We had discussed earlier the notion of potential energy as being the energy stored in the body at its given position. If the position of the particle changes on account of forces acting on it, then the change in its potential energy is just the amount of work done on the body by the force. As we had discussed earlier, forces for which the work done is independent of the path are the conservative forces.
The force of gravity is a conservative force and we can calculate the potential energy of a body arising out of this force, called the gravitational potential energy. Consider points close to the surface of earth, at distances from the surface much smaller than the radius of the earth. In such cases, the force of gravity is practically a constant equal to mg, directed towards the centre of the earth. If we consider a point at a height h1 from the surface of the earth and another point vertically above it at a height $h_2$ from the surface, the work done in lifting the particle of mass m from the first to the second position is denoted by $W_{12}$
$W_{12}$ = Force × displacement
$W_{12}=mg(h_2-h_1)$ (1)
If we associate a potential energy W(h) at a point at a height h above the surface such that
$W(h)=mgh+W_0$ (2)
(where $W_o$ = constant);
then it is clear that
$W_{12}=W(h_2)-W(h_1)$ (3)
The work done in moving the particle is just the difference of potential energy between its final and initial positions. Observe that the constant $W_o$ cancels out in Eq. (3). Setting h = 0 in the last equation, we get W(h = 0) = $W_o$. h = 0 means points on the surface of the earth. Thus, Wo is the potential energy on the surface of the earth.
If we consider points at arbitrary distance from the surface of the earth, the result just derived is not valid since the assumption that the gravitational force mg is a constant is no longer valid. However, from our discussion we know that a point outside the earth, the force of gravitation on a particle directed towards the centre of the earth is
$F=\frac{GM_Em}{r^2}$ (4)
where $M_E$ = mass of earth, m = mass of the particle and r its distance from the centre of the earth. If we now calculate the work done in lifting a particle from $r = r_1$ to $r = r_2$ ($r_2 > r_1$) along a vertical path, we get instead of Eq. (1)
$W_{12}=\int_{r_1}^{r_2}\frac{GMm}{r^2}dr$
$=GM_Em\left(\frac{1}{r_2}-\frac{1}{r_1}\right)$ (5)
In place of Eq. (2), we can thus associate a potential energy W(r) at a distance r, such that
$W(r)=-\frac{GM_Em}{r}+W_1$
valid for r > R ,
so that once again $W_{12}=W(h_2)-W(h_1)$ The work done in moving the particle is. Setting r = infinity in the last equation, we get W (r = infinity) = $W_1$. Thus, W1 is the potential energy at infinity. One should note that only the difference of potential energy between two points has a definite meaning from Eqs. (3) and (5). One conventionally sets $W_1$ equal to zero, so that the potential energy at a point is just the amount of work done in displacing the particle from infinity to that point.
We have calculated the potential energy at a point of a particle due to gravitational forces on it due to the earth and it is proportional to the mass of the particle. The gravitational potential due to the gravitational force of the earth is defined as the potential energy of a particle of unit mass at that point. From the earlier discussion, we learn that the gravitational potential energy associated with two particles of masses $m_1$ and $m_2$ separated by distance by a distance r is given by
$V=-\frac{Gm_1m_2}{r}$ (if we choose V = 0 as r → ∞)
It should be noted that an isolated system of particles will have the total potential energy that equals the sum of energies (given by the above equation) for all possible pairs of its constituent particles. This is an example of the application of the superposition principle.
Example 1
Find the potential energy of a system of four particles placed at the vertices of a square of side l. Also obtain the potential at the centre of the square.
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| Fig.1 |
Answer
Consider four masses each of mass m at the corners of a square of side l; See Fig. 1. We have four mass pairs at distance l and two diagonal pairs at distance $\sqrt{2}l$
Hence,
$W(r)=-4\frac{Gm^2}{l}-2\frac{Gm^2}{\sqrt{2}l}$
$=-\frac{-2Gm^2}{l}\left(2+\frac{1}{\sqrt{2}}\right)=-5.41\frac{Gm^2}{l}$
The gravitational potential at the centre of the square ($r=\sqrt{2}l/2$) is
$U(r)=-4\sqrt{2}\frac{Gm}{l}$
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