Gravitational Potential Energy

Gravitational Potential Energy is the energy an element has due to its position in a gravitational field, mainly determined by its height above the ground. The elevated the element, the more energy it stores because gravity can pull it down with more force. You see GPE in everyday situations like a ball held high before falling, water stored in a dam ready to generate power, or objects on a shelf waiting to drop. In short, GPE shows how energy is stored in objects based on their height and how it can be converted into motion or work when released.

1.0Gravitational Potential

Gravitational potential at a point is the work done by an external force in moving a unit mass from infinity to that point without altering its kinetic energy.

$V_P=\frac{W_{\infty \Rightarrow P}}{m}$

Note: The motion of a unit mass is considered slow, ensuring no energy is used to change the system's kinetic energy. Gravitational potential is assumed to be zero at infinity.

2.0Gravitational Potential Due To A Point Mass

If a particle of mass m is at distance x from given particle of mass M, then, work done to displace mass m  by distance dx towards M under equilibrium

$d{W}_{ext}=\frac{GMm}{x^2}(-dx)Cos(180^o)$

Negative sign of dx indicates that displacement is towards mass M

$d{W}_{ext}=\frac{GMm}{x^2}(-dx)Cos(-1)\Rightarrow d{W}_{ext}=\frac{GMm}{x^2}dx$

$d{W}_{ext}=\frac{GMm}{x^2}dx$

total work done in bringing the particle of mass m from infinity to a separation r from mass M is-

$W_{\text{ext}}={\int }_{\infty }^r\frac{GMm}{x^2}dx={\left[\frac{GMm}{r}\right]}_{\infty }^r=\left(-\frac{GMm}{r}\right)-\left(\frac{GMm}{\infty }\right)$

$W_{ext}=-\frac{GMm}{r}$

As per Definition,

$V_P=\frac{W_{\infty \Rightarrow P}}{m}$

$V_P=\frac{W_{ext}}{m}=-\frac{GM}{r}$

At r distance from the particle of mass M

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

Note: Negative sign in potential indicates attractive nature of force.

3.0Relation Between Potential and Intensity

$I=-\frac{dV}{dr}$

Note:

1. If V is fixed in a region I=0
2. This relation is valid for all conservative fields.

4.0Gravitational Potential Due to a Spherical Shell

Spherical Shell (M, R)

Case-1. r > R (outside the sphere) $V_{out}=-\frac{GM}{r}$

Case-2. r = R (on the surface) $V_{Surface}=-\frac{GM}{r}$

Case-3. r < R (Inside the sphere, Potential is same everywhere and is equal to its value at the surface)

$V_{in}=-\frac{GM}{r}$

5.0Gravitational Potential Due to a Solid Sphere

Solid Sphere (M, R)

Case-1. r > R (outside the sphere) $V_{out}=-\frac{GM}{r}$

Case-2. r = R (on the surface) $V_{Surface}=-\frac{GM}{r}$

Case-3. r < R (Inside the sphere,Potential is same everywhere and is equal to its value at the surface)

$V_{in}=-\frac{GM}{2R^3}(3R^2-r^2)$

6.0Relation Between Gravitational Field And Potential

In gravitational Field,Field strength E and potential V are different at different points.So they are function of position.

Conversion of V function into E function

To convert V function into E function differentiation is required

1. More than one variable

$E=-\mathbf{gradient}V=-\left[\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}\right]$

$E=-\left[\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}\right]$

$\frac{მV}{მx}$ is called partial differentiation of V with respect to x      

2. Only one variable

$E=-\frac{dV}{dx}=-\frac{dV}{dr}$ or $E=(-SlopeofV-x)$ or $(-slopeofV-r)$

Conversion of E function into V function

In this case Integration is required

1. More than one variable

$dV=-\mathbf{E}\cdot d\mathbf{r}$

${\int }_a^{b}dV=-{\int }_a^b\mathbf{E}\cdot d\mathbf{r}$

$V_b-V_a=-{\int }_a^b\mathbf{E}\cdot d\mathbf{r}$

$d\mathbf{r}=dx\hat{i}+dy\hat{j}+dz\hat{k}$

7.0Gravitational Potential Energy

The potential energy of a system associated with a conservative force is defined as

${\int }_{U_i}^{U_f}dU=-\int \vec{F}\cdot d\vec{r}$

$\vec{F}=\frac{G{m}_1{m}_2}{r^2}$

$dW=\vec{F}\cdot d\vec{r}=-\frac{G{m}_1{m}_2}{r^2}dr$

$dU=-dW=\int \frac{G{m}_1{m}_2}{r^2}dr$

${\int }_{r_1}^{r_2}dU=-G{m}_1{m}_2{\int }_{r_1}^{r_2}\frac{dr}{r^2}=G{m}_1{m}_2{\left[-\frac{1}{r}\right]}_{r_1}^{r_2}$

$U(r_2)-U(r_1)=G{m}_1{m}_2\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$

The potential energy of the two particle system to be zero when the distance between them is infinity

$U(\infty )=0$

$U(r)=U(r)-U(\infty )$

$r_1=r$ and $r_2=\infty$

$U(\infty )-U(r)=G{m}_1{m}_2(\frac{1}{r}-\frac{1}{\infty })$

$U(r)=-\frac{G{m}_1{m}_2}{r}$

8.0Gravitational Potential Energy For Three Particle System

In a system of multiple particles, the overall  gravitational potential energy is the addition of the potential energies of all pairs.

$U_{\text{System}}=\left(-\frac{G{m}_1{m}_2}{r_1}\right)+\left(-\frac{G{m}_2{m}_3}{r_2}\right)+\left(-\frac{G{m}_1{m}_3}{r_3}\right)=-\frac{G{m}_1{m}_2}{r_1}-\frac{G{m}_2{m}_3}{r_2}-\frac{G{m}_1{m}_3}{r_3}$

If Gravitational Potential at a point P is $V_P$

Then GPE of a particle of mass m situated at P is : $U_P=m(V_P)$

Note: At infinity, $V_{\infty }=0\Rightarrow U_{\infty }=0$

9.0Sample Questions On Gravitational Potential Energy

Q1. Find gravitational potential at the centroid.

Solution:

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

$V_{\text{net}}=-\frac{Gm}{r}-\frac{Gm}{r}-\frac{Gm}{r}$

$r\cos 30^{\circ }=\frac{a}{2}$

$r\times \frac{\sqrt{3}}{2}=\frac{a}{2}=\frac{a}{\sqrt{3}}$

$V_{\text{net}}=-\frac{3Gm}{\frac{a}{\sqrt{3}}}=-3\sqrt{3}\frac{Gm}{a}$

Q2. An element of mass M is placed at the center of a spherical shell of the same mass and radius R. Determine the gravitational potential at a point located at a distance of R/3  from the center of the shell.

Solution:      

$V_P=V_{Sphere}+V_{particle}$

$V_P=-(\frac{GM}{R}+\frac{GM}{\frac{R}{3}})$

$V_P=-\frac{GM}{R}(1+3)=-4\frac{GM}{R}$

Q3. If the given system of three particles, each of mass m, on the vertices of an equilateral triangle side a is to be changed to side of 2a, then find the work done on the system.

Solution: 

$U_{\text{System}}=-\frac{G{m}^2}{a}-\frac{G{m}^2}{a}-\frac{G{m}^2}{a}=-\frac{3G{m}^2}{a}$

$\text{Now a is changed by }2a$

$U_{\text{System}}=-\frac{G{m}^2}{2a}-\frac{G{m}^2}{2a}-\frac{G{m}^2}{2a}=-\frac{3G{m}^2}{2a}$

$U_i=-\frac{3G{m}^2}{a}\text{and}{U}_f=-\frac{3G{m}^2}{2a}$

$\text{Work Done }=\Delta U$

$W=U_f-U_i$

$W=-\frac{3G{m}^2}{2a}-\left(-\frac{3G{m}^2}{a}\right)=-\frac{3G{m}^2}{a}\left[\frac{1}{2}-1\right]$

$W=\frac{3G{m}^2}{2a}$

Q4. Find gravitational potential energy of a particle mass 'm' placed at a height R above surface of solid sphere of mass M and radius 'R'.

Solution:

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

$V=-\frac{GM}{(R+h)}=-\frac{GM}{R+R}=-\frac{GM}{2R}$

$\text{Now potential energy }(U)=V(m)$

$U=-\frac{GM}{2R}(m)\Rightarrow U=-\frac{GMm}{2R}$