Question

Two bodies of masses $\mathrm{m}$ and $4\mathrm{m}$ are placed at a distance of $\mathrm{r}$. The gravitational potential at a point on the line joining them where the gravitational field is zero is

• A. $-\frac{4\mathrm{Gm}}{\mathrm{r}}$
• B. $-\frac{6\mathrm{Gm}}{\mathrm{r}}$
• C. $-\frac{9\mathrm{Gm}}{\mathrm{r}}$
• D. Zero

Answer 1

The correct option is C

$-\frac{9\mathrm{Gm}}{\mathrm{r}}$

Step 1: Given data

Mass of a first body $\mathrm{A}$ is = $\mathrm{m}$

Mass of a second body $\mathrm{B}$ is = $4\mathrm{m}$

Both masses are separated by a distance = $\mathrm{r}$

If a point mass of one unit is placed along the axis connecting the centers of the two objects. Therefore, $\mathrm{C}$can be regarded as a neutral point, located on this line and at which the point mass is not subject to the effects of gravity.

Then this neutral point $\mathrm{C}$is situated at, $\mathrm{x}$ distance from $\mathrm{A}$ and $\mathrm{r}-\mathrm{x}$ distance from $\mathrm{B}$.

We have to find the net gravitational potential.

Step 2: Formula to be used

The gravitational force between a given mass and a unit mass is called the gravitational field intensity.

If the force of gravity between two objects is determined by,

$\mathrm{F}=\frac{{\mathrm{Gm}}_1{\mathrm{m}}_2}{{\mathrm{r}}^2}$

Here, ${\mathrm{m}}_1$ and ${\mathrm{m}}_2$ are the mass of two objects and $\mathrm{r}$ is the distance between their centers and $\mathrm{G}$ is the universal gravitational constant.

The gravitation field intensity ${\mathrm{E}}_{\mathrm{g}}$ of an object, consider ${\mathrm{m}}_2=1$, we get

${\overrightarrow{\mathrm{E}}}_{\mathrm{g}}=\frac{\mathrm{Gm}}{{\mathrm{r}}^2}$

Because of mass $\mathrm{A}$, the gravitational field is,

${\overrightarrow{\mathrm{E}}}_{\mathrm{gA}}=\frac{\mathrm{Gm}}{{\mathrm{x}}^2}$

Because of mass $\mathrm{B}$, the gravitational field at point $\mathrm{C}$ is,

${\overrightarrow{\mathrm{E}}}_{\mathrm{gB}}=\frac{4\mathrm{Gm}}{{\left(\mathrm{r}-\mathrm{x}\right)}^2}$

So, the total gravitational field at this point is zero, we get,

${\overrightarrow{\mathrm{E}}}_{\mathrm{gA}}+{\overrightarrow{\mathrm{E}}}_{\mathrm{gB}}=0$

Step 3: Find the point lying on the line joining the point masses where the net gravitational force is zero.

Because the directions of the fields produced at $\mathrm{C}$ by the two masses are opposite, the outcome of adding them together is zero.

So,

$\frac{4\mathrm{Gm}}{{\left(\mathrm{r}-\mathrm{x}\right)}^2}=\frac{\mathrm{Gm}}{{\mathrm{x}}^2}$

$\frac{4}{{\left(\mathrm{r}-\mathrm{x}\right)}^2}=\frac{1}{{\mathrm{x}}^2}$

The above expression can be written as,

$\frac{2}{\left(\mathrm{r}-\mathrm{x}\right)}=\frac{1}{\mathrm{x}}$

$\mathrm{r}-\mathrm{x}=2\mathrm{x}$

$\mathrm{r}=3\mathrm{x}$

$\mathrm{x}=\frac{\mathrm{r}}{3}$
And

$\mathrm{r}-\mathrm{x}=\frac{2\mathrm{r}}{3}$

Step 4: Calculating the net gravitational potential

The gravitational potential at this point ${\mathrm{V}}_{\mathrm{C}}$, is

${\mathrm{V}}_{\mathrm{C}}={\mathrm{V}}_{\mathrm{A}}+{\mathrm{V}}_{\mathrm{B}}$
We know that,
$\mathrm{V}=-\frac{\mathrm{Gm}}{\mathrm{r}}$
The gravitational potential of $\mathrm{A}$ at $\mathrm{C}$ is,
${\mathrm{V}}_{\mathrm{A}}=-\frac{\mathrm{Gm}}{\mathrm{x}}$
Substitute the value of $\mathrm{x}$, we get,
${\mathrm{V}}_{\mathrm{A}}=-\frac{\mathrm{Gm}}{{\displaystyle \frac{\mathrm{r}}{3}}}$

$=-\frac{3\mathrm{Gm}}{\mathrm{r}}$
The gravitational potential of $\mathrm{B}$ at point $\mathrm{C}$ is,
${\mathrm{V}}_{\mathrm{B}}=-\frac{4\mathrm{Gm}}{\mathrm{r}-\mathrm{x}}$
Substituting the value of $\mathrm{x}$, we get,

${\mathrm{V}}_{\mathrm{B}}=-\frac{4\mathrm{Gm}}{\mathrm{r}-{\displaystyle \frac{\mathrm{r}}{3}}}$

$=-\frac{3\times 4\mathrm{Gm}}{2\mathrm{r}}$
$=-\frac{6\mathrm{Gm}}{\mathrm{r}}$
Therefore,
${\mathrm{V}}_{\mathrm{C}}=-\frac{3\mathrm{Gm}}{\mathrm{r}}-\frac{6\mathrm{Gm}}{\mathrm{r}}$
$=-\frac{9\mathrm{Gm}}{\mathrm{r}}$
Hence, option (C) is the correct answer.