Gravitational Potential MCQ - Practice Questions with Answers

In a gravitational field potential V at a point, P is defined as the negative of work done per unit mass in changing the position of a test mass from some reference point to the given point.

Note-usually reference point is taken as infinity and potential at infinity is taken as Zero.

We know that $W=\int \overrightarrow{F}\cdot \overrightarrow{dr}$
${}^{\text{So }}V=-\frac{W}{m}=-\int \frac{\overrightarrow{F}\cdot \overrightarrow{dr}}{m}$
And $\overrightarrow{I}=\frac{\overrightarrow{F}}{m}$
$V=-\int \overrightarrow{I}\cdot \overrightarrow{dr}$
$V\to$ Gravitational potential
$I\to$ Field Intensity
$dr\to$ small distance
We can also write $I=-\frac{dV}{dr}$

This means a negative gradient of potential gives the intensity of the field.

The negative sign indicates that in the direction of intensity, the potential decreases.

• It is a scalar quantity.

Unit $\to$ Joule $/\mathrm{kg}$ or ${\mathrm{m}}^2/{\sec }^2$
Dimension : $\left[M^0{L}^2{T}^{-2}\right]$

• Gravitational Potential at a distance 'r'

If the field is produced by a point mass then 

$\begin{array}{rl}& I=\frac{GM}{r^2}\\ & V=-\int \overrightarrow{I}\cdot \overrightarrow{dr}\\ & \text{ So }\\ & V=-\frac{GM}{r}\\ & \text{ at }r=\mathrm{\infty }V=0=V_{max}\end{array}$

• Gravitational Potential difference

In the gravitational field, the work done to move a unit mass from one position to the other is known as Gravitational Potential difference.

            If the point mass M is producing the field

Points A and B are shown in the figure.

$V_A=$ Gravitational potential at point A
$V_B=$ Gravitational potential at point B

$r_B\to$ the distance of mass at $B$
$r_A\to$ distance of mass at $A$
$\mathrm{\Delta }V=$ The gravitational potential difference in bringing mass m from point A to point B in the gravitational field produced by M .

$$\begin{array}{rl}& \mathrm{\Delta }V=V_B-V_A=\frac{W_{A\to B}}{m}\\ & \mathrm{\Delta }V=-GM[\frac{1}{r_B}-\frac{1}{r_A}]\end{array}$$
 

• Superposition of Gravitational potential

The net gravitational potential at a given point due to different point masses (M₁, M₂, M₃…) can be calculated by doing a scalar sum of their individual Gravitational potential.

   $\begin{array}{rl}V& =V_1+V_2+V_3\cdots \\ & =-\frac{G{M}_1}{r_1}-\frac{G{M}_2}{r_2}-\frac{G{M}_3}{r_3}\cdots \\ V& =-G\sum _{i=1}^{i=n}\frac{M_i}{r_i}\\ M_i& \to \text{ mass }\\ r_i& \to \text{ distances }\end{array}$

• Point of zero potential

           Let m₁ and m₂ be separated at a distance d from each other

And P is the point where net Gravitational potential $V=V_1+V_2=0$
Then P is the point of zero Gravitational potential
Let point $P$ is at distance $\times$ from $m_1$
Then For point $P$

$$\begin{array}{rl}& V=V_1+V_2=0\\ & -\frac{G{m}_1}{r_1}-\frac{G{m}_2}{r_2}=0\\ & -\frac{G{m}_1}{x}-\frac{G{m}_2}{d-x}=0\\ & x=\frac{m_{1}d}{m_1-m_2}\end{array}$$
 

For Uniform circular ring

$r=$ distance from ring
$a\to$ radius of Ring
$V\to$ Potential

At a point on its Axis

$$V=-\frac{GM}{\sqrt{a^2+r^2}}$$

At the center

$$V=-\frac{GM}{a}$$
 

For Uniform disc

$a\to$ Radius of disc
M-mass of dise
- At the center of the disc

$$V=-\frac{2GM}{a}$$

- At a point on its axis

$$V=-\frac{2GM}{a^2}(\sqrt{a^2+x^2}-x)$$
 

 For Spherical shell

$R\to$ Radius of shell
$r\to$ distance from the center of the shell
- Inside the surface

$$\begin{array}{c}r<R\\ V=-\frac{GM}{R}\end{array}$$

- on the surface

$$\begin{array}{rl}r& =R\\ V& =-\frac{GM}{R}\end{array}$$

- Outside the surface

$$\begin{array}{rl}r& >R\\ V& =-\frac{GM}{r}\end{array}$$
 

Uniform solid sphere

$R\to$ Radius of sphere
$M\to$ Mass of sphere
$r\to$ distance from the center of the sphere
- Inside the surface

$$\begin{array}{rl}& r<R\\ & V=-\frac{GM}{2R}[3-{\left(\frac{r}{R}\right)}^2]\end{array}$$

- on the surface

$$V_{\text{sur face }}=-\frac{GM}{R}$$

- Outside the surface

$$V=-\frac{GM}{r}$$

- Tip-V centre $=\frac{3}{2}$ V sur face