JEE Main & Advanced Physics Gravitation / गुरुत्वाकर्षण Gravitational Potential

At a point in a gravitational field potential \[V\] is defined as negative of work done per unit mass in shifting a test mass from some reference point (usually at infinity) to the given point i.e.,

\[V=-\frac{W}{m}\]\[=-\int{\frac{\overrightarrow{F}.d\overrightarrow{r\,}}{m}}\] \[=-\int{\overrightarrow{I\,}.d\overrightarrow{r\,}}\]              [As \[\frac{F}{m}=I\]]

\[\therefore \] \[h=0\]     

i.e., negative gradient of potential gives intensity of field or potential is a scalar function of position whose space derivative gives intensity. Negative sign indicates that the direction of intensity is in the direction where the potential decreases.

(i) It is a scalar quantity because it is defined as work done per unit mass.

(ii) Unit : Joule/kg or \[{{m}^{2}}/{{\sec }^{2}}\]

(iii) Dimension : \[[{{M}^{0}}{{L}^{2}}{{T}^{-2}}]\]

(iv) If the field is produced by a point mass then

\[V=-\int_{{}}^{{}}{I\ dr}\]\[T=2\pi \sqrt{\frac{{{r}^{3}}}{GM}}=2\pi \sqrt{\frac{{{R}^{3}}}{g{{R}^{2}}}}=2\pi \sqrt{\frac{R}{g}}\]                   

[As \[I=-\frac{GM}{{{r}^{2}}}\]]

\[\therefore \] \[V=-\frac{GM}{r}+c\] [Here c = constant of integration]

Assuming reference point at \[\infty \] and potential to be zero there we get

\[0=-\frac{GM}{\infty }+c\,\Rightarrow c=0\]

\[\therefore \]  Gravitational potential \[V=-\frac{GM}{r}\]

(v) Gravitational potential difference : It is defined as the work done to move a unit mass from one point to the other in the gravitational field. The gravitational potential difference in bringing unit test mass m from point A to point B under the gravitational influence of source mass M is

(vi) Potential due to large numbers of particle is given by scalar addition of all the potentials.

\[V={{V}_{1}}+{{V}_{2}}+{{V}_{3}}+..........\]

\[=-\frac{GM}{{{r}_{1}}}-\frac{GM}{{{r}_{2}}}-\frac{GM}{{{r}_{3}}}........\]

\[=-G\sum\limits_{i=1}^{i=n}{\frac{{{M}_{i}}}{{{r}_{i}}}}\]