Relation Between Gravitational Field And Potential

Relation Between Gravitational Field and Potential

The relationship between the gravitational field and gravitational potential is a cornerstone in the study of gravitation. The gravitational field at a point in space represents the force per unit mass that would be exerted on a small object placed at that point. On the other hand, the gravitational potential at a point is the amount of work required to bring a unit mass from infinity to that point without any acceleration.

Gravitational field and potential are related as
$\overrightarrow{E}=-\frac{dV}{dr}$
Where E is the Gravitational field
And $V$ is the Gravitational potential
And $r$ is the position vector
A negative sign indicates that in the direction of intensity, the potential decreases.
If $\overrightarrow{r}=x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}$

Then

$E_x=\frac{\delta V}{dx},E_y=\frac{\delta V}{dy},E_z=\frac{\delta V}{dz}$

Proof

Let the gravitational field at a point r due to a given mass distribution be E.

If a test mass m is placed inside a uniform gravitational field E.

Then force on a particle m when it is at r is $\overrightarrow{F}=m\overrightarrow{E}$ as shown in figure

As the particle is displaced from $r$ to $r+dr$ the work done by the gravitational force on it is

$dW=\overrightarrow{F}\cdot \overrightarrow{r}=m\overrightarrow{E}\cdot d\overrightarrow{r}$
The change in potential energy during this displacement is

$dU=-dW=-\overrightarrow{F}\cdot \overrightarrow{r}=-m\overrightarrow{E}\cdot d\overrightarrow{r}$
And we know that Relation between Potential and Potential energy
As $U=mV$
So $dV=\frac{dU}{m}=-\overrightarrow{E}\cdot d\overrightarrow{r}$
Integrating between $r_1$, and $r_2$

We get

$V\left(\overrightarrow{r_2}\right)-V\left(\overrightarrow{r_1}\right)={\int }_{r_1}^{r_2}-\overrightarrow{E}\cdot d\overrightarrow{r}$
If $\underset{―}{r_1=r_0}$, is taken at the reference point, $V\left(r_0\right)=0$.
Then the potential $V(r_2=r)$ at any point $r$ is

$V(\overrightarrow{r})={\int }_{r_0}^r-\overrightarrow{E}\cdot d\overrightarrow{r}$

in Cartesian coordinates, we can write
$\begin{array}{rl}& \overrightarrow{E}=E_x\overrightarrow{i}+E_y\overrightarrow{j}+E_z\overrightarrow{k}\\ & \text{If}\overrightarrow{r}=x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}\end{array}$
Then $d\overrightarrow{r}=dx\overrightarrow{i}+dy\overrightarrow{j}+dz\overrightarrow{k}$
So

$\begin{array}{rl}& \overrightarrow{E}\cdot d\overrightarrow{r}=-dV=E_{x}dx+E_{y}dy+E_{z}dz\\ & dV=-E_{x}dx-E_{y}dy-E_{z}dz\end{array}$
If y and z remain constant, $\mathrm{dy}=\mathrm{dz}=0$

Thus

$E_x=\frac{dV}{dx}$
Similarly

$E_y=\frac{dV}{dy},E_z=\frac{dV}{dz}$

Recommended Topic Video

Solved Examples Based on Relation Between Gravitational Field And Potential

Example 1: The gravitational field in a region is given by $\overrightarrow{E}=-(20)(\hat{i}+\hat{j})N-k{g}^{-1}$. Find the gravitational potential at the origin $(0,0)$ in $J{\mathrm{kg}}^{-1}$.

1) 0

2) 3

3) 2

4) -1

Solution:

The relation between the gravitational field and potential as

$\begin{array}{rl}& V=-\int E\cdot dr=-[\int E_x\cdot dx+\int E_y\cdot dy]=20x+20y\\ & \Rightarrow V=0\text{at the origin}(0,0)\end{array}$

Hence, the answer is the option (1).

Example 2: The gravitational field in a region is given by $\overrightarrow{g}=5N/\mathrm{kg}\hat{i}+12\mathrm{N}/\mathrm{kg}\hat{j}$. The change in the gravitational potential energy ( in joules) of a particle of mass 2 kg when it is taken from the origin to a point (7 m,-3 m) is :

1) 2

2) 13

3) -71

4) 71

Solution:

The relation between gravitational field and potential is given by
$\begin{array}{rl}& \overrightarrow{E}=-\frac{dV}{dr}\\ & \text{So}=-\int E\cdot dr\end{array}$

and $\mathrm{U}=\mathrm{mV}$

$\begin{array}{rl}\mathrm{\Delta }U& =-\int mE\cdot dr\\ \mathrm{\Delta }U& =-2\ast \int (5\overrightarrow{i}+12\overrightarrow{j})\cdot (dx\overrightarrow{i}+dy\overrightarrow{j})\mathrm{\Delta }U=-2\ast {\int }_0^{7}5dx-24{\int }_0^{-3}dy\\ \text{so}\mathrm{\Delta }U& =-2[5\ast (7-0)+12\ast (-3)]=2J\end{array}$

Hence, the answer is the option (1).

Example 3: The gravitational field in a region is given by: $\overrightarrow{E}=(5\mathrm{N}/\mathrm{kg})\hat{i}+(12\mathrm{N}/\mathrm{kg})\hat{j}$ If the potential at the origin is taken to be zero, then the ratio of the potential at the points $(12\mathrm{m},0)$ and $(0,5\mathrm{m})$ is

1) Zero
2) 1
3) $\frac{144}{25}$
4) $\frac{25}{144}$

Solution:

The gravitational field in a region is given by
$\overrightarrow{E}=(5N/kg)\hat{i}+(12N/kg)\hat{j}$

(Potential at origin is O ) $\to$ given
and we know that $E=\frac{dv}{dr}\Rightarrow (dv=Edr)$
In vector form, the position vector is written as

$\begin{array}{rl}& \overrightarrow{r_1}=12i+oj\\ & \overrightarrow{r_2}=0\hat{i}+5\hat{j}\\ & d{V}_1=E\cdot d{r}_1\\ & =(5\hat{i}+12\hat{j})\cdot (12\hat{i}+0\hat{j})=12\times 5\\ & d{V}_2=E\cdot d{r}_2\\ & =(5\hat{i}+12\hat{j})\cdot (0\hat{i}+5\hat{j})=5\times 12\\ & \frac{d{V}_1}{d{V}_2}=1\end{array}$

Hence, the answer is the option (2).

Example 4: On the x-axis and at a distance x from the origin, the gravitational field due to mass distribution is given by $\frac{Ax}{{(x^2+a^2)}^{\frac{3}{2}}}$ in the x-direction. The magnitude of gravitational potential on the x-axis at a distance x, taking its value to be zero at infinity, is:

1) $\frac{A}{{(x^2+a^2)}^{\frac{1}{2}}}$
2) $\frac{A}{{(x^2+a^2)}^{\frac{3}{2}}}$
3) $A{(x^2+a^2)}^{\frac{1}{2}}$
4) $A{(x^2+a^2)}^{\frac{3}{2}}$

Solution:

Given

$E_G=\frac{Ax}{{(x^2+a^2)}^{3/2}},V_{\mathrm{\infty }}=0$
Using

$\begin{array}{rl}& {\int }_{{\mathrm{V}}_{\mathrm{\infty }}}^{{\mathrm{V}}_x}\mathrm{dV}=-{\int }_{\mathrm{\infty }}^{\mathrm{x}}{\overrightarrow{\mathrm{E}}}_{\mathrm{G}}\cdot {\overrightarrow{\mathrm{d}}}_{\mathrm{x}}\\ & {\mathrm{V}}_{\mathrm{x}}-{\mathrm{V}}_{\mathrm{\infty }}=-{\int }_{\mathrm{\infty }}^{\mathrm{x}}\frac{\mathrm{Ax}}{{({\mathrm{x}}^2+{\mathrm{a}}^2)}^{3/2}\mathrm{dx}}\end{array}$

put $x^2+a^2=z$

$2\mathrm{xd}x=\mathrm{dz}$
So

$\begin{array}{c}{\mathrm{V}}_{\mathrm{x}}-0=-{\int }_{\mathrm{\infty }}^{\mathrm{x}}\frac{\mathrm{Adz}}{2(\mathrm{z}{)}^{3/2}}={\left[\frac{\mathrm{A}}{{\mathrm{z}}^{1/2}}\right]}_{\mathrm{\infty }}^{\mathrm{x}}={\left[\frac{\mathrm{A}}{{({\mathrm{x}}^2+{\mathrm{a}}^2)}^{1/2}}\right]}_{\mathrm{\infty }}^{\mathrm{x}}\\ {\mathrm{V}}_{\mathrm{x}}=\frac{\mathrm{A}}{{({\mathrm{x}}^2+{\mathrm{a}}^2)}^{1/2}}-0=\frac{\mathrm{A}}{{({\mathrm{x}}^2+{\mathrm{a}}^2)}^{1/2}}\end{array}$

Hence, the answer is the option (1).

Example 5: What is the relationship between gravitational field strength and gravitational potential?

1) They are of the same size but opposite in direction

2) Gravitational potential is a derivative of gravitational field strength.

3) The intensity of the gravitational field is derived from the gravitational potential.

4) There is no relationship between the given two quantities.

Solution:

Gravitational potential (V) is defined as the amount of work done per unit mass in bringing an object from infinity to a point in space and is given by the formula $V=-\frac{Gm}{r}$.

A negative sign means that work is done against the force of gravity.

The gravitational field (g) is related to the gravitational potential by the formula $g=-\frac{dV}{dr}$, where $\frac{dV}{dr}$ is the derivative of the gravitational potential with respect to distance.

Hence, the answer is the option(3).

Summary

The gravitational field and gravitational potential are closely linked, with the field representing the force per unit mass and the potential representing the work done to move a unit mass from infinity to a point in space. The field is the negative gradient of the potential, meaning it points in the direction of decreasing potential. This relationship is fundamental in understanding gravitational interactions, from celestial orbits to everyday phenomena like falling objects.