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Gravitational Potential Energy - Practice Questions & MCQ

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

Quick Facts

  • Gravitational Potential Energy (U) is considered one the most difficult concept.

  • 27 Questions around this concept.

Solve by difficulty

A particle of mass 10 g is kept on the surface of a uniform sphere of mass 100 Kg and radius  10 cm  Find the work to be done against the gravitational force between them to take the particle far away from the sphere.

\left ( You\: might\: take\: G= 6.67\times 10^{-11}Nm^{2}/kg^{2} \right )

Energy required to move a body of mass  from  an orbit of radius  2R  to  3R  is:

A body of mass 'm' is taken from the earth's surface to a height equal to twice the radius (R) of the earth. The change in potential energy of the body will be

Concepts Covered - 1

Gravitational Potential Energy (U)

It is the amount of work done in bringing a body from  \infty  to that point against gravitational force.

  • It is Scalar quantity

  • SI Unit: Joule

  • Dimension : \left[ ML^{2}T^{-2}\right ]

  • Gravitational Potential energy at a point

             If the point mass M is producing the field

             

         Then gravitational force on test mass m at a distance r from M is given by F=\frac{GMm}{r^2}

        And the amount of work done in bringing a body from \infty to r

                   = W=\int_{\infty}^{r}\frac{GMm}{x^2}dx=-\frac{GMm}{r}

         And this is equal to gravitational potential energy

               So U=-\frac{GMm}{r}

U \rightarrow gravitational potential energy 

M \rightarrow Mass of source-body

m \rightarrow mass of test body

r \rightarrow distance between two

Note- U is always negative in the gravitational field because Force is attractive in nature.

  Means As the distance r increases U becomes less negative 

I.e U will increase as r increases

And for r=\infty, U=o which is maximum

  • Gravitational Potential energy of discrete distribution of masses

  U=-G\left [ \frac{m_{1}m_{2}}{r_{12}}+\frac{m_{2}m_{3}}{r_{23}}+\cdot \cdot \cdot \right ]

U \rightarrow Net Gravitational Potential Energy

r_{12},r_{23}\rightarrow The distance of masses from each other

 

  • Change of potential energy

if a body of mass m is moved from  r_{1 } to r_{2 }

Then Change of potential energy is given as

 \Delta U=GMm\left [ \frac{1}{r_{1}}-\frac{1}{r_{2}} \right ]

 

\Delta U \rightarrow change of energy

r_{1},r_{2}\rightarrow distances

If r_{1}>r_{2} then the change in potential energy of the body will be negative.

              I.e To decrease potential energy of a body we have to bring that body closer to the earth.

  • The relation between Potential and Potential energy

As U=\frac{-GMm}{r}=m\left [ \frac{-GM}{r} \right ]

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

So U=mV

Where V\rightarrow Potential

U\rightarrow Potential energy

r\rightarrow distance

 

  • Gravitational Potential Energy at the center of the earth relative to infinity

           {U_{centre}=mV_{centre}}\\ {V_{centre}\rightarrow Potential\: at\: centre}

           U=m\left ( -\frac{3}{2}\frac{GM}{R} \right )

               m \rightarrow mass of body

            M \rightarrow Mass of earth

 

  • The gravitational potential energy at height 'h' from the earth's surface

U_{h}=-\frac{GMm}{R+h}

Using GM=gR^2

U_{h}=-\frac{gR^{2}m}{R+h}

 

U_{h}=-\frac{mgR}{1+\frac{h}{R}}

 

U_{h}\rightarrow The potential energy at the height h

R\rightarrow Radius of earth

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Gravitational Potential Energy (U)

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