# 4.3 Gravitational potential energy  (Page 4/4)

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${U}_{13}=-\frac{G{m}_{1}{m}_{3}}{{r}_{13}}$

Similarly, the potential energy due to second particle is equal to the negative of work by gravitational force due to it :

${U}_{23}=-\frac{G{m}_{2}{m}_{3}}{{r}_{23}}$

Thus, potential energy of three particles at given positions is algebraic sum of negative of gravitational work in (i) bringing first particle (ii) bringing second particle in the presence of first particle and (iii) bringing third particle in the presence of first two particles :

$⇒U=-G\left(\frac{{m}_{1}{m}_{2}}{{r}_{12}}+\frac{{m}_{1}{m}_{3}}{{r}_{13}}+\frac{{m}_{2}{m}_{3}}{{r}_{23}}\right)$

Induction of forth particle in the system will involve work by gravitation in assembling three particles as given by the above expression plus works by the individual gravitation of three already assembled particles when fourth particle is brought from the infinity.

$⇒U=-G\left(\frac{{m}_{1}{m}_{2}}{{r}_{12}}+\frac{{m}_{1}{m}_{3}}{{r}_{13}}+\frac{{m}_{2}{m}_{3}}{{r}_{23}}+\frac{{m}_{1}{m}_{4}}{{r}_{14}}+\frac{{m}_{2}{m}_{4}}{{r}_{24}}+\frac{{m}_{3}{m}_{4}}{{r}_{34}}\right)$

Proceeding in this fashion, we can calculate potential energy of a system of particles. We see here that this process resembles the manner in which a system of particles like a rigid body is constituted bit by bit. As such, this potential energy of the system represents the “energy of constitution” and is called “self energy” of the rigid body or system of particles. We shall develop alternative technique (easier) to measure potential energy and hence “self energy” of regular geometric shapes with the concept of gravitational potential in a separate module.

## Examples

Problem 1: Find the work done in bringing three particles, each having a mass of 0.1 kg from large distance to the vertices of an equilateral triangle of 10 cm in a gravity free region. Assume that no change of kinetic energy is involved in bringing particles.

Solution : We note here that all three particles have same mass. Hence, product of mass in the expression of gravitational potential energy reduces to square of mass. The gravitational potential energy of three particles at the vertices of the equilateral triangle is :

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

where “a” is the side of the equilateral triangle.

Putting values,

$⇒U=-\frac{3X6.67X{10}^{-11}X0.12}{0.1}=-3X6.67X{10}^{-10}X0.01=-20X{10}^{-12}\phantom{\rule{1em}{0ex}}J$

$⇒U=-2X{10}^{-11}\phantom{\rule{1em}{0ex}}J$

Hence, work done by external force in bringing three particles from large distance is :

$⇒W=U=-2X{10}^{-11}J$

## Work and energy

An external force working on a system brings about changes in the energy of system. If change in energy is limited to mechanical energy, then work by external force will be related to change in mechanical energy as :

${W}_{F}=\Delta E=\Delta U+\Delta K$

A change in gravitational potential energy may or may not be accompanied with change in kinetic energy. It depends on the manner external force works on the system. If we work on the system in such a manner that we do not impart kinetic energy to the particles of the system, then there is no change in kinetic energy. In that case, the work by external force is equal to the change in gravitational potential energy alone.

There can be three different situations :

Case 1 : If there is change in kinetic energy, then work by external force is equal to the change in potential and kinetic energy:

${W}_{F}=\Delta U+\Delta K$

Case 2 : If there is no change in kinetic energy, then work by external force is equal to the change in potential energy alone :

$\Delta K=0$

Putting in the expression of work,

${W}_{F}=\Delta U$

Case 3 : If there is no external force, then work by external force is zero. The change in one form of mechanical energy is compensated by a corresponding negative change in the other form. This means that mechanical energy of the system is conserved. Here,

${W}_{F}=0$

Putting in the expression of work,

$⇒\Delta U+\Delta K=0$

We shall, now, work with two illustrations corresponding to following situations :

• Change in potential energy without change in kinetic energy
• Change in potential energy without external force

## Change in potential energy without change in kinetic energy

Problem 2: Three particles, each having a mass of 0.1 kg are placed at the vertices of an equilateral triangle of 10 cm. Find the work done to change the positions of particles such that side of the triangle is 20 cm. Assume that no change of kinetic energy is involved in changing positions.

Solution : The work done to bring the particles together by external force in gravitational field is equal to potential energy of the system of particles. This means that work done in changing the positions of the particles is equal to change in potential energy due to change in the positions of particles. For work by external force,

${W}_{F}=\Delta U+\Delta K$

Here, ΔK = 0

${W}_{F}=\Delta U$

Now, we have seen that :

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

Hence, change in gravitational potential energy is :

$⇒\Delta U=-\frac{3G{m}^{2}}{{a}_{2}}-\left(-\frac{3G{m}^{2}}{{a}_{1}}\right)$

$⇒\Delta U=3G{m}^{2}\left[-\frac{1}{{a}_{2}}+\frac{1}{{a}_{1}}\right]$

Putting values, we have : $⇒\Delta U=3X6.67X{10}^{-11}X{0.1}^{2}\left[-\frac{1}{0.2}+\frac{1}{0.1}\right]$

$⇒\Delta U=3X6.67X{10}^{-11}X0.12X5$

$⇒\Delta U=1.00X{10}^{-11}\phantom{\rule{1em}{0ex}}J$

## Change in potential energy without external force

Problem 3: Three identical solid spheres each of mass “m” and radius “R” are released from positions as shown in the figure (assume no external gravitation). What would be the speed of any of three spheres just before they collide.

Solution : Since no external force is involved, the mechanical energy of the system at the time of release should be equal to mechanical energy just before the collision. In other words, the mechanical energy of the system is conserved. The initial potential energy of system is given by,

${U}_{i}=-\frac{3G{m}^{2}}{a}$

Let “v” be the speed of any sphere before collision. The configuration just before the collision is shown in the figure. We can see that linear distance between any two centers of two identical spheres is “2R”. Hence, potential energy of the configuration before collision is,

${U}_{f}=-\frac{3G{m}^{2}}{2R}$

Applying conservation of mechanical energy,

${K}_{i}+{U}_{i}={K}_{f}+{U}_{f}$

$⇒0-\frac{3G{m}^{2}}{a}=\frac{1}{2}m{v}^{2}-\frac{3G{m}^{2}}{2R}$

$v=\sqrt{\left\{Gm\left(\frac{1}{R}-\frac{2}{a}\right)\right\}}$

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