\frac{GMm}{r^2}dr(since\hat{r}\cdot \overrightarrow{dr}=dr)$

$orU=-GMm{\left(\frac{1}{r}\right)}_{\infty }^r=-GMm\left(\frac{1}{r}-\frac{1}{\infty }\right)=-\frac{GMm}{r}$

GRAVITATIONAL SELF ENERGY

It is the energy possessed by a body due to the interaction forces inside the body. This can be defined as the work done in assembling all the particles of a body in a definite shape and size or it is the work done in creating a body.

INTERACTION ENERGY OF A SYSTEM OF PARTICLES

In a system of N particles with masses m₁, m­₂ ... m_n separated from each other by a distance r₁₂, r₁₃ where r₁₂ is the separation between m₁ and m₂ and so on.

The total interaction energy of system is,

$U=-\frac{1}{2}\sum _{i=1}^N\sum _{j=1}^N\frac{G{m}_i{m}_j}{r_{ij}}$

The factor 1/2 is taken because the interaction energy for each possible pair of system is taken twice during summation as for masses m₁ and m₃

$U=-\frac{G{m}_1{m}_3}{r_{13}}=-\frac{G{m}_3{m}_1}{r_{31}}$ (i $\ne$ j)

Illustration.1 Gravitational Self energy of a Uniform Sphere (star)

Solution: $U_{shell}=-G\frac{\left(\frac{4}{3}\pi r^3\rho \right)\left(4\pi r^{2}dr\rho \right)}{r}where\rho =\frac{M}{\left(\frac{4}{3}\right)\pi r^3}$

$=-\frac{1}{3}G{\left(4\pi \rho \right)}^2{r}^{4}dr$

$U_{star}=-\frac{1}{3}G{\left(4\pi \rho \right)}^2\underset{0}{\overset{R}{\int }}{r}^{4}dr=-\frac{1}{3}G{\left(4\pi \rho \right)}^2{\left[\frac{r^5}{5}\right]}_0^R=-\frac{3}{5}G{\left(\frac{4\pi }{3}{R}^3\rho \right)}^2\frac{1}{R}$

$\therefore U_{star}=-\frac{3}{5}\frac{G{M}^2}{R}$

GRAVITATIONAL POTENTIAL

Gravitational potential $\text{V= }\frac{\text{U}}{\text{m}}$

Dimensional Formula of gravitational potential $=\frac{Work}{mass}=\frac{M{L}^2{T}^2}{M}=\left[M^0{L}^2{T}^{-2}\right].$

Illustration 2:

At a point above the surface of earth, the gravitational potential is – 5.12 x 10⁷ J/kg and the acceleration due to gravity is 6.4 m/s². Assuming the mean radius of the earth to be 6400 km, calculate the height of this point above the earth's surface.

Solution: Let r be the distance of the given point from the centre of the earth. Then

Gravitational potential = $-\frac{GM}{r}$

= – 5.12 x x10⁷ ….(1)

and acceleration due to gravity,

g = $\frac{Gm}{r^2}=6.4$ . . . (2)

Dividing (1) by (2), we get

r = $\frac{5.12x10^7}{6.4}=8x10^{6}m=8000km$

$\therefore$ height of the point from earth's surface

= r – R = 8000 – 6400 = 1600 km

GRAVITATIONAL POTENTIAL & FIELD DUE TO VARIOUS OBJECTS

RELATION BETWEEN GRAVITATIONAL POTENTIAL (V) AND INTENSITY (I OR E)

$\vec{E}=-$gradient V = – grad V

OR $E=-\frac{dv}{dr}$

RELATION BETWEEN g AND g

On the surface of earth, $g=\frac{GM}{R^2}$

Substituting the values of G, M, R we get g = 9.81 m/s².

WORK DONE IN DISPLACEMENT OF A BODY IN GRAVITATIONAL FIELD

If a body of mass m is displaced in a gravitational field from point A to B, then work done by external agent can be given as

$W=U_B-U_A$

= $\left(m{V}_{gB}\right)-\left(m{V}_{gA}\right)$

= $m\left(V_{gB}-V_{gA}\right)$

Thus work done in displacing a body in gravitational field can be given as

Work done = mass of body ´ gravitational potential difference of the terminal points.

Illustration 3:

Find work done in shifting a body of mass m from a height h above the earth's surface to a height 2h above the earth's surface.

Solution:

The gravitational potential at a height h and 2h above the earth surface is given as:

$V_h=-\frac{G{M}_e}{R_e+h}{V}_{2h}=-\frac{G{M}_e}{R_e+2h}$

If a body of mass m is shifted from h to 2h, work done in the process is

$W=m\left(V_{2h}-V_h\right)=m\left[\frac{G{M}_e}{R_e+h}-\frac{G{M}_e}{R_e+2h}\right]$