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Gravitational Potential - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

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

  • 22 Questions around this concept.

Solve by difficulty

A particle of mass M is situated at the centre of a spherical shell of same mass and radius a. The gravitational potential at a point situated at \frac{a}{2} distance from the centre, will be:

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 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<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<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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