Relation Between Gravitational Field And Potential MCQ - Practice Questions with Answers

 

Gravitational field and potential are related as

$\overrightarrow{E}=-\frac{dV}{dr}$

Where E is Gravitational field

And V is 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 ${\mathbf{r}}_1$, and ${\mathbf{r}}_2$
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 ${\mathrm{r}}_1={\mathrm{r}}_0$, is taken at the reference point, $\mathrm{V}\left({\mathrm{r}}_0\right)=0$.
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  

$$\begin{array}{c}\overrightarrow{E}\cdot d\overrightarrow{r}=-dV=E_{x}dx+E_{y}dy+E_{z}dz\\ dV=-E_{x}dx-E_{y}dy-E_{z}dz\end{array}$$

If $y$ and $z$ remain constant, $dy=dz=0$

Thus

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

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