Gravitational Potential Energy

What is Gravitational Potential Energy?

Gravitational Potential Energy (GPE) is the energy an object possesses due to its position in a gravitational field. Specifically, it's the energy stored in an object as a result of its height above a reference point, typically the Earth's surface. The amount of gravitational potential energy depends on the object's mass, the height at which it is positioned, and the strength of the gravitational field.

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

• It is a Scalar quantity

• SI Unit: Joule

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

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 $F=\frac{G M m}{r^2}$

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

$=W=\int_{\infty}^r \frac{G M m}{x^2} d x=-\frac{G M m}{r}$

And this is equal to gravitational potential energy

So $U=-\frac{G M m}{r}$

$U \rightarrow$ 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.
This means As the distance $r$ increases $U$ becomes less negative
I.e $U$ will increase as $r$ increases

And for $r=\infty_2 \mathrm{U}=\mathrm{o}$ 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}}+\cdots\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, the Change of potential energy is given as

$
\Delta U=G M m\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 the potential energy of the body will be negative.

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

The Relation Between Potential and Potential Energy

${ }_{\text {As }} U=\frac{-G M m}{r}=m\left[\frac{-G M}{r}\right]$
So $U=m V$
Where $V \rightarrow$ Potential
$U \rightarrow$ Potential energy
$r \rightarrow$ distance

Gravitational Potential Energy at the Centre of the Earth Relative to Infinity

$\begin{gathered}U_{\text {centre }}=m V_{\text {centre }} \\ V_{\text {centre }} \rightarrow \text { Potential at centre } \\ U=m\left(-\frac{3}{2} \frac{G M}{R}\right) \\ m \rightarrow \text { mass of body } \\ M \rightarrow \text { Mass of earth }\end{gathered}$

The Gravitational Potential Energy at Height 'h' from the Earth's Surface

$\begin{aligned} & U_h=-\frac{G M m}{R+h} \\ & \text { Using } G M=g R^2 \\ & U_h=-\frac{g R^2 m}{R+h} \\ & U_h=-\frac{m g R}{1+\frac{h}{R}} \\ & U_h \rightarrow \text { The potential energy at the height } h \\ & R \rightarrow \text { Radius of earth }\end{aligned}$

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Solved Examples Based on Gravitational Potential Energy

Example 1: Two hypothetical planets of masses m₁ and m₂ are at rest when they are infinite distance apart. Because of the gravitational force, they move towards each other along the line joining their centres. What is their speed when their separation is ‘d’? (Speed of m₁ is $v$₁ and that of m₂ is $v$₂)

1) $v_1=v_2$
2)

$
\begin{aligned}
v_1 & =m_2 \sqrt{\frac{2 G}{d\left(m_1+m_2\right)}} \\
v_2 & =m_1 \sqrt{\frac{2 G}{d\left(m_1+m_2\right)}}
\end{aligned}
$

3)

$
\begin{aligned}
& v_1=m_1 \sqrt{\frac{2 G}{d\left(m_1+m_2\right)}} \\
& v_2=m_2 \sqrt{\frac{2 G}{d\left(m_1+m_2\right)}}
\end{aligned}
$

4)

$
\begin{aligned}
& v_1=m_2 \sqrt{\frac{2 G}{m_1}} \\
& v_2=m_1 \sqrt{\frac{2 G}{m_2}}
\end{aligned}
$

Solution:

The initial energy of the system = 0

Final energy $=\frac{1}{2} M_1 V_1^2+\frac{1}{2} M_2 V_2^2-\frac{G M_1 M_2}{d}$
From the conservation of energy

$
\frac{1}{2} M_1 V_1^2+\frac{1}{2} M_2 V_2^2=\frac{G M_1 M_2}{d}-\cdots-\cdots-1
$
From the conservation of Linear Momentum

$
\begin{aligned}
& \quad m_1 v_1^2+m_2\left(\frac{-m_1 v_1}{m_2}\right)^2=\frac{2 G m_1 m_2}{d} \\
& \frac{m_1 m_2 v_1^2+m_1^2 v_1^2}{m_2}=\frac{2 G m_1 m_2}{d}=v_1=m_2 \sqrt{\frac{2 G}{d\left(m_1+m_2\right)}} \\
& \quad v_2=m_1 \sqrt{\frac{2 G}{d\left(m_1+m_2\right)}} \\
& \text { Similarly }
\end{aligned}
$

Hence, the answer is the option (3).

Example 2: A tunnel is dug along the diameter of the earth (Radius R & mass M). There is a particle of mass 'm' at the centre of the tunnel. The minimum velocity given to the particle so that it just reaches the surface of the earth is :

1) $\sqrt{\frac{G M}{R}}$
2) $\sqrt{\frac{G M}{2 R}}$
3) $\sqrt{\frac{2 G M}{R}}$
4) it will reach with the help of negligible velocity.

Solution:

Gravitational Potential Energy at the centre of the earth relative to infinity -
$
\begin{aligned}
& U=m\left(-\frac{3}{2} \frac{G M}{R}\right) \\
& m \rightarrow \text { mass of body } \\
& M \rightarrow \text { Mass of earth }
\end{aligned}
$

wherein

$
\text { Ucentre }=m \text { Vcentre }
$

$V$ centre $\rightarrow$ Potential at centre

Gravitational Potential energy at a point

$\begin{aligned} & W=-\frac{G M m}{r} \\ & U=-\frac{G M m}{r} \\ & U \rightarrow \text { gravitational potential energy } \\ & M \rightarrow \text { Mass of source body } \\ & m \rightarrow \text { mass of test body } \\ & r \rightarrow \text { distance between two }\end{aligned}$

wherein

Always negative in the gravitational field because Force is attractive in nature.

Let the minimum speed imparted to the particle of mass m so that it just reaches the surface of the earth is v.

Applying conservation of energy

$
\frac{1}{2} m v^2+\left[-\frac{3}{2} \frac{G M}{R} m\right]=-\frac{G M}{R} m+0
$
Solving we get $V=\sqrt{\frac{G M}{R}}$

Hence, the answer is the option (1).

Example 3: Four particles, each of which is mass m, are placed at the vertices of the square of side a. What is the potential energy of the system?

1) $\frac{-\sqrt{2} G m^2}{a}\left(2-\frac{1}{\sqrt{2}}\right)$
2) $\frac{-2 G m^2}{a}\left(2+\frac{1}{\sqrt{2}}\right)$
3) $\frac{-\sqrt{2} G m^2}{a}\left(\sqrt{2}-\frac{1}{\sqrt{2}}\right)$
4) $\frac{-\sqrt{2} G m^2}{a}\left(\sqrt{2}+\frac{1}{\sqrt{2}}\right)$

Solution:

Gravitational Potential energy of discrete distribution of masses
$
\begin{aligned}
& U=-G\left[\frac{m_1 m_2}{r_{12}}+\frac{m_2 m_3}{r_{23}}+\cdots\right] \\
& U \rightarrow \text { Gravitational Potential Energy }
\end{aligned}
$

$r_{12}, r_{23} \rightarrow$ Distance of masses from each other
wherein

$
\text { if } r=\infty
$

$U$ becomes Zero (maximum)

$
\text { P.E. }=\frac{-G m_1 m_2}{r}
$

The short trick to calculate the pair
$
\begin{aligned}
& =\frac{n(n-1)}{2} \\
& =\frac{4 * 3}{2}=6
\end{aligned}
$
Pair calculation $(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)$
$\begin{array}{lllllll}\text { Distance } & a & \sqrt{2} a & a & a & \sqrt{2} a & a\end{array}$
Four pairs of distances $=a$
Two pairs of distances $=\sqrt{2} a$

Hence P.E. of the system

$
\begin{aligned}
& \text { Hence P.E. of the system }=\frac{-4 G m^2}{a}+\frac{-2 G m^2}{\sqrt{2} a} \\
& P . E=\frac{-2 G m^2}{a}\left[2+\frac{1}{\sqrt{2}}\right]
\end{aligned}
$

Hence, the answer is the option (2).

Example 4: 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. $\left(\right.$ You might take $\left.G=6.67 \times 10^{-11} \mathrm{Nm}^2 / \mathrm{kg}^2\right)$

1) $6.67 \times 10^{-9} \mathrm{~J}$
2) $6.67 \times 10^{-10} \mathrm{~J}$
3) $13.34 \times 10^{-10} \mathrm{~J}$
4) $3.33 \times 10^{-10} \mathrm{~J}$

Solution:

Change of potential energy

$\begin{aligned} & \Delta U=G M m\left[\frac{1}{r_1}-\frac{1}{r_2}\right] \\ & r_1>r_2 \\ & \Delta U \rightarrow \text { change of energy } \\ & r_1, r_2 \rightarrow \text { distances } \\ & \text { wherein } \\ & \text { if the body is moved from } r_1 \text { to } r_2 \text { use this equation } \\ & \text { dw= } \int d r=\frac{G m_1 m_2}{r^2} d r \\ & \int d w=G m_1 m_2 \int_r^{\infty} \frac{d r}{r^2}=G m_1 m_2\left[\frac{1}{r}\right]_r^{\infty} \\ & \mathrm{w}=\frac{G m_1 m_2}{r} \\ & \mathrm{w}=\frac{\left(6.67 \times 10^{-11}\right)(100 \times)\left(10 \times 10^{-3}\right)}{10 \times 10^{-2}} \\ & =6.67 \times 10^{-10} \mathrm{~J}\end{aligned}$

Hence, the answer is the option (2).

Example 5: Energy required to move a body of mass m from an orbit of radius 2R to 3R is:

1) $G M m / 12 R^2$
2) $G M m / 3 R^2$
3) $G M m / 8 R$
4) $G M m / 6 R$

Solution:

$\begin{aligned} & \mathrm{E}=(P \cdot E)_{3 R}-(P \cdot E)_{2 R} \\ & =-\frac{G m M}{3 R}-\left(-\frac{G m M}{2 R}\right)=+\frac{G m M}{6 R}\end{aligned}$

Hence, the answer is the option (4).

Summary

Gravitational Potential Energy (GPE) is the energy stored in an object due to its position in a gravitational field, determined by its mass, height, and the gravitational constant. It plays a critical role in various phenomena, such as objects falling or celestial bodies interacting. The energy is always negative, reflecting the attractive nature of gravity, and changes based on the object's position relative to other masses. Examples and calculations demonstrate how GPE influences the motion and energy transformations of objects in different scenarios.