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

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 

A 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 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  $\vec{F}=m \vec{E}$ 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 $U=m V$
So $d V=\frac{d U}{m}=-\vec{E} \cdot d \vec{r}$
Integrating between $\mathbf{r}_1$, and $\mathbf{r}_2$
we get $V\left(\vec{r}_2\right)-V\left(\vec{r}_1\right)=\int_{r_1}^{r_2}-\vec{E} \cdot d \vec{r}$
If $\mathrm{r}_1=\mathrm{r}_0$, is taken at the reference point, $\mathrm{V}\left(\mathrm{r}_0\right)=0$.
Then the potential $V\left(r_2=r\right)$ at any point $r$ is

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

in Cartesian coordinates, we can write

$$
\begin{aligned}
& \vec{E}=E_x \vec{i}+E_y \vec{j}+E_z \vec{k} \\
& \text { If } \vec{r}=x \vec{i}+y \vec{j}+z \vec{k}
\end{aligned}
$$
Then $d \vec{r}=d x \vec{i}+d y \vec{j}+d z \vec{k}$

So  

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

If $y$ and $z$ remain constant, $d y=d z=0$

Thus

$$
E_x=\frac{d V}{d x}
$$

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