JEE Main 2025 Syllabus PDF - Subject-Wise Detailed Syllabus

# Gravitational Potential - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Gravitational Potential due to Uniform solid sphere is considered one of the most asked concept.

• 16 Questions around this concept.

## Solve by difficulty

Which of the following most closely depicts the correct variation of the gravitation potential V(r) due to a large planet of radius R and uniform mass density ? (figures are not drawn to scale)

## Concepts Covered - 5

Gravitational Potential

In a gravitational field potential V at a point, P is defined as 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 $W =\int {\overrightarrow{F}\cdot \overrightarrow{dr}}$

So $V=-\frac{W}{m}=-\int \frac{\overrightarrow{F}\cdot \overrightarrow{dr}}{m}$

And $\vec{I}=\frac{\vec{F}}{m}$

$V=-\int \overrightarrow{I}\cdot \overrightarrow{dr}$

$V\rightarrow$ Gravitational potential

$I\rightarrow$ Field Intensity

$dr\rightarrow$ small distance

We can also write $I=-\frac{dV}{dr}$

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.

• $Unit \to \; Joule/kg \; or \; m^{2}/sec^{2}$

• $Dimension\: :\; \left [ M^{0}L^{2}T^{-2} \right ]$

• Gravitational Potential at a distance 'r'

If the field is produced by a point mass then

$I= \frac{GM}{r^{2}}$

So $V=-\int \overrightarrow{I}\cdot \overrightarrow{dr}$

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

at $r=\infty$       $V=0=V_{max}$

• 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

Point A and B are shown in the figure.

$V_{A}$=Gravitational potential at point A

$V_{B}$=Gravitational potential at point B

$r_{B}\rightarrow$ the distance of mass at $B$

$r_{A}\rightarrow$ distance of mass at $A$

$\Delta V$=The gravitational potential difference in bringing unit mass m from point A to point B in the gravitational field produced by M.

$\Delta V=V_{B}-V_{A}=\frac{W_{A\rightarrow B}}{m}$

$\Delta V=-GM\left [ \frac{1}{r_{B}}-\frac{1}{r_{A}} \right ]$

• 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 individuals Gravitational potential.

$V=V_{1}+V_{2}+V_{3}\cdot \cdot \cdot \cdot$

$=-\frac{GM_1}{r_{1}}-\frac{GM_2}{r_{2}}-\frac{GM_3}{r_{3}}\cdot \cdot \cdot \cdot$

$V=-G\: \sum_{i=1}^{i=n} \frac{M_{i}}{r_{i}}$

$M_{i}\rightarrow$ mass

$r_{i}\rightarrow$ distances

• Point of zero potential

Let m1 and m2 are 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 x from m1

Then For point P

$V=V_{1}+V_{2} =0$

$-\frac{Gm_1}{r_1}-\frac{Gm_2}{r_2}=0$

$-\frac{Gm_1}{x}-\frac{Gm_2}{d-x}=0$

So $x=\frac{m_1d}{m_1-m_2}$

Gravitational potential due to Uniform circular ring

For Uniform circular ring

$r=$ distance from ring

$a\rightarrow$ radius of Ring

$V\rightarrow$ Potential

At a point on its Axis

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

At the center

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

Gravitational Potential due to Uniform disc

For Uniform disc

$a \rightarrow$ Radius of disc

M-mass of disc

• At the center of the disc

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

• At a point on its axis

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

Gravitational Potential due to spherical shell

For Spherical shell

$R\rightarrow$ Radius of shell

$r\rightarrow$ distance from the center of the shell

• Inside the surface

$r

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

• on the surface

$r=R$

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

• Outside the surface

$r>R$

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

Gravitational Potential due to Uniform solid sphere

Uniform solid sphere

$R\rightarrow$ Radius of sphere

$M\rightarrow$ Mass of sphere

$r\rightarrow$ distance from the center of sphere

• Inside the surface

$r

$V=-\frac{GM}{2R}\left[3-\left ( \frac{r}{R} \right )^{2}\right]$

• on the surface

$V_{surface}=-\frac{GM}{R}$

• Outside the surface

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

• Tip-$Vcentre=\frac{3}{2}Vsurface$

## Study it with Videos

Gravitational Potential
Gravitational potential due to Uniform circular ring
Gravitational Potential due to Uniform disc

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