Relation Between Gravitational Field And Potential - Practice Questions & MCQ

Gravitational field and potential are related as

$
\vec{E}=-\frac{d V}{d r}
$

Where E is Gravitational field
And V is Gravitational potential
And r is the position vector
And Negative sign indicates that in the direction of intensity the potential decreases.
If $\vec{r}=x \vec{i}+y \vec{j}+z \vec{k}$

Then

$
E_x=\frac{\delta V}{d x}, E_y=\frac{\delta V}{d y}, E_z=\frac{\delta V}{d z}
$

Proof-
Let gravitational field at a point $r$ due to a given mass distribution is 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 $\vec{F}=m \vec{E}$ as shown in figure

As the particle is displaced from r to r + dr the

work done by the gravitational force on it is

$
d W=\vec{F} \cdot \vec{r}=m \vec{E} \cdot d \vec{r}
$

The change in potential energy during this displacement is

$
d U=-d W=-\vec{F} \cdot \vec{r}=-m \vec{E} \cdot d \vec{r}
$

And we know that Relation between Potential and Potential energy
As $U=m V$

$
\mathrm{So}^{d V}=\frac{d U}{m}=-\vec{E} \cdot d \vec{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}-\vec{E} \cdot d \vec{r}
$

If $r_1=r_0$, is taken at the reference point, $V\left(r_0\right)=0$.
Then the potential $\mathrm{V}\left(\mathrm{r}_2=\mathrm{r}\right)$ at any point r is

$
V(\vec{r})=\int_{r_0}^r-\vec{E} \cdot d \vec{r}
$

in Cartesian coordinates, we can write

$
\vec{E}=E_x \vec{i}+E_y \vec{j}+E_z \vec{k}
$

If $\vec{r}=x \vec{i}+y \vec{j}+z \vec{k}$
Then $d \vec{r}=d x \vec{i}+d y \vec{j}+d z \vec{k}$

So  

$
\begin{aligned}
& \vec{E} \cdot d \vec{r}=-d V=E_x d x+E_y d y+E_z d z \\
& d V=-E_x d x-E_y d y-E_z d z
\end{aligned}
$

If y and z remain constant, $\mathrm{dy}=\mathrm{dz}=0$
Thus $E_x=\frac{d V}{d x}$

Similarly

$
E_y=\frac{d V}{d y}, E_z=\frac{d V}{d z}
$